Split (1)

train · 4.29k rows

text stringlengths 790 2.88k
.71: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. 2.3 Time, Velocity, and Speed 5. (a) Calculate Earth's average speed relative to the Sun. (b) What is its average velocity over a period of one year? 6. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter's frame of reference. (b) What is its average velocity over one revolution? 7. The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present? 8. Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant? 9. On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world's nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s? 10. Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84×106 m (1%)? This content is available for free at http://cnx.org/content/col11844/1.13 11. A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0º
south of east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip? 12. The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance? 13. Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person's voice was so loud in the astronaut's space helmet that it was picked up by the astronaut's microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s). 14. A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion. 15. The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10−10 m in diameter. (a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron's average velocity? 2.4 Acceleration 16. A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration? 17. Professional Application Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp
rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of (9.80 m/s2) by taking its ratio to the acceleration of gravity. 18. A commuter backs her car out of her garage with an acceleration of 1.40 m/s2. (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration? 19. Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of (9.80 m/s2)? Chapter 2 \| Kinematics 89 2.5 Motion Equations for Constant Acceleration in One Dimension 20. An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2. (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period. 21. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10−3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball? 22. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105 m/s2 for 8.10×10−4 s. What is its muzzle velocity (that is, its final velocity)? 23. (a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed? (c) In emergencies
the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2? 24. While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car's final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly. 25. At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2. (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense? 26. Professional Application: Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat? 27. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10−2 s, calculate the distance over which the puck accelerates. 28. A powerful motorcycle can accelerate from rest to 26.8 m/ s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time? 29. Freight trains can produce only relatively small accelerations and decelerations. (
a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case? 30. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration. 31. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne? (b) How long does this take? 32. Professional Application: A woodpecker's brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker's head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in m/s2 and in multiples of. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain's deceleration, expressed in multiples of? = 9.80 m/s2 33. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last? 34. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few lifethreatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceler
ation to be relatively small. If we assume that a pilot's speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m. 35. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem. 36. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves? 37. Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any 90 Chapter 2 \| Kinematics information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity. known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation. 38. A bicycle racer sprints at the end of a
race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish? 39. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course? 40. (a) A world record was set for the men's 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race? 2.7 Falling Objects Assume air resistance is negligible unless otherwise stated. 41. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be 0 = 0. 42. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d)
2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water. 43. A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball? 44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. 45. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a This content is available for free at http://cnx.org/content/col11844/1.13 46. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/ s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water? 47. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed? 48. A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m
, and he is 1.80 m tall? 49. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down? 50. A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air? 51. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can't see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head? 52. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground. 53. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day. 54. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball's initial velocity? 55. Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound
to travel up the well. The speed of sound is 332.00 m/s in this well. 56. A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5 s). (d) How much did the ball Chapter 2 \| Kinematics 91 compress during its collision with the floor, assuming the floor is absolutely rigid? 57. A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground. 58. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3 s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? 2.8 Graphical Analysis of One Dimensional Motion Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you. 59. (a) By taking the slope of the curve in Figure 2.72, verify that the velocity of the jet car is 115 m/s at = 20 s. (b) By taking the slope of the curve at any point in Figure 2.73, verify that the jet car's acceleration is 5.0 m/s2. Figure 2.74 61. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures. 62. By taking the slope of the curve
in Figure 2.75, verify that the acceleration is 3.2 m/s2 at = 10 s. Figure 2.72 Figure 2.75 63. Construct the displacement graph for the subway shuttle train as shown in Figure 2.30(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure. 64. (a) Take the slope of the curve in Figure 2.76 to find the jogger's velocity at = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.77. Figure 2.73 60. Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at = 10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures. Figure 2.76 92 Chapter 2 \| Kinematics Figure 2.77 Figure 2.80 Figure 2.78 65. A graph of () is shown for a world-class track sprinter in a 100-m race. (See Figure 2.79). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at = 5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race? Figure 2.79 66. Figure 2.80 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs. Test Prep for AP® Courses 2.1 Displacement 1. Which of the following statements comparing position, distance, and displacement is correct? This content is available for free at http://cnx.org/content/col11844/1.13 a. An object may record a distance of zero while recording a non-zero displacement. b. An object may record a non-zero distance while recording a displacement of zero. c. An object may record a non-zero distance while maintaining a position of zero. Chapter 2 \| Kinematics 93 d. An object may record a non-zero displacement while maintaining a position of zero. velocity v as a function of time t is shown in the graph. The five labeled points divide the graph into four sections. 2.2 Vectors, Scalars, and Coordinate Systems 2. A student is trying to determine the
acceleration of a feather as she drops it to the ground. If the student is looking to achieve a positive velocity and positive acceleration, what is the most sensible way to set up her coordinate system? a. Her hand should be a coordinate of zero and the upward direction should be considered positive. b. Her hand should be a coordinate of zero and the downward direction should be considered positive. c. The floor should be a coordinate of zero and the upward direction should be considered positive. d. The floor should be a coordinate of zero and the downward direction should be considered positive. 2.3 Time, Velocity, and Speed 3. A group of students has two carts, A and B, with wheels that turn with negligible friction. The two carts travel along a straight horizontal track and eventually collide. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together. a. Describe an experimental procedure to determine the velocities of the carts before and after the collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure. b. There will be sources of error in the measurements taken in the experiment both before and after the collision. Which velocity will be more greatly affected by this error: the velocity prior to the collision or the velocity after the collision? Or will both sets of data be affected equally? Justify your answer. 2.4 Acceleration 4. Figure 2.81 Graph showing Velocity vs. Time of a cart. A cart is constrained to move along a straight line. A varying net force along the direction of motion is exerted on the cart. The cart's Which of the following correctly ranks the magnitude of the average acceleration of the cart during the four sections of the graph? a. aCD > aAB > aBC > aDE b. aBC > aAB > aCD > aDE c. aAB > aBC > aDE > aCD d. aCD > aAB > aDE > aBC 5. Push a book across a table and observe it slow to a stop. Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered positive. Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered negative.
2.5 Motion Equations for Constant Acceleration in One Dimension 6. A group of students is attempting to determine the average acceleration of a marble released from the top of a long ramp. Below is a set of data representing the marble's position with respect to time. Position (cm) Time (s) 0.0 0.3 1.25 2.8 5.0 7.75 11.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Use the data table above to construct a graph determining the acceleration of the marble. Select a set of data points from the table and plot those points on the graph. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points. Using the best-fit line, determine the value of the marble's acceleration. 2.7 Falling Objects 7. Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 100 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid? a. 0 m/s2 b. 0.03 m/s2 c. 0.20 m/s2 d. 0.65 m/s2 e. 33 m/s2 94 Chapter 2 \| Kinematics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 95 3 TWO-DIMENSIONAL KINEMATICS Figure 3.1 Everyday motion that we experience is, thankfully, rarely as tortuous as a rollercoaster ride like this—the Dragon Khan in Spain's Universal Port Aventura Amusement Park. However, most motion is in curved, rather than straight-line, paths. Motion along a curved path is two- or threedimensional motion, and can be described in a similar fashion to one-dimensional motion. (credit: Boris23/Wikimedia Commons) Chapter Outline 3.1. Kinematics in Two Dimensions: An Introduction 3.2. Vector Addition and Subtraction: Graphical Methods 3.3. Vector Addition and Subtraction: Analytical Methods 3.4. Projectile Motion
3.5. Addition of Velocities Connection for AP® Courses Most instances of motion in everyday life involve changes in displacement and velocity that occur in more than one direction. For example, when you take a long road trip, you drive on different roads in different directions for different amounts of time at different speeds. How can these motions all be combined to determine information about the trip such as the total displacement and average velocity? If you kick a ball from ground level at some angle above the horizontal, how can you describe its motion? To what maximum height does the object rise above the ground? How long is the object in the air? How much horizontal distance is covered before the ball lands? To answer questions such as these, we need to describe motion in two dimensions. Examining two-dimensional motion requires an understanding of both the scalar and the vector quantities associated with the motion. You will learn how to combine vectors to incorporate both the magnitude and direction of vectors into your analysis. You will learn strategies for simplifying the calculations involved by choosing the appropriate reference frame and by treating each dimension of the motion separately as a one-dimensional problem, but you will also see that the motion itself occurs in the same way regardless of your chosen reference frame (Essential Knowledge 3.A.1). 96 Chapter 3 \| Two-Dimensional Kinematics This chapter lays a necessary foundation for examining interactions of objects described by forces (Big Idea 3). Changes in direction result from acceleration, which necessitates force on an object. In this chapter, you will concentrate on describing motion that involves changes in direction. In later chapters, you will apply this understanding as you learn about how forces cause these motions (Enduring Understanding 3.A). The concepts in this chapter support: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.1 An observer in a particular reference frame can describe the motion of an object using such quantities as position, displacement, distance, velocity, speed, and acceleration. 3.1 Kinematics in Two Dimensions: An Introduction Learning Objectives By the end of this section, you will be able to: • Observe that motion in two dimensions consists of horizontal and vertical components. • Understand the independence of horizontal and vertical vectors in two-dimensional motion. The information presented in this section supports the following AP® learning objectives and science practices: •
