Pearson Physics Level 20 Unit II Dynamics: Chapter 3 Solutions

Transcription

1 Example 3.1 Practice Problems Pearson Physics Level 20 Unit II Dynamics: Chapter 3 Solutions Student Book page There is no forward force acting on the car. The free-body diagram below shows four forces. 1

2 2. The free-body diagram below shows four forces. Concept Check The net force on an object is equal to zero when the vector sum of the forces is zero. For two forces, the net force is zero when the forces are equal in magnitude and opposite in direction. An example would be an object that is suspended from a single string. The tension in the string equals the weight of the object. Because the string exerts an upward force, the net force on the object is zero. 2

3 Student Book pages Example 3.2 Practice Problems 1. Given A = 200 N [forward] B app = 100 N [backward] f net force on sled ( net ) Draw a free-body diagram for the sled. = 150 N [forward] = 60 N [backward] Add the force vectors shown in the vector addition diagram. net = A + B + app + f F net = F A + F B + F app + F f = 200 N N + ( 100 N) + ( 60 N) = 200 N N 100 N 60 N = 190 N net = 190 N [forward] The net force on the sled is 190 N [forward]. 2. Given A = 5000 N [E] B = 4000 N [E] C = 4500 N [W] D = 3500 N [W] F f = 1000 N net force on load ( net ) Draw a free-body diagram for the load. 3

4 To determine the direction of f, first determine net if there were no friction. net no fric = A + B + C + D F net no fric = F A + F B + F C + F D = 5000 N N + ( 4500 N) + ( 3500 N) = 5000 N N 4500 N 3500 N = 1000 N net no fric = 1000 N [E] If there were no friction, net would be directed east. So f must be directed west. Calculate net for the situation in this problem. net = net no fric + f F net = F net no fric + F f = 1000 N + ( 1000 N) = 1000 N 1000 N = 0 N net = 0 N (a) The net force on the load is 0 N. (b) If the load is initially at rest, it will not move. Example 3.3 Practice Problems 1. Given T 1 = 60.0 N [on rope] T2 Student Book page = = 20.0 net force on canoe ( net ) = 60.0 N [on rope] 4

5 Draw a free-body diagram for the canoe. Resolve all forces into x and y components. Vector x component y component T 1 (60.0 N)(cos 40.0 ) (60.0 N)(sin 40.0 ) T 2 (60.0 N)(cos 20.0 ) (60.0 N)(sin 20.0 ) Add the x and y components of all force vectors in the vector addition diagram. x direction net x = T1 x + 2 T x y direction net y = T1 y + 2 F net x = F T1 x + F T2 x F net y = F T1 y + F T2 y = (60.0 N)(cos 40.0 ) + (60.0 N)(cos 20.0 ) = (60.0 N)(sin 40.0 ) + { (60.0 N)(sin 20.0 )} = N = N Use the Pythagorean theorem to find the magnitude of net. F net = 2 + F 2 net x net y = (102.3 N) 2 + (18.05 N) 2 = N Use the tangent function to find the direction of net. tan = opposite adjacent T y 5

6 tan = N N = = tan 1 (0.1763) = 10.0 From the vector addition diagram, this angle is between net x-axis. net = N [10.0 ] The net force on the canoe is N [10.0 ]. 2. Given T 1 = 65.0 N [30.0 ] T = 70.0 N [300 ] 2 net force on sled ( net ) Draw a free-body diagram for the sled. and the positive Resolve all forces into x and y components. Vector x component y component T 1 (65.0 N)(cos 30.0 ) (65.0 N)(sin 30.0 ) T 2 (70.0 N)(cos 60.0 ) (70.0 N)(sin 60.0 ) 6

7 Add the x and y components of all force vectors in the vector addition diagram. x direction = net x T x F net x = T1 x T x y direction net y = T1 y + 2 F + F T2 x F net y = F T1 y + F T2 y = (65.0 N)(cos 30.0 ) + (70.0 N)(cos 60.0 ) = (65.0 N)(sin 30.0 ) + { (70.0 N)(sin 60.0 )} = N = N Use the Pythagorean theorem to find the magnitude of net. F net = 2 + F 2 net x net y = (91.29 N) 2 + ( N) 2 = 95.5 N Use the tangent function to find the direction of net. tan = opposite adjacent = N N = = tan 1 (0.3080) = 17.1 From the vector addition diagram, this angle is between net and the positive x-axis. So the direction of net measured counterclockwise from the positive x-axis is = 343. net = 95.5 N [343 ] T y 7

8 The net force on the sled is 95.5 N [343 ]. Student Book page 135 Example 3.4 Practice Problems 1. From Example 3.4, F T weight and F 1 T F 2 T. 1 If the weight is halved, then the magnitude of T would be half, which in turn 1 would make the magnitude of T half. The directions of 2 T and 1 T would not 2 change. 2. Given g = 245 N [down] 1 = = 90.0 forces ( T and 1 T ) 2 Draw a free-body diagram for the sign. Resolve all forces into x and y components. Vector x component y component T 1 T 1 T 2 T2 F cos 40.0 FT sin F 0 g 0 F g Since the sign is at equilibrium, the net force in both the x and y directions is zero. F net x = F net y = 0 N Add the x and y components of all force vectors separately. x direction = net x T x T x y direction = 1 net y T y + g 8