3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) • 3.A.1.2 The student is able to design an experimental investigation of the motion of an object. (S.P. 4.2) • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Figure 3.2 Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow roads and sidewalks, making two-dimensional, zigzagged paths. (credit: Margaret W. Carruthers) Two-Dimensional Motion: Walking in a City Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 3.3. Figure 3.3 A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 97 The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a twodimensional path, such as the one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line distance? An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem, 2 + 2 = 2, can be used to find the straight-line distance. Figure 3.4 The Pythagorean theorem relates the length of the legs of a right triangle, labeled and, with the hypotenuse, labeled. The. This can be rewritten, solving for : = 2 + 2 relationship is given by: 2 + 2 = 2. The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is (9 blocks)2+ (5 blocks)2= 10.3 blocks, considerably shorter than the 14 blocks
you walked. (Note that we are using three significant figures in the answer. Although it appears that “9” and “5” have only one significant digit, they are discrete numbers. In this case “9 blocks” is the same as “9.0 or 9.00 blocks.” We have decided to use three significant figures in the answer in order to show the result more precisely.) Figure 3.5 The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are square and the same size. The fact that the straight-line distance (10.3 blocks) in Figure 3.5 is less than the total distance walked (14 blocks) is one example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.) As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector's magnitude. The arrow's length is indicated by hash marks in Figure 3.3 and Figure 3.5. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 3.5. The first represents a 9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.) The Independence of Perpendicular Motions The person taking the path shown in Figure 3.5 walks east and then
north (two perpendicular directions). How far he or she walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward. Independence of Motion The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the horizontal direction does not affect motion in the vertical direction, and vice versa. 98 Chapter 3 \| Two-Dimensional Kinematics This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving movement in two directions at once. For example, let's compare the motions of two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A stroboscope has captured the positions of the balls at fixed time intervals as they fall. Figure 3.6 This shows the motions of two identical balls—one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are independent. Applying the Science Practices: Independence of Horizontal and Vertical Motion or Maximum Height and Flight Time Choose one of the following experiments to design: Design an experiment to confirm what is shown in Figure 3.6, that the vertical motion of the two balls is independent of the horizontal motion. As you think about your experiment, consider the following questions: • How will you measure the horizontal and vertical positions of each ball over time? What equipment will this require? • How will you measure the time interval between each of your position measurements? What equipment will this require? If you were to create separate graphs of the horizontal velocity for each ball versus time, what do you predict it would look like? Explain. If you were to compare graphs of the vertical velocity for each ball versus time, what do you predict it would look like? Explain. If there is a significant amount of air resistance, how will that affect each of your graphs? • • • Design a two-dimensional ballistic motion experiment that demonstrates the relationship between the maximum height reached by an object and the object's time of flight. As you think about your
experiment, consider the following questions: • How will you measure the maximum height reached by your object? • How can you take advantage of the symmetry of an object in ballistic motion launched from ground level, reaching maximum height, and returning to ground level? • Will it make a difference if your object has no horizontal component to its velocity? Explain. • Will you need to measure the time at multiple different positions? Why or why not? • Predict what a graph of travel time versus maximum height will look like. Will it be linear? Parabolic? Horizontal? Explain the shape of your predicted graph qualitatively or quantitatively. If there is a significant amount of air resistance, how will that affect your measurements and your results? • It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, and not by any horizontal forces.) Careful examination of the ball thrown horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is thrown. This result means that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal conditions. In the real world, air resistance will affect the speed of the balls in both directions. The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve (break) it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. We shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas of physics. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 99 PhET Explorations: Ladybug Motion 2D Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and
playback the motion to analyze the behavior. Figure 3.7 Ladybug Motion 2D (http://cnx.org/content/m54779/1.2/ladybug-motion-2d_en.jar) 3.2 Vector Addition and Subtraction: Graphical Methods By the end of this section, you will be able to: Learning Objectives • Understand the rules of vector addition, subtraction, and multiplication. • Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects. The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Figure 3.8 Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai'i to Moloka'i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey. (credit: US Geological Survey) Vectors in Two Dimensions A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector's magnitude and pointing in the direction of the vector. Figure 3.9 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics,, and its direction by. 100 Chapter 3 \| Two-Dimensional Kinematics Vectors in this Text In this text, we will
represent a vector with a boldface variable. For example, we will represent the quantity force with the vector F, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as, and the direction of the variable will be given by an angle. Figure 3.9 A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1° north of east. Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total displacement vector D. Using a protractor, draw a line at an angle relative to the east-west axis. The length of the arrow is proportional to the vector's magnitude and is measured along the line with a ruler. In this example, the magnitude of the vector is 10.3 units, and the direction is 29.1° north of east. Vector Addition: Head-to-Tail Method The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 101 Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant vector D. The length of the arrow D is proportional to the vector's magnitude and is measured to be 10.3 units. Its direction, described as the angle with respect to the east (or horizontal axis) is measured with a protractor to be 29.1°. Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. Figure 3.
12 Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector. Figure 3.13 Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail. Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors. 102 Chapter 3 \| Two-Dimensional Kinematics Figure 3.14 Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.) Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.) The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors. Example 3.1 Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0° north of east. Then, she walks 23.0 m heading 15.0° north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east. Strategy Represent each displacement vector graphically with an arrow, labeling the first A, the second B, and the third C, making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted R. Solution (1) Draw the three displacement vectors. Figure 3.15 (2) Place the vectors head to tail retaining both their initial magnitude and direction. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 103 Figure 3.16 (3) Draw the
resultant vector, R. Figure 3.17 (4) Use a ruler to measure the magnitude of R, and a protractor to measure the direction of R. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector. Figure 3.18 In this case, the total displacement R is seen to have a magnitude of 50.0 m and to lie in a direction 7.0° south of east. By using its magnitude and direction, this vector can be expressed as = 50.0 m and = 7.0° south of east. Discussion The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 3.19 and we will still get the same solution. 104 Chapter 3 \| Two-Dimensional Kinematics Figure 3.19 Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order. A + B = B + A. (3.1) (This is true for the addition of ordinary numbers as well—you get the same result whether you add 2 + 3 or 3 + 2, for example). Vector Subtraction Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract B from A, written A – B, we must first define what we mean by subtraction. The negative of a vector B is defined to be –B ; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 3.20. In other words, B has the same length as –B, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction. Figure 3.20 The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So B is the negative of –B ; it has the same length but opposite direction. The subtraction of vector B
from vector A is then simply defined to be the addition of –B to A. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results. A – B = A + (–B). (3.2) This is analogous to the subtraction of scalars (where, for example, 5 – 2 = 5 + (–2) ). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates. Example 3.2 Subtracting Vectors Graphically: A Woman Sailing a Boat A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0° north of east from her current location, and then travel 30.0 m in a direction 112° north of east (or 22.0° west of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 105 north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock. Figure 3.21 Strategy We can represent the first leg of the trip with a vector A, and the second leg of the trip with a vector B. The dock is located at a location A + B. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance (30.0 m) in the direction 180° – 112° = 68° south of east. We represent this as –B, as shown below. The vector –B has the same magnitude as B but is in the opposite direction. Thus, she will end up at a location A + (–B), or A – B. Figure 3.22 We will perform vector addition to compare the location of the dock, A + B, with the location at which the woman mistakenly arrives, A + (–B). Solution (1) To determine the location at which the woman arrives by accident, draw vectors A and –B. (2) Place the vectors head to tail. (3) Draw the resultant vector R. (4) Use a ruler and protractor to measure the magnitude and direction of R. Figure 3.23
In this case, = 23.0 m and = 7.5° south of east. 106 Chapter 3 \| Two-Dimensional Kinematics (5) To determine the location of the dock, we repeat this method to add vectors A and B. We obtain the resultant vector R': Figure 3.24 In this case = 52.9 m and = 90.1° north of east. We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip. Discussion Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition. Multiplication of Vectors and Scalars If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 × 27.5 m, or 82.5 m, in a direction 66.0° north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same. If the scalar is negative, then multiplying a vector by it changes the vector's magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector A is multiplied by a scalar, • • • the magnitude of the vector becomes the absolute value of, if is positive, the direction of the vector does not change, if is negative, the direction is reversed. In our case, = 3 and = 27.5 m. Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1. Resolving a Vector into Components In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south and east-west components. For example,
we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0° north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton's Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components. PhET Explorations: Maze Game Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 107 Figure 3.25 Maze Game (http://cnx.org/content/m54781/1.2/maze-game_en.jar) 3.3 Vector Addition and Subtraction: Analytical Methods Learning Objectives By the end of this section, you will be able to: • Understand the rules of vector addition and subtraction using analytical methods. • Apply analytical methods to determine vertical and horizontal component vectors. • Apply analytical methods to determine the magnitude and direction of a resultant vector. The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can
be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known. Resolving a Vector into Perpendicular Components Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like A in Figure 3.26, we may wish to find which two perpendicular vectors, A and A, add to produce it. Figure 3.26 The vector A, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, A and A. These vectors form a right triangle. The analytical relationships among these vectors are summarized below. A and A are defined to be the components of A along the x- and y-axes. The three vectors A, A, and A form a right triangle: A + Ay = A. (3.3) Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if A = 3 m east, A = 4 m north, and A = 5 m north-east, then it is true that the vectors A + Ay = A. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is3.4) Thus, 108 Chapter 3 \| Two-Dimensional Kinematics If the vector A is known, then its magnitude (its length) and its angle (its direction) are known. To find and, its xand y-components, we use the following relationships for a right triangle. + ≠ (3.5) and = cos = sin. (3.6) (3.7) Figure 3.27 The magnitudes of the vector components A and A can be related to the resultant vector A and the angle with trigonometric identities. Here we see that = cos and = sin. Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods. Figure 3.28 We can use the relationships = cos and = sin to determine the magnitude of the horizontal and vertical component vectors in this example. Then = 10.3 blocks and =
29.1º, so that = cos = = sin = 10.3 blocks cos 29.1º sin 29.1º 10.3 blocks = 9.0 blocks = 5.0 blocks. (3.8) (3.9) Calculating a Resultant Vector If the perpendicular components A and A of a vector A are known, then A can also be found analytically. To find the magnitude and direction of a vector from its perpendicular components A and A, we use the following relationships: = 2 + 2 = tan−1( / ). (3.10) (3.11) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 109 Figure 3.29 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components and have been determined. Note that the equation = hypotenuse. For example, if and are 9 and 5 blocks, respectively, then = 92 +52=10.3 blocks, again consistent with the example of the person walking in a city. Finally, the direction is = tan–1(5/9)=29.1º, as before. 2 is just the Pythagorean theorem relating the legs of a right triangle to the length of the 2 + Determining Vectors and Vector Components with Analytical Methods Equations = cos and = sin are used to find the perpendicular components of a vector—that is, to go 2 and = tan–1( / ) are used to find a vector from its from and to and. Equations = perpendicular components—that is, to go from and to and. Both processes are crucial to analytical methods of vector addition and subtraction. 2 + Adding Vectors Using Analytical Methods To see how to add vectors using perpendicular components, consider Figure 3.30, in which the vectors A and B are added to produce the resultant R. Figure 3.30 Vectors A and B are two legs of a walk, and R is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R. If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-
direction. Those paths are the x- and y-components of the resultant, R and R. If we know R 2 and = tan–1( / ). When you use the analytical and R, we can find and using the equations = method of vector addition, you can determine the components or the magnitude and direction of a vector. 2 + Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations = cos and = sin to find the components. In Figure 3.31, 110 Chapter 3 \| Two-Dimensional Kinematics these components are,,, and. The angles that vectors A and B make with the x-axis are A and B, respectively. Figure 3.31 To add vectors A and B, first determine the horizontal and vertical components of each vector. These are the dotted vectors A, A, B and B shown in the image. Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 3.32, and = + = +. (3.12) (3.13) Figure 3.32 The magnitude of the vectors A and B add to give the magnitude of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors A and B add to give the magnitude of the resultant vector in the vertical direction. Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R are known, its magnitude and direction can be found. Step 3. To get the magnitude of the resultant, use the Pythagorean theorem: Step 4. To get the direction of the resultant: = 2. 2 + = tan−1( / ). The following example illustrates this technique for adding vectors using perpendicular components. (3.14) (3.15) This content is available for free at http://cnx.org/content
/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 111 Example 3.3 Adding Vectors Using Analytical Methods Add the vector A to the vector B shown in Figure 3.33, using perpendicular components along the x- and y-axes. The xand y-axes are along the east–west and north–south directions, respectively. Vector A represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0º north of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0º north of east. Figure 3.33 Vector A has magnitude 53.0 m and direction 20.0 º north of the x-axis. Vector B has magnitude 34.0 m and direction 63.0º north of the x-axis. You can use analytical methods to determine the magnitude and direction of R. Strategy The components of A and B along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant. Solution Following the method outlined above, we first find the components of A and B along the x- and y-axes. Note that = 53.0 m, A = 20.0º, = 34.0 m, and B = 63.0º. We find the x-components by using = cos, which gives and = cos A = (53.0 m)(cos 20.0º) = (53.0 m)(0.940) = 49.8 m = cos B = (34.0 m)(cos 63.0º) = (34.0 m)(0.454) = 15.4 m. Similarly, the y-components are found using = sin A : and = sin A = (53.0 m)(sin 20.0º) = (53.0 m)(0.342) = 18.1 m = sin B = (34.0 m)(sin 63.0 º ) = (34.0 m)(0.891) = 30.3 m. The x- and y-components of the resultant are thus and = + = 49.8 m + 15.4 m = 65.2 m = + = 18.1 m+30.3 m = 48.4 m. Now we can find the magnitude of the
resultant by using the Pythagorean theorem: = 2 + 2 = (65.2)2 + (48.4)2 m so that = 81.2 m. (3.16) (3.17) (3.18) (3.19) (3.20) (3.21) (3.22) (3.23) 112 Chapter 3 \| Two-Dimensional Kinematics Finally, we find the direction of the resultant: Thus, = tan−1( / )=+tan−1(48.4 / 65.2). = tan−1(0.742) = 36.6 º. (3.24) (3.25) Figure 3.34 Using analytical methods, we see that the magnitude of R is 81.2 m and its direction is 36.6º north of east. Discussion This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector. Subtraction of vectors is accomplished by the addition of a negative vector. That is, A − B ≡ A + (–B). Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of –B are the negatives of the components of B. The x- and y-components of the resultant A − B = R are thus and = + – = + – and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.) (3.26) (3.27) Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics. Figure 3.35 The subtraction of the two vectors shown in Figure 3.30. The components of –B are the negatives of the components of B. The method of subtraction is the same as that for addition. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 113 PhET Explorations: Vector Addition Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats
. Figure 3.36 Vector Addition (http://cnx.org/content/m54783/1.2/vector-addition_en.jar) 3.4 Projectile Motion By the end of this section, you will be able to: Learning Objectives • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. • Determine the location and velocity of a projectile at different points in its trajectory. • Apply the principle of independence of motion to solve projectile motion problems. The information presented in this section supports the following AP® learning objectives: • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.37 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s
, x, and y. (Note that in the last section we used the notation A to represent a vector with components A and A. If we continued this format, we would call displacement s with components s and s. However, to simplify the notation, we will simply represent the component vectors as x and y.) Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: = – = – 9.80 m/s2. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, = 0. Both accelerations are constant, so the kinematic equations can be used. Review of Kinematic Equations (constant ) = ( − 0). (3.28) (3.29) (3.30) (3.31) (3.32) 114 Chapter 3 \| Two-Dimensional Kinematics Figure 3.37 The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is, and it makes an angle with the horizontal. Given these assumptions, the following steps are then used to analyze projectile motion: Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so = cos and = sin are used. The magnitude of the components of displacement s along these axes are and The magnitudes of the components of the velocity v are = cos and = sin θ, where is the magnitude of the velocity and is its direction, as shown in Figure 3.38. Initial values are denoted with a subscript 0, as usual. Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms: Horizontal Motion( = 0) = 0 + = 0 = = velocity is a constant. Vertical Motion(assuming positive is up = − = −9.80m/s2) (( − 0). (3.33)
(3.34) (3.35) (3.36) (3.37) (3.38) (3.39) (3.40) Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below. Step 4. Recombine the two motions to find the total displacement s and velocity v. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing = displacement s and is the direction of the velocity v : 2 and = tan−1( / ) in the following form, where is the direction of the 2 + Total displacement and velocity = 2 + 2 = tan−1( / ) 2 2 + = = tan−1( / ). (3.41) (3.42) (3.43) (3.44) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 115 Figure 3.38 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because = 0 and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory. Example 3.4 A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3.39. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when
it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which = 0 and = –. We can then define 0 and 0 to be zero and solve for the desired quantities. Solution for (a) 116 Chapter 3 \| Two-Dimensional Kinematics By “height” we mean the altitude or vertical position above the starting point. The highest point in any trajectory, called the apex, is reached when = 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find : 2 = 0 2 − 2( − 0). (3.45) Figure 3.39 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally. Because 0 and are both zero, the equation simplifies to Solving for gives 3.46) (3.47) Now we must find 0, the component of the initial velocity in the y-direction. It is given by 0 = 0 sin, where 0 is the initial velocity of 70.0 m/s, and 0 = 75.0° is the initial angle. Thus, 0 = 0 sin 0 = (70.0 m/s)(sin 75°) = 67.6 m/s. and is so that Discussion for (a) = (67.6 m/s)2 2(9.80 m/s2), = 233m. (3.48) (3.49) (3.50) Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. Solution for (b) As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the
easiest method is to use = 0 + 1 (0 + ). Because 0 is zero, this equation reduces to simply 2 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics Note that the final vertical velocity,, at the highest point is zero. Thus, = 1 2 (0 + ). = 2 (0y + ) = 2(233 m) (67.6 m/s) = 6.90 s. 117 (3.51) (3.52) Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using = 0 + 0 − 1 the quadratic equation for.) 2 2, and solving Solution for (c) Because air resistance is negligible, = 0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by = 0 +, where 0 is equal to zero: =, where is the x-component of the velocity, which is given by = 0 cos 0. Now, = 0 cos 0 = (70.0 m/s)(cos 75.0°) = 18.1 m/s. The time for both motions is the same, and so is = (18.1 m/s)(6.90 s) = 125 m. Discussion for (c) (3.53) (3.54) (3.55) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below. In solving part (a) of the preceding example, the expression we found for is valid for any projectile motion where air resistance is negligible. Call the maximum height = ; then, = 2 0 2. (3.56) This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. Defining a Coordinate System It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the and positions. Often, it is convenient to choose the initial position of the object as the origin such that 0 = 0 and 0 =
0. It is also important to define the positive and negative directions in the and directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration,, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, takes a positive value. Example 3.5 Calculating Projectile Motion: Hot Rock Projectile Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0° above the horizontal, as shown in Figure 3.40. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock's velocity at impact? 118 Chapter 3 \| Two-Dimensional Kinematics Figure 3.40 The trajectory of a rock ejected from the Kilauea volcano. Strategy Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain and at the final time determined in the first part of the example. Solution for (a) While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using = 0 + 0 − 1 2 2. (3.57) If we take the initial position 0 to be zero, then the final position is = −20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from 0 = 0 sin 0 = ( 25.0
m/s )( sin 35.0° ) = 14.3 m/s. Substituting known values yields Rearranging terms gives a quadratic equation in : −20.0 m = (14.3 m/s) − 4.90 m/s2 2. 4.90 m/s2 2 − (14.3 m/s) − (20.0 m) = 0. (3.58) (3.59) This expression is a quadratic equation of the form 2 + + = 0, where the constants are = 4.90, = – 14.3, and = – 20.0. Its solutions are given by the quadratic formula3.60) This equation yields two solutions: = 3.96 and = – 1.03. (It is left as an exercise for the reader to verify these solutions.) The time is = 3.96 s or – 1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus, = 3.96 s. (3.61) Discussion for (a) The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Solution for (b) From the information now in hand, we can find the final horizontal and vertical velocities and and combine them to find the total velocity and the angle 0 it makes with the horizontal. Of course, is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore: = 0 cos 0 = (25.0 m/s)(cos 35°) = 20.5 m/s. The final vertical velocity is given by the following equation: = 0 − (3.62) (3.63) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 119 where 0y was found in part (a) to be 14.3 m/s. Thus, so that = 14.3 m/s − (9.80 m/s2)(3.96 s) = −24.5 m/s. To find the magnitude of the final velocity we combine
its perpendicular components, using the following equation: = 2 + 2 = (20.5 m/s)2 + ( − 24.5 m/s)2, which gives The direction is found from the equation: = 31.9 m/s. = tan−1( / ) so that Thus, Discussion for (b) = tan−1( − 24.5 / 20.5) = tan−1( − 1.19). = −50.1 °. (3.64) (3.65) (3.66) (3.67) (3.68) (3.69) (3.70) The negative angle means that the velocity is 50.1° below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.40.) One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further. Figure 3.41 Trajectories of projectiles on level ground. (a) The greater the initial speed 0, the greater the range for a given initial angle. (b) The effect of initial angle 0 on the range of a projectile with a given initial speed. Note that the range is the same for 15° and 75°, although the maximum heights of those paths are different. 120 Chapter 3 \| Two-Dimensional Kinematics How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed 0, the greater the range, as shown in Figure 3.41(a). The initial angle 0 also has a dramatic effect on the range, as illustrated in Figure 3.41(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with 0 = 45°. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38°. Interestingly, for
every initial angle except 45°, there are two angles that give the same range—the sum of those angles is 90°. The range also depends on the value of the acceleration of gravity. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range of a projectile on level ground for which air resistance is negligible is given by = 2 sin 2 0 0, (3.71) where 0 is the initial speed and 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-ofchapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.42.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic. Figure 3.42 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved. PhET Explorations: Projectile Motion Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target. Figure
3.43 Projectile Motion (http://cnx.org/content/m54787/1.2/projectile-motion_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 121 3.5 Addition of Velocities Learning Objectives By the end of this section, you will be able to: • Apply principles of vector addition to determine relative velocity. • Explain the significance of the observer in the measurement of velocity. The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2) • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1) Relative Velocity If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 3.44. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 3.45. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways. Figure 3.44 A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore. 122 Chapter 3 \| Two-Dimensional Kinematics Figure 3.45 An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow). In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium
has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 3.44 and Figure 3.45. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means. How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the velocity of the ball is 35 m/s relative to the stationary, profusely sweating goalkeeper standing in front of the goal. In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity ( and ) and its components ( and ) along the x- and y-axes of an appropriately chosen coordinate system: = cos = sin 2 2 + = = tan−1( / ). (3.72) (3.73) (3.74) (3.75) Figure 3.46 The velocity,, of an object traveling at an angle to the horizontal axis is the sum of component vectors v and v. These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 123 Take-Home Experiment: Relative Velocity of a Boat Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of
the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat. Example 3.6 Adding Velocities: A Boat on a River Figure 3.47 A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the boat relative to the shore? Refer to Figure 3.47, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat's velocity relative to an observer on the shore, vtot. The velocity of the boat, vboat, is 0.75 m/s in the direction relative to the river and the velocity of the river, vriver, is 1.20 m/s to the right. Strategy We start by choosing a coordinate system with its -axis parallel to the velocity of the river, as shown in Figure 3.47. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the -axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations tot = = tan−1( / ) directly. 2 + 2 and Solution The magnitude of the total velocity is where and Thus, yielding tot = 2, 2 + = river = 1.20 m/s = boat = 0.750 m/s. tot = (1.20 m/s)2 + (0.750 m/s)2 tot = 1.42 m/s. The direction of the total velocity is given by: (3.76) (3.77) (3.78) (3.79) (3.80) 124 Chapter 3 \| Two-Dimensional Kinematics This equation gives Discussion = tan−1( / ) = tan−1(0.750 / 1.20). = 32.0º. (3.81) (3.82) Both the magnitude and the direction of the total velocity are consistent with Figure 3.47. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 32.0º )
the total velocity has relative to the riverbank. Example 3.7 Calculating Velocity: Wind Velocity Causes an Airplane to Drift Calculate the wind velocity for the situation shown in Figure 3.48. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0º west of north. Figure 3.48 An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind? Strategy In this problem, somewhat different from the previous example, we know the total velocity vtot and that it is the sum of two other velocities, vw (the wind) and vp (the plane relative to the air mass). The quantity vp is known, and we are asked to find vw. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of vw, then we can combine them to solve for its magnitude and direction. As shown in Figure 3.48, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to vp ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods.) Solution Because vtot is the vector sum of the vw and vp, its x- and y-components are the sums of the x- and y-components of the wind and plane velocities. Note that the plane only has vertical component of velocity so p = 0 and p = p. That is, tot = w (3.83) and This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 125 tot = w + p. We can use the first of these two equations to find w : Because tot = 38.0 m / s and cos 110º = – 0.342 we have w = tot = totcos 110º. w = (38.0 m/s)(–0.342)=–13.0 m/s. The minus sign indicates motion
west which is consistent with the diagram. Now, to find w we note that tot = w + p Here tot = totsin 110º ; thus, w = (38.0 m/s)(0.940) − 45.0 m/s = −9.29 m/s. (3.84) (3.85) (3.86) (3.87) (3.88) This minus sign indicates motion south which is consistent with the diagram. Now that the perpendicular components of the wind velocity w and w are known, we can find the magnitude and direction of vw. First, the magnitude is w = w 2 2 + w = ( − 13.0 m/s)2 + ( − 9.29 m/s)2 w = 16.0 m/s. = tan−1(w / w) = tan−1( − 9.29 / −13.0) = 35.6º. so that The direction is: giving Discussion (3.89) (3.90) (3.91) (3.92) The wind's speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 3.48. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction. Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity. Relative Velocities and Classical Relativity When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to each other measure the same phenomenon. Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the
20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than 3,000 km/s. Most things we encounter in daily life move slower than this speed. Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 3.49.) To the observer on shore, the 126 Chapter 3 \| Two-Dimensional Kinematics binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in Figure 3.49. Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer. Figure 3.49 Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.) Example 3.8 Calculating Relative Velocity: An Airline Passenger Drops a Coin An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin
when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth? Figure 3.50 The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the coin move almost horizontally. Strategy Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/ This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 127 s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes. Solution for (a) Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation: 2 = 0 2 − 2( − 0). Substituting known values into the equation, we get 2 = 02 − 2(9.80 m/s2)( − 1.50 m − 0 m) = 29.4 m2 /s2 yielding = −5.42 m/s. (3.93) (3.94) (3.95) We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane. Solution for (b) Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is = − 5.42 m/s, the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and = 260 m/
s. The x- and y-components of velocity can be combined to find the magnitude of the final velocity: Thus, yielding The direction is given by: so that Discussion = 2 + 2. = (260 m/s)2 + ( − 5.42 m/s)2 = 260.06 m/s. = tan−1( / ) = tan−1( − 5.42 / 260) = tan−1( − 0.0208) = −1.19º. (3.96) (3.97) (3.98) (3.99) (3.100) In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not (260 – 5.42) m/s ; rather, it is 260.06 m/s. The velocity's magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure 3.50.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path. Making Connections: Relativity and Einstein Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more surprises await. 128 Chapter 3 \| Two-Dimensional Kinematics
PhET Explorations: Motion in 2D Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). Figure 3.51 Motion in 2D (http://cnx.org/content/m54798/1.2/motion-2d_en.jar) Glossary air resistance: a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero analytical method: the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities classical relativity: the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s commutative: refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum component (of a 2-d vector): a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components direction (of a vector): the orientation of a vector in space head (of a vector): the end point of a vector; the location of the tip of the vector's arrowhead; also referred to as the “tip” head-to-tail method: a method of adding vectors in which the tail of each vector is placed at the head of the previous vector kinematics: the study of motion without regard to mass or force magnitude (of a vector): the length or size of a vector; magnitude is a scalar quantity motion: displacement of an object as a function of time projectile: an object that travels through the air and experiences only acceleration due to gravity projectile motion: the motion of an object that is subject only to the acceleration of gravity range: the maximum horizontal distance that a projectile travels relative velocity: the velocity of an object as observed from a particular reference frame relativity: the study of how different observers moving relative to each other measure the same phenomenon resultant: the sum of two or more vectors resultant vector: the vector sum of two or more vectors scalar: a quantity with magnitude but no direction tail: the start point of a vector
; opposite to the head or tip of the arrow trajectory: the path of a projectile through the air vector: a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction vector addition: the rules that apply to adding vectors together velocity: speed in a given direction This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 129 Section Summary 3.1 Kinematics in Two Dimensions: An Introduction • The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and vertical components. • The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa. 3.2 Vector Addition and Subtraction: Graphical Methods • The graphical method of adding vectors A and B involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector R is defined such that A + B = R. The magnitude and direction of R are then determined with a ruler and protractor, respectively. • The graphical method of subtracting vector B from A involves adding the opposite of vector B, which is defined as −B. In this case, A – B = A + (–B) = R. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector R. • Addition of vectors is commutative such that A + B = B + A. • The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. If a vector A is multiplied by a scalar quantity, the magnitude of the product is given by. If is positive, the direction of the product points in the same direction as A ; if is negative, the direction of the product points in the opposite direction as A. • 3.3 Vector Addition and Subtraction: Analytical Methods • The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector. • The steps to add vectors A and B using the analytical method are as follows: Step 1
: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations and = cos = cos = sin = sin. Step 2: Add the horizontal and vertical components of each vector to determine the components and of the resultant vector, R : and = + = +. Step 3: Use the Pythagorean theorem to determine the magnitude,, of the resultant vector R : Step 4: Use a trigonometric identity to determine the direction,, of R : = tan−1( / ). = 2. 2 + 3.4 Projectile Motion • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. • To solve projectile motion problems, perform the following steps: 1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities and, and the components of the velocity v are given by = cos and = sin, where is the magnitude of the velocity and is its direction. 2. Analyze the motion of the projectile in the horizontal direction using the following equations: 130 Chapter 3 \| Two-Dimensional Kinematics Horizontal motion( = 0) = 0 + 3. Analyze the motion of the projectile in the vertical direction using the following equations: = 0 = vx = velocity is a constant. Vertical motion(Assuming positive direction is up; = − = −9.80 m/s20 + ) 2 = 0 4. Recombine the horizontal and vertical components of location and/or velocity using the following equations( − 0). = 2 + 2 = tan−1( / ) = 2 2 + v = tan−1( / ). • The maximum height of a projectile launched with initial vertical velocity 0 is given by = 2 0 2. • The maximum horizontal distance traveled by a projectile is called the range. The range of a projectile on level ground launched at an angle 0 above the horizontal with initial speed 0 is given by = 2 sin 2 0 0. 3.5 Addition of Velocities • Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as = cos = sin 2 2 + = = tan−1( / ). • Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. • Relativity is the study of how different observers measure the same phenomenon
, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s). Conceptual Questions 3.2 Vector Addition and Subtraction: Graphical Methods 1. Which of the following is a vector: a person's height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth's population, the acceleration of gravity? 2. Give a specific example of a vector, stating its magnitude, units, and direction. 3. What do vectors and scalars have in common? How do they differ? 4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 131 Figure 3.52 5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure 3.53. What other information would he need to get to Sacramento? Figure 3.53 6. Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A + B the sum of the lengths of the two steps? 7. Explain why it is not possible to add a scalar to a vector. 8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more? 3.3 Vector Addition and Subtraction: Analytical Methods 9. Suppose you add two vectors A and B. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude? 10. Give an example of a nonzero vector that has a component of zero. 11.
Explain why a vector cannot have a component greater than its own magnitude. 12. If the vectors A and B are perpendicular, what is the component of A along the direction of B? What is the component of B along the direction of A? 3.4 Projectile Motion 13. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90° ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at = 0? (d) Can the speed ever be the same as the initial speed at a time other than at = 0? 132 Chapter 3 \| Two-Dimensional Kinematics 14. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90° ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity? 15. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory? 16. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time. 3.5 Addition of Velocities 17. What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane? 18. A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn't he need to keep his eyes on the ball? 19. If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at
the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it? 20. The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger's frame of reference. Draw its path as viewed by a stationary observer. 21. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 3 \| Two-Dimensional Kinematics 133 Problems & Exercises 3.2 Vector Addition and Subtraction: Graphical Methods Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits. 1. Find the following for path A in Figure 3.54: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish. Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side. 2. Find the following for path B in Figure 3.54: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish. 3. Find the north and east components of the displacement for the hikers shown in Figure 3.52. 4. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.55, then this problem asks you to find their sum R = A + B.) Figure 3.56 6. Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B, which is 20.0 m in a direction exactly 40° south of west, and then leg A, which is 12.0 m in a direction exactly 20° west of north. (This problem shows that A
+ B = B + A.) 7. (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0° north of east (which is equivalent to subtracting B from A —that is, to finding R′ = A − B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0° south of west and then 12.0 m in a direction 20.0° east of south (which is equivalent to subtracting A from B —that is, to finding R′′ = B - A = - R′ ). Show that this is the case. 8. Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.) 9. Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24. 10. Find the magnitudes of velocities A and B in Figure 3.57 Figure 3.55 The two displacements A and B add to give a total displacement R having magnitude and direction. 5. Suppose you first walk 12.0 m in a direction 20° west of north and then 20.0 m in a direction 40.0° south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.56, then this problem finds their sum R = A + B.) Figure 3.57 The two velocities vA and vB add to give a total vtot. 11. Find the components of tot along the x- and y-axes in Figure 3.57. 134 Chapter 3 \| Two-Dimensional Kinematics 12. Find the components of tot along a set of perpendicular axes rotated 30° counterclockwise relative to those in Figure 3.57. 3.3 Vector Addition and Subtraction: Analytical Methods 13. Find the following for path C in Figure 3.58: (a)
the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side. 14. Find the following for path D in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. 15. Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.59. Figure 3.59 16. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.60, then this problem asks you to find their sum R = A + B.) This content is available for free at http://cnx.org/content/col11844/1.13 Figure 3.60 The two displacements A and B add to give a total displacement R having magnitude and direction. Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique. 17. Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path. 18. You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order. 19. Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to
25.0 m straight south. (This is equivalent to subtracting B from A —that is, finding R′ = A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A = B + C. Is that consistent with your result?) 20. A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result? Chapter 3 \| Two-Dimensional Kinematics 135 Figure 3.61 21. You fly 32.0 km in a straight line in still air in the direction 35.0º south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º. 22. A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A B and C in Figure 3.62, and then correctly calculates the length and orientation of the fourth side D. What is his result? Figure 3.62 23. In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island? 24. Suppose a pilot flies 40.0 km in a direction 60º north of east
and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.63. Find her total distance from the starting point and the direction of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass. Figure 3.63 3.4 Projectile Motion 25. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0° above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the and distances from where the projectile was launched to where it lands? 26. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball? 27. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground? 28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.) 29. An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull
's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch? 30. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take? 31. Verify the ranges for the projectiles in Figure 3.41(a) for = 45° and the given initial velocities. 32. Verify the ranges shown for the projectiles in Figure 3.41(b) for an initial velocity of 50 m/s at the given initial angles. 33. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here? 136 Chapter 3 \| Two-Dimensional Kinematics 34. An arrow is shot from a height of 1.5 m toward a cliff of height. It is shot with a velocity of 30 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff? 35. In the standing broad jump, one squats and then pushes off with
the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity,. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.) 36. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions. 37. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net? 38. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25° relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release? 39. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance. 40. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water. 41. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west
and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0° below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m. 42. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40° above the horizontal. 43. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent's goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s. 44. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial This content is available for free at http://cnx.org/content/col11844/1.13 speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems. 45. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0° above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45° when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38° will give a longer range than 45° in the shot put.) 46. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the
basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket? 47. A football player punts the ball at a 45.0° angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally? 48. Prove that the trajectory of a projectile is parabolic, having the form = + 2. To obtain this expression, solve the equation = 0 for and substitute it into the expression for = 0 – (1 / 2) 2 (These equations describe the and positions of a projectile that starts at the origin.) You should obtain an equation of the form = + 2 where and are constants. 49. Derive = 2 sin 2θ0 0 for the range of a projectile on level ground by finding the time at which becomes zero and substituting this value of into the expression for − 0, noting that = − 0 50. Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon. 51. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball's needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen. Chapter 3 \| Two-Dimensional Kinematics 137 3.5 Addition of Velocities 52. Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a
) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45º south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass? 53. A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If the bird turns around and flies with the wind, how long will he take to return 6.00 km? (c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind. 54. Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line? 55. Verify that the coin dropped by the airline passenger in the Example 3.8 travels 144 m horizontally while falling 1.50 m in the frame of reference of the Earth. 56. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0º relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback? 57. A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0º north of east. What is the velocity of the ship relative to the Earth? 58. (a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0º south of west. It
is in the jet stream, which is blowing at 35.0 m/s in a direction 15º south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane's path. 59. (a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 7.00 m/s? (b) What would its speed be relative to the Earth? 60. (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20º south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is 45.0º south of west, while its direction of travel relative to the air is 5.00º south of west. What is the airplane's speed relative to the air mass? (b) What is the airplane's speed relative to the Earth? 61. A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck. 62. The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0º east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0º south of west relative to the Earth. What is the velocity of the wind relative to the water? 63. The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. Figure 3.64 illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5. The
results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information. Figure 3.64 Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 Mly across. 64. (a) Use the distance and velocity data in Figure 3.64 to find the rate of expansion as a function of distance. (b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively. 65. An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground? 66. A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00º west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.) 67. An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a 90.0º angle relative to his path as shown in Figure 3.65. What angle must the puck's velocity make relative to the player (in his frame of reference) to hit the center of the goal? 138 Chapter 3 \| Two-Dimensional Kinematics Figure 3.65 An ice hockey player moving across the rink must shoot
backward to give the puck a velocity toward the goal. 68. Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height? (d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. 69. Unreasonable Results A commercial airplane has an air speed of 280 m/s due east and flies with a strong tailwind. It travels 3000 km in a direction 5º south of east in 1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the magnitude and direction of the tailwind's velocity. (c) What is unreasonable about both of these velocities? (d) Which premise is unreasonable? 70. Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway. Test Prep for AP® Courses 3.1 Kinematics in Two Dimensions: An Introduction 1. A ball is thrown at an angle of 45 degrees above the horizontal. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground? a. Always in the same direction as the motion, initially b. positive and gradually dropping to zero by the time it hits the ground Initially positive in the upward direction, then zero at maximum height, then negative from there until it hits the ground c. Always in the opposite direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground d. Always in the downward direction with the same constant value 2. In an experiment, a student launches a ball with an initial horizontal velocity at an elevation 2 meters above ground. The ball follows a parabolic trajectory until
it hits the ground. Which of the following accurately describes the graph of the ball's vertical acceleration versus time (taking the downward direction to be negative)? a. A negative value that does not change with time b. A gradually increasing negative value (straight line) This content is available for free at http://cnx.org/content/col11844/1.13 c. An increasing rate of negative values over time (parabolic curve) d. Zero at all times since the initial motion is horizontal 3. A student wishes to design an experiment to show that the acceleration of an object is independent of the object's velocity. To do this, ball A is launched horizontally with some initial speed at an elevation 1.5 meters above the ground, ball B is dropped from rest 1.5 meters above the ground, and ball C is launched vertically with some initial speed at an elevation 1.5 meters above the ground. What information would the student need to collect about each ball in order to test the hypothesis? 3.2 Vector Addition and Subtraction: Graphical Methods 4. A ball is launched vertically upward. The vertical position of the ball is recorded at various points in time in the table shown. Chapter 3 \| Two-Dimensional Kinematics 139 Table 3.1 Height (m) Time (sec) 0.490 0.882 1.176 1.372 1.470 1.470 1.372 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Which of the following correctly describes the graph of the ball's vertical velocity versus time? a. Always positive, steadily decreasing b. Always positive, constant c. Initially positive, steadily decreasing, becoming negative at the end Initially zero, steadily getting more and more negative d. 5. Table 3.2 Height (m) Time (sec) 0.490 0.882 1.176 1.372 1.470 1.470 1.372 0.1 0.2 0.3 0.4 0.5 0.6 0.7 A ball is launched at an angle of 60 degrees above the horizontal, and the vertical position of the ball is recorded at various points in time in the table shown, assuming the ball was at a height of 0 at time t = 0. a. Draw a graph of the ball's vertical velocity versus time. b. Describe the graph of the ball's horizontal velocity. c. Draw a graph of the ball's vertical acceleration versus time. 3.
4 Projectile Motion 6. In an experiment, a student launches a ball with an initial horizontal velocity of 5.00 meters/sec at an elevation 2.00 meters above ground. Draw and clearly label with appropriate values and units a graph of the ball's horizontal velocity vs. time and the ball's vertical velocity vs. time. The graph should cover the motion from the instant after the ball is launched until the instant before it hits the ground. Assume the downward direction is negative for this problem. 140 Chapter 3 \| Two-Dimensional Kinematics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 141 4 DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION Figure 4.1 Newton’s laws of motion describe the motion of the dolphin’s path. (credit: Jin Jang) Chapter Outline 4.1. Development of Force Concept 4.2. Newton's First Law of Motion: Inertia 4.3. Newton's Second Law of Motion: Concept of a System 4.4. Newton's Third Law of Motion: Symmetry in Forces 4.5. Normal, Tension, and Other Examples of Force 4.6. Problem-Solving Strategies 4.7. Further Applications of Newton's Laws of Motion 4.8. Extended Topic: The Four Basic Forces—An Introduction Connection for AP® Courses Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular motion, such as that of a jumping dolphin, a leaping pole vaulter, a bird in flight, or an orbiting satellite. The study of motion is kinematics, but kinematics only describes the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of principles under which nature functions. They are also universal laws in that they apply to situations on Earth as well as in space. Isaac Newton’s (1642–1727) laws of motion were just one part of the monumental work that has made him legendary. The development of Newton’s laws marks the transition from the Renaissance into the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. For
many centuries natural philosophers had debated the nature of the universe based largely on certain rules of logic, with great weight given to the thoughts of earlier 142 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion classical philosophers such as Aristotle (384–322 BC). Among the many great thinkers who contributed to this change were Newton and Galileo Galilei (1564–1647). Figure 4.2 Isaac Newton’s monumental work, Philosophiae Naturalis Principia Mathematica, was published in 1687. It proposed scientific laws that are still used today to describe the motion of objects. (credit: Service commun de la documentation de l'Université de Strasbourg) Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than “logical” argument. Galileo’s use of the telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons orbiting Jupiter and made other observations that were inconsistent with certain ancient ideas and religious dogma. For this reason, and because of the manner in which he dealt with those in authority, Galileo was tried by the Inquisition and punished. He spent the final years of his life under a form of house arrest. Because others before Galileo had also made discoveries by observing the nature of the universe and because repeated observations verified those of Galileo, his work could not be suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the church and scientific communities. Galileo also contributed to the formulation of what is now called Newton’s first law of motion. Newton made use of the work of his predecessors, which enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great contributions to the theories of light and color. It is amazing that many of these developments were made by Newton working alone, without the benefit of the usual interactions that take place among scientists today. Newton’s laws are introduced along with Big Idea 3, that interactions can be described by forces. These laws provide a theoretical basis for studying motion depending on interactions between the objects. In particular, Newton's laws are applicable to all forces in inertial frames of references (Enduring Understanding 3.A). We will find that all forces are vectors; that is, forces always have both a magnitude and a direction (Essential Knowledge 3.A.2). Furthermore, we will learn that all forces are a result of interactions between two or more objects (Essential Knowledge 3.A.3
). These interactions between any two objects are described by Newton's third law, stating that the forces exerted on these objects are equal in magnitude and opposite in direction to each other (Essential Knowledge 3.A.4). We will discover that there is an empirical cause-effect relationship between the net force exerted on an object of mass m and its acceleration, with this relationship described by Newton's second law (Enduring Understanding 3.B). This supports Big Idea 1, that inertial mass is a property of an object or a system. The mass of an object or a system is one of the factors affecting changes in motion when an object or a system interacts with other objects or systems (Essential Knowledge 1.C.1). Another is the net force on an object, which is the vector sum of all the forces exerted on the object (Essential Knowledge 3.B.1). To analyze this, we use free-body diagrams to visualize the forces exerted on a given object in order to find the net force and analyze the object's motion (Essential Knowledge 3.B.2). Thinking of these objects as systems is a concept introduced in this chapter, where a system is a collection of elements that could be considered as a single object without any internal structure (Essential Knowledge 5.A.1). This will support Big Idea 5, that changes that occur to the system due to interactions are governed by conservation laws. These conservation laws will be the focus of later chapters in this book. They explain whether quantities are conserved in the given system or change due to transfer to or from another system due to interactions between the systems (Enduring Understanding 5.A). Furthermore, when a situation involves more than one object, it is important to define the system and analyze the motion of a whole system, not its elements, based on analysis of external forces on the system. This supports Big Idea 4, that interactions between systems cause changes in those systems. All kinematics variables in this case describe the motion of the center of mass of the system (Essential Knowledge 4.A.1, Essential Knowledge 4.A.2). The internal forces between the elements of the system do not affect the velocity of the center of mass (Essential Knowledge 4.A.3). The velocity of the center of mass will change only if there is a net external force exerted on the system (Enduring Understanding 4.A). We will learn that some of these interactions can be explained by the existence of fields extending through space, supporting
Big Idea 2. For example, any object that has mass creates a gravitational field in space (Enduring Understanding 2.B). Any material object (one that has mass) placed in the gravitational field will experience gravitational force (Essential Knowledge 2.B.1). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 143 Forces may be categorized as contact or long-distance (Enduring Understanding 3.C). In this chapter we will work with both. An example of a long-distance force is gravitation (Essential Knowledge 3.C.1). Contact forces, such as tension, friction, normal force, and the force of a spring, result from interatomic electric forces at the microscopic level (Essential Knowledge 3.C.4). It was not until the advent of modern physics early in the twentieth century that it was discovered that Newton’s laws of motion produce a good approximation to motion only when the objects are moving at speeds much, much less than the speed of light and when those objects are larger than the size of most molecules (about 10–9 m in diameter). These constraints define the realm of classical mechanics, as discussed in Introduction to the Nature of Science and Physics. At the beginning of the twentieth century, Albert Einstein (1879–1955) developed the theory of relativity and, along with many other scientists, quantum theory. Quantum theory does not have the constraints present in classical physics. All of the situations we consider in this chapter, and all those preceding the introduction of relativity in Special Relativity, are in the realm of classical physics. The development of special relativity and empirical observations at atomic scales led to the idea that there are four basic forces that account for all known phenomena. These forces are called fundamental (Enduring Understanding 3.G). The properties of gravitational (Essential Knowledge 3.G.1) and electromagnetic (Essential Knowledge 3.G.2) forces are explained in more detail. Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Essential Knowledge 1.C.1 Inertial mass is the property of an object or a system that determines how its motion changes when it interacts with other objects or systems. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.A A field associates a value of some physical quantity with every point in space. Field models are
useful for describing interactions that occur at a distance (long-range forces) as well as a variety of other physical phenomena. Essential Knowledge 2.A.1 A vector field gives, as a function of position (and perhaps time), the value of a physical quantity that is described by a vector. Essential Knowledge 2.A.2 A scalar field gives the value of a physical quantity. Enduring Understanding 2.B A gravitational field is caused by an object with mass. Essential Knowledge 2.B.1 A gravitational field g at the location of an object with mass m causes a gravitational force of magnitude mg to be exerted on the object in the direction of the field. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.2 Forces are described by vectors. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Essential Knowledge 3.A.4 If one object exerts a force on a second object, the second object always exerts a force of equal magnitude on the first object in the opposite direction. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using = ∑ /. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing the forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.1 Gravitational force describes the interaction of one object that has mass with another object that has mass. Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). Enduring Understanding 3.G Certain types of forces are considered fundamental. Essential Knowledge 3.G.1 Gravitational forces are exerted at all scales and dominate at the largest distance and mass scales. Essential Knowledge 3.G.2 Electromagnetic forces are
exerted at all scales and can dominate at the human scale. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.A The acceleration of the center of mass of a system is related to the net force exerted on the system, where = ∑ /. Essential Knowledge 4.A.1 The linear motion of a system can be described by the displacement, velocity, and acceleration of its center of mass. Essential Knowledge 4.A.2 The acceleration is equal to the rate of change of velocity with time, and velocity is equal to the rate of change of position with time. 144 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Essential Knowledge 4.A.3 Forces that systems exert on each other are due to interactions between objects in the systems. If the interacting objects are parts of the same system, there will be no change in the center-of-mass velocity of that system. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Essential Knowledge 5.A.1 A system is an object or a collection of objects. The objects are treated as having no internal structure. 4.1 Development of Force Concept By the end of this section, you will be able to: • Understand the definition of force. Learning Objectives The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) • 3.A.3.2 The student is able to challenge a claim that an object can exert a force on itself. (S.P. 6.1) • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) • 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Dynamics is the study of the forces that cause
objects and systems to move. To understand this, we need a working definition of force. Our intuitive definition of force—that is, a push or a pull—is a good place to start. We know that a push or pull has both magnitude and direction (therefore, it is a vector quantity) and can vary considerably in each regard. For example, a cannon exerts a strong force on a cannonball that is launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea. Our everyday experiences also give us a good idea of how multiple forces add. If two people push in different directions on a third person, as illustrated in Figure 4.3, we might expect the total force to be in the direction shown. Since force is a vector, it adds just like other vectors, as illustrated in Figure 4.3(a) for two ice skaters. Forces, like other vectors, are represented by arrows and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in Two-Dimensional Kinematics. By definition, force is always the result of an interaction of two or more objects. No object possesses force on its own. For example, a cannon does not possess force, but it can exert force on a cannonball. Earth does not possess force on its own, but exerts force on a football or on any other massive object. The skaters in Figure 4.3 exert force on one another as they interact. No object can exert force on itself. When you clap your hands, one hand exerts force on the other. When a train accelerates, it exerts force on the track and vice versa. A bowling ball is accelerated by the hand throwing it; once the hand is no longer in contact with the bowling ball, it is no longer accelerating the bowling ball or exerting force on it. The ball continues moving forward due to inertia. Figure 4.3 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. Figure 4.3(b) is our first example of a free-body diagram, which is a technique used to illustrate all the external forces acting on a body. The body is represented by a single
isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. (These forces are the only ones shown, because only external forces acting on the body affect its motion. We can ignore any internal forces within the body.) Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton’s laws of motion. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 145 A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a standard distance. One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure 4.4, and use the force it exerts to pull itself back to its relaxed shape—called a restoring force—as a standard. The magnitude of all other forces can be stated as multiples of this standard unit of force. Many other possibilities exist for standard forces. (One that we will encounter in Magnetism is the magnetic force between two wires carrying electric current.) Some alternative definitions of force will be given later in this chapter. Figure 4.4 The force exerted by a stretched spring can be used as a standard unit of force. (a) This spring has a length when undistorted. (b) When stretched a distance Δ, the spring exerts a restoring force, Frestore, which is reproducible. (c) A spring scale is one device that uses a spring to measure force. The force Frestore is exerted on whatever is attached to the hook. Here Frestore has a magnitude of 6 units in the force standard being employed. Take-Home Experiment: Force Standards To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a hook. Find a small household item that could be attached to the rubber band using a paper clip, and use this item as a weight to investigate the stretch of the rubber band. Measure the amount of stretch produced in the rubber band with one, two, and four of these (identical) items suspended from the rubber band. What is the relationship between the number of items and the amount of stretch? How large a stretch would you expect for the same number of items suspended from two rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached)
if the weights are also pushed to the side with a pencil? 4.2 Newton's First Law of Motion: Inertia Learning Objectives By the end of this section, you will be able to: • Define mass and inertia. • Understand Newton's first law of motion. Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following: Newton’s First Law of Motion There exists an inertial frame of reference such that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion. The first law of motion postulates the existence of at least one frame of reference which we call an inertial reference frame, relative to which the motion of an object not subject to forces is a straight line at a constant speed. An inertial reference frame is any reference frame that is not itself accelerating. A car traveling at constant velocity is an inertial reference frame. A car slowing down for a stoplight, or speeding up after the light turns green, will be accelerating and is not an inertial reference frame. Finally, when the car goes around a turn, which is due to an acceleration changing the direction of the velocity vector, it is not an inertial reference frame. Note that Newton’s laws of motion are only valid for inertial reference frames. Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net external force) for there to be any change in velocity (either a change in magnitude or direction) in an inertial reference frame. We will define net external force in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down? The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother
by rubbing 146 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object will slow down. Friction is an external force. Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian fashion, “That is the nature of the beast.” True perhaps, but not a useful insight. Mass The property of a body to remain at rest or to remain in motion with constant velocity is called inertia. Newton’s first law is often called the law of inertia. As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass. An object with a small mass will exhibit less inertia and be more affected by other objects. An object with a large mass will exhibit greater inertia and be less affected by other objects. This inertial mass of an object is a measure of how difficult it is to alter the uniform motion of
the object by an external force. Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the masses of objects are determined by comparison with the standard kilogram. Check Your Understanding Which has more mass: a kilogram of cotton balls or a kilogram of gold? Solution They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them are volume and density. 4.3 Newton's Second Law of Motion: Concept of a System Learning Objectives By the end of this section, you will be able to: • Define net force, external force, and system. • Understand Newton’s second law of motion. • Apply Newton’s second law to determine the weight of an object. Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned. First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration. Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an external force acts from outside the system of interest. For example, in Figure 4.5(a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between
elements of the system. Again looking at Figure 4.5(a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 147 When we describe the acceleration of a system, we are modeling the system as a single point which contains all of the mass of that system. The point we choose for this is the point about which the system’s mass is evenly distributed. For example, in a rigid object, this center of mass is the point where the object will stay balanced even if only supported at this point. For a sphere or disk made of homogenous material, this point is of course at the center. Similarly, for a rod made of homogenous material, the center of mass will be at the midpoint. For the rider in the wagon in Figure 4.5, the center of mass is probably between the rider’s hips. Due to internal forces, the rider’s hand or hair may accelerate slightly differently, but it is the acceleration of the system’s center of mass that interests us. This is true whether the system is a vehicle carrying passengers, a bowl of grapes, or a planet. When we draw a free-body diagram of a system, we represent the system’s center of mass with a single point and use vectors to indicate the forces exerted on that center of mass. (See Figure 4.5.) Figure 4.5 Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows representing all external forces are shown. The system of interest is the wagon and its rider. The weight w of the system and the support of the ground N are also shown for completeness and are assumed to cancel.
The vector f represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force, Fnet. The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration ( a′ > a ) when an adult pushes the child. Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 4.5. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight w and the support of the ground N, and the horizontal force f represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure 4.5(b) shows how vectors representing the external forces add together to produce a net force, Fnet. To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality a ∝ Fnet (4.1) where the symbol ∝ means “proportional to,” and Fnet is the net external force. (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in Two-Dimensional Kinematics.) This proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the
myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification 148 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 4.6, the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as a ∝ 1 (4.2) where is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force. Figure 4.6 The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (The effect of gravity on the ball is ignored.) (b) The same player exerts an identical force on a stalled SUV and produces a far smaller acceleration (even if friction is negligible). (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for the free-body diagram will emerge as you do more problems. Both of these proportionalities have been experimentally verified repeatedly and consistently, for a broad range of systems and scales. Thus, it has been experimentally found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion. Applying the Science Practices: Testing the Relationship Between Mass, Acceleration, and Force Plan three simple experiments using objects you have at home to test relationships between mass, acceleration, and force. (a) Design an experiment to test the relationship between mass and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure force is constant? (b) Design a similar experiment to test the relationship between mass and force. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure acceleration is constant? (c) Design a similar experiment to test the relationship between force and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? Will you have any
trouble ensuring that the mass is constant? What did you learn? Newton’s Second Law of Motion The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton’s second law of motion is This is often written in the more familiar form a = Fnet. Fnet = a. When only the magnitude of force and acceleration are considered, this equation is simply net =. (4.3) (4.4) (4.5) Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 149 Applying the Science Practices: Systems and Free-Body Diagrams First, consider a person on a sled sliding downhill. What is the system in this situation? Try to draw a free-body diagram describing this system, labeling all the forces and their directions. Which of the forces are internal? Which are external? Next, consider a person on a sled being pushed along level ground by a friend. What is the system in this situation? Try to draw a free-body diagram describing this system, labelling all the forces and their directions. Which of the forces are internal? Which are external? Units of Force Fnet = a is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1m/s2. That is, since Fnet = a, 1 N = 1 kg ⋅ m/s2. (4.6) While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Weight and the Gravitational Force When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If
air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight w. Weight can be denoted as a vector w because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration. Using Galileo’s result and Newton’s second law, we can derive an equation for weight. Consider an object with mass falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude. Newton’s second law states that the magnitude of the net external force on an object is net =. Since the object experiences only the downward force of gravity, net =. We know that the acceleration of an object due to gravity is, or =. Substituting these into Newton’s second law gives Weight This is the equation for weight—the gravitational force on a mass : =. Since = 9.80 m/s2 on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: = = (1.0 kg)(9.80 m/s2 ) = 9.8 N. (4.7) (4.8) Recall that can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight. When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object. The acceleration due to gravity varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only 1.67 m/s2. A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon. The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as
Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “freefall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness. It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and 150 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons. Common Misconceptions: Mass vs. Weight Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object ( ) multiplied by the acceleration due to gravity ( ). Like any other force, weight is measured in terms of newtons (or pounds in English units). Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is 1.67 m/s2 (which is much less than the acceleration due to gravity on Earth, 9.80 m/s2 ). If you measured your weight on Earth and then measured your weight on the Moon, you would find that you “weigh” much
less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing “mass” (which in turn causes them to weigh less). Take-Home Experiment: Mass and Weight What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon? Example 4.1 What Acceleration Can a Person Produce when Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration? Figure 4.7 The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? Strategy Since Fnet and are given, the acceleration can be calculated directly from Newton’s second law as stated in Fnet = a. Solution The magnitude of the acceleration is = net. Entering known values gives = 51 N 24 kg (4.9) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Substituting the units kg ⋅ m/s2 for N yields = 51 kg ⋅ m/s2 24 kg = 2.1 m/s2. 151 (4.10) Discussion The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted
by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would soon be reached. Example 4.2 What Rocket Thrust Accelerates This Sled? Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure 4.8. The sled’s initial acceleration is 49 m/s2, opposing the motion is known to be 650 N. the mass of the system is 2100 kg, and the force of friction Figure 4.8 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T. As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight, w. The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale. Strategy Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure. Solution Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with net =, (4.11) where net is the net force along the horizontal direction. We can see from Figure 4.8 that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force
is net = 4 −. (4.12) 152 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Substituting this into Newton’s second law gives net = = 4 −. Using a little algebra, we solve for the total thrust 4T: 4 = +. Substituting known values yields 4 = + = (2100 kg)(49 m/s2 ) + 650 N. So the total thrust is and the individual thrusts are Discussion 4 = 1.0×105 N, = 1.0×105 N 4 = 2.6×104 N. (4.13) (4.14) (4.15) (4.16) (4.17) The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45's. (Recall that, the acceleration due to gravity, is 9.80 m/s2. When we say that an acceleration is 45's, it is 459.80 m/s2, which is approximately 440 m/s2.) While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious. Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion. Applying the Science Practices: Sums of Forces Recall that forces are vector quantities, and therefore the net force acting on a system should be the vector sum of the forces. (a) Design an experiment to test this hypothesis. What sort of a system would be appropriate and convenient to have multiple forces applied to it? What features of the system should be held constant? What could be varied? Can forces be arranged in multiple directions so that, while the hypothesis is still tested, the resulting calculations are not too inconvenient? (b) Another group of students has done such
an experiment, using a motion capture system looking down at an air hockey table to measure the motion of the 0.10-kg puck. The table was aligned with the cardinal directions, and a compressed air hose was placed in the center of each side, capable of varying levels of force output and fixed so that it was aimed at the center of the table. Table 4.1 Forces Measured acceleration (magnitudes) 3 N north, 4 N west 5 N south, 12 N east 48 ± 4 m/s2 132 ± 6 m/s2 6 N north, 12 N east, 4 N west 99 ± 3 m/s2 Given the data in the table, is the hypothesis confirmed? What were the directions of the accelerations? 4.4 Newton's Third Law of Motion: Symmetry in Forces Learning Objectives By the end of this section, you will be able to: • Understand Newton's third law of motion. • Apply Newton's third law to define systems and solve problems of motion. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 153 The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) • 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) • 3.A.4.1 The student is able to construct explanations of physical situations involving the interaction of bodies using Newton's third law and the representation of action-reaction pairs of forces. (S.P. 1.4, 6.2) • 3.A.4.2 The student is able to use Newton's third law to make claims and predictions about the action-reaction pairs of forces when two objects interact. (S.P. 6.4, 7.2) • 3.
A.4.3 The student is able to analyze situations involving interactions among several objects by using free-body diagrams that include the application of Newton's third law to identify forces. (S.P. 1.4) • 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) • 4.A.2.1 The student is able to make predictions about the motion of a system based on the fact that acceleration is equal to the change in velocity per unit time, and velocity is equal to the change in position per unit time. (S.P. 6.4) • 4.A.2.2 The student is able to evaluate using given data whether all the forces on a system or whether all the parts of a system have been identified. (S.P. 5.3) • 4.A.3.1 The student is able to apply Newton's second law to systems to calculate the change in the center-of-mass velocity when an external force is exerted on the system. (S.P. 2.2) There is a passage in the musical Man of la Mancha that relates to Newton’s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, “Of course I hit her back, Your Grace, but she’s a lot harder than me and you know what they say, ‘Whether the stone hits the pitcher or the pitcher hits the stone, it’s going to be bad for the pitcher.’” This is exactly what happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in Newton’s third law of motion. Newton’s Third Law of Motion Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the force exerted
is the action and the force experienced as a consequence is the reaction. Newton’s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system. We can readily see Newton’s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.9. She pushes against the pool wall with her feet and accelerates in the direction opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not because they act on different systems. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then Fwall on feet is an external force on this system and affects its motion. The swimmer moves in the direction of Fwall on feet. In contrast, the force Ffeet on wall acts on the wall and not on our system of interest. Thus Ffeet on wall does not directly affect the motion of the system and does not cancel Fwall on feet. Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction. 154 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Figure 4.9 When the swimmer exerts a force Ffeet on wall on the wall, she accelerates in the direction opposite to that of her push. This means the net external force on her is in the direction opposite to Ffeet on wall. This opposition occurs because, in accordance with Newton’s third law of motion, the wall exerts a force Fwall on feet on her, equal in magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates the system of interest. Note that Ffeet on wall does not act on this system (the swimmer) and, thus, does not cancel Fwall on feet. Thus the free-body diagram shows only Fwall on feet, w, the gravitational force, and BF, the buoyant force of the water supporting the swimmer’s weight. The vertical forces w and BF cancel since there is no vertical motion. Similarly, when a person stands on Earth, the Earth exerts a force on the person, pulling the
person toward the Earth. As stated by Newton’s third law of motion, the person also exerts a force that is equal in magnitude, but opposite in direction, pulling the Earth up toward the person. Since the mass of the Earth is so great, however, and =, the acceleration of the Earth toward the person is not noticeable. Other examples of Newton’s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho’s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent’s body. Example 4.3 Getting Up To Speed: Choosing the Correct System A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 4.10. Her mass is 65.0 kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N. This content is available for free
at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 155 Figure 4.10 A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for f, since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for Example 4.4, since it asks for the acceleration of the entire group of objects. Only Ffloor and f are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for this example so that Fprof will be an external force and enter into Newton’s second law. Note that the free-body diagrams, which allow us to apply Newton’s second law, vary with the system chosen. Strategy Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 4.10. The professor pushes backward with a force Ffoot of 150 N. According to Newton’s third law, the floor exerts a forward reaction force Ffloor of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of Ffloor. Note that we do not include the forces Fprof or Fcart because these are internal forces, and we do not include Ffoot because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton’s second law to find the acceleration as requested. See the free-body diagram in the figure. Solution Newton’s second law is given by = net. The net external force on System 1 is deduced from Figure 4.10 and the discussion above to be net = floor − = 150 N − 24.0 N = 126 N. The mass of System 1 is = (65.0 + 12.0 + 7.0) kg = 84 kg. These values of net and produce an acceleration of net = = 126 N 84 kg = 1.5 m/s2. Discussion (4.18
) (4.19) (4.20) (4.21) 156 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion None of the forces between components of System 1, such as between the professor’s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem. Example 4.4 Force on the Cart—Choosing a New System Calculate the force the professor exerts on the cart in Figure 4.10 using data from the previous example if needed. Strategy If we now define the system of interest to be the cart plus equipment (System 2 in Figure 4.10), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, Fprof, is an external force acting on System 2. Fprof was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2. Solution Newton’s second law can be used to find Fprof. Starting with = net and noting that the magnitude of the net external force on System 2 is net = prof −, we solve for prof, the desired quantity: prof = net +. (4.22) (4.23) (4.24) The value of is given, so we must calculate net net. That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that net =, (4.25) where the mass of System 2 is 19.0 kg ( = 12.0 kg + 7.0 kg) and its acceleration was found to be = 1.5 m/s2 in the previous example. Thus, Now we can find the desired force: net =, net = (19.0 kg)(1.5 m/s2 ) = 29 N. prof = net +, prof = 29 N+24.0 N = 53 N. (4.26) (4.27) (4.28
) (4.29) Discussion It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing). PhET Explorations: Gravity Force Lab Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 157 Figure 4.11 Gravity Force Lab (http://cnx.org/content/m54849/1.2/gravity-force-lab_en.jar) 4.5 Normal, Tension, and Other Examples of Force Learning Objectives By the end of this section, you will be able to: • Define normal and tension forces. • Apply Newton's laws of motion to solve problems involving a variety of forces. • Use trigonometric identities to resolve weight into components. The information presented in this section supports the following AP® learning objectives and science practices: • 2.B.1.1 The student is able to apply = to calculate the gravitational force on an object with mass m in a gravitational field of strength g in the context of the effects of a net force on objects and systems. (S.P. 2.2, 7.2) • 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) • 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) • 3.A.4.1 The student is able to construct explanations of physical situations involving the interaction of bodies using Newton's third law and the
representation of action-reaction pairs of forces. (S.P. 1.4, 6.2) • 3.A.4.2 The student is able to use Newton's third law to make claims and predictions about the action-reaction pairs of forces when two objects interact. (S.P. 6.4, 7.2) • 3.A.4.3 The student is able to analyze situations involving interactions among several objects by using free-body diagrams that include the application of Newton's third law to identify forces. (S.P. 1.4) • 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) • 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text. Normal Force Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 4.12(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 4.12(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until the restoring force becomes as large
as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it. 158 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Figure 4.12 (a) The person holding the bag of dog food must supply an upward force Fhand equal in magnitude and opposite in direction to the weight of the food w. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N equal in magnitude and opposite in direction to the weight of the load. We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol N. (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example. Common Misconception: Normal Force (N) vs. Newton (N) In this section we have introduced the quantity normal force, which is represented by the variable N. This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force ( N ) happen to be newtons (N). For example, the normal force N that the floor exerts on a chair might be N = 100 N. One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work ( ) and the unit watts (W). Example 4.5 Weight on an Incline, a Two-Dimensional Problem Consider the skier on a slope shown in Figure 4.13. Her mass including equipment is 60.0 kg.
(a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 159 Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely w⊥ and w ∥. N is equal in magnitude to w⊥, so that there is no motion perpendicular to the slope, but is less than ∥, so that there is a downslope acceleration (along the parallel axis). Strategy This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols ⊥ and ∥ to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w, f, and N in Figure 4.13. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining ∥ to be the component of weight parallel to the slope and ⊥ the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope. Solution The magnitude of the component of the weight parallel to the slope is ∥ = sin (25º) = sin (25º)
, and the magnitude of the component of the weight perpendicular to the slope is ⊥ = cos (25º) = cos (25º). (a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope ∥ and friction. Using Newton’s second law, with subscripts to denote quantities parallel to the slope, where net ∥ = ∥ = sin (25º), assuming no friction for this part, so that ∥ = net ∥ ∥ = net ∥ = sin (25º) (9.80 m/s2)(0.4226) = 4.14 m/s2 = sin (25º) (4.30) (4.31) (4.32) is the acceleration. (b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now net ∥ = ∥ −, (4.33) and substituting this into Newton’s second law, ∥ = net ∥, gives 160 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion ∥ = net ∣ ∣ = ∥ − = sin (25º) −. We substitute known values to obtain ∥ = (60.0 kg)(9.80 m/s2)(0.4226) − 45.0 N 60.0 kg, which yields ∥ = 3.39 m/s2, (4.34) (4.35) (4.36) which is the acceleration parallel to the incline when there is 45.0 N of opposing friction. Discussion Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is = sin, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).
Resolving Weight into Components Figure 4.14 An object rests on an incline that makes an angle θ with the horizontal. When an object rests on an incline that makes an angle with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, w⊥, and a force acting parallel to the plane, w ∥. The perpendicular force of weight, w⊥, is typically equal in magnitude and opposite in direction to the normal force, N. The force acting parallel to the plane, w ∥, causes the object to accelerate down the incline. The force of friction, f, opposes the motion of the object, so it acts upward along the plane. It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle to the horizontal, then the magnitudes of the weight components are and ∥ = sin () = sin () ⊥ = cos () = cos (). (4.37) (4.38) Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between w and w⊥. Knowing this property, you can use trigonometry to determine the magnitude of the weight components: cos () = ⊥ ⊥ = cos () = cos () ∥ sin () = ∥ = sin () = sin () (4.39) (4.40) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion 161 Take-Home Experiment: Force Parallel To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show? Tension A tension is a force along the length of a medium, especially a force carried by
a flexible medium, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope. Consider a person holding a mass on a rope as shown in Figure 4.15. Figure 4.15 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope. Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus, net = − = 0, (4.41) where and are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass: For a 5.00-kg mass, then (neglecting the mass of the rope) we see that = = (5.00 kg)(9.80 m/s2 ) = 49.0 N. = =. (4.42) (4.43) If we cut the rope and insert a spring,
the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope. Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure 4.16 (a) and (b). 162 Chapter 4 \| Dynamics: Force and Newton's Laws of Motion Figure 4.16 (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force’s direction, but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed. Example 4.6 What Is the Tension in a Tightrope? Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17. Figure 4.17 The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing. Strategy As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w and the two tensions TL (left tension) and TR (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions L and R must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are L and. Thus, the magnitude of those forces must be equal

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.