9 F net x = F T 1x + F T2 x F net y = F T 1y + F g 0 = FT 1 cos F T 0 = F 2 T sin ( F 1 g ) Fg F T 2 = FT 1 cos = F T 1 sin 40.0 Fg F T 1 = sin N = sin 40.0 Substitute F T into the equation for F 1 T. 2 F T 2 = FT 1 cos 40.0 = (381.2 N)(cos 40.0 ) = N T 2 = N [0 ] From the free-body diagram, the direction of T1 the positive x-axis is = 140. = N [140 ] T 1 = kg m s 2 = N measured counterclockwise from The forces in the ropes are N [140 ] and N [0 ] respectively. 3. (a) Use Example 3.4 to write the equation for F T in terms of 1 1. F F = g T 1 sin 1 If 1 decreases, sin 1 would approach zero and the magnitude of T1 increase. Since F T F 2 T, the magnitude of 1 T would also increase. 2 (b) would From the free-body diagrams in Example 3.4 and in Practice Problem 2, F T1 y = F g ; otherwise, the sign would accelerate in the vertical direction. So 1 can never equal zero. Fg This conclusion agrees with what you observe from the equation F T = 1 sin, 1 where sin 1 0. As sin 1 approaches zero, the magnitude of T would 1 increase, until eventually the rope breaks. 9

10 3.1 Check and Reflect Student Book page 136 Knowledge 1. (a) Force is a push or a pull on an object. It is a vector quantity with both magnitude and direction. The SI unit of force is the newton (N). (b) Force is a dynamics quantity and not a kinematics quantity because force explains the causes of motion while kinematics only describes motion. Applications 2. (a) 10

11 (b) 11

12 3. (a) (b) Given f = 500 N [forward] g air = 200 N [backward] N net force on biker-bike combination ( net ) = 1800 N [down] = 1800 N [up] 12

13 Since the biker-bike combination is not accelerating up or down, the net force in the vertical direction is zero. Add the horizontal force vectors shown in the vector addition diagram in part (a). net = f + air F net = F f + F air = 500 N + ( 200 N) = 300 N net = 300 N [forward] The net force on the biker-bike combination is 300 N [forward]. 4. (a) An angle of 0 between forces 1 and 2 will maximize the net force acting on the object as shown below. For maximum net force, forces must be parallel. (b) An angle of 180 between forces 1 and 2 will minimize the net force acting on the object as shown below. For minimum net force, forces must be antiparallel. 5. Given T 1 = 80.0 N [45.0 ] T2 net force on tree ( net ) = 90.0 N [15 ] 13

14 Draw a free-body diagram for the tree. Resolve all forces into x and y components. Vector x component y component T 1 (80.0 N)(cos 45.0 ) (80.0 N)(sin 45.0 ) T 2 (90.0 N)(cos 15 ) (90.0 N)(sin 15 ) Add the x and y components of all force vectors in the vector addition diagram. 14

15 x direction = net x T x T x y direction = F net x = F T1 x + F T2 x F net y = F T1 y + F T2 y = (80.0 N)(cos 45.0 ) + (90.0 N)(cos 15 ) = (80.0 N)(sin 45.0 ) + { (90.0 N)(sin 15 )} = N = N Use the Pythagorean theorem to find the magnitude of net. F net = 2 + F 2 net x net y = (143.5 N) 2 + (33.27 N) 2 = N Use the tangent function to find the direction of net. tan = opposite adjacent = N N = = tan 1 (0.2319) = 13 From the vector addition diagram, this angle is between net and the positive x-axis. net = N [13 ] The net force on the tree is N [13 ]. 6. Given 1 = 65 N [30.0 ] 2 = 80 N [115 ] 3 = 105 N [235 ] net force on object ( net ) Draw a free-body diagram for the object. net y T y T y 15

16 Resolve all forces into x and y components. Vector x component y component 1 (65 N)(cos 30.0 ) (65 N)(sin 30.0 ) 2 (80 N)(cos 65.0 ) (80 N)(sin 65.0 ) 3 (105 N)(cos 55.0 ) (105 N)(sin 55.0 ) 16

17 Add the x and y components of all force vectors in the vector addition diagram. x direction net x y direction net y = 1x + 2 x + 3 x F net x = F 1x + F 2 x + F 3 x = (65 N)(cos 30.0 ) + { (80 N)(cos 65.0 )} + { (105 N)(cos 55.0 )} = 37.7 N = 1y + 2 y + 3 y F net y = F 1y + F 2 y + F 3 y = (65 N)(sin 30.0 ) + (80 N)(sin 65.0 ) + { (105 N)(sin 55.0 )} = 19.0 N Use the Pythagorean theorem to find the magnitude of net. F net = 2 + F 2 net x net y = ( 37.7 N) 2 + (19.0 N) 2 = 42 N 17

18 Use the tangent function to find the direction of net. tan = opposite adjacent = 19.0 N 37.7 N = = tan 1 (0.5032) = 26.7 From the vector addition diagram, this angle is between net and the negative x-axis. So the direction of net measured counterclockwise from the positive x- axis is = 153. net = 42 N [153 ] The net force on the object is 42 N [153 ]. Extensions 7. Given g = 245 N [down] 1 = 2 = 55.0 forces ( T and 1 T ) 2 Draw a free-body diagram for the sign. Resolve all forces into x and y components. Vector x component y component T 1 T 1 T 2 T2 F cos 55.0 FT sin F cos 55.0 FT sin g 0 F g 18

19 Since the sign is at equilibrium, F net x = 0. So = net x T x = T1 x T x F + F T2 x F T1 x = F T 2x Add the y components of all force vectors in the vector addition diagram. y direction net y = T 1y + T2 y + g F net y = F T 1y + F T2 y + F g 0 = FT sin F 1 T sin ( F 2 g ) Fg 0 = F T + F 1 T 2 sin 55.0 Fg F T 1 + F T = 2 sin N = sin 55.0 = kg m s 2 = N Since both ropes form the same angle with the ceiling, F T = 1 T2 F T 2 = 1 2 (299.1 N) = N T 2 = N [55 ] From the free-body diagram, the direction of T1 from the positive x-axis is = 125. = N [125 ] T 1 F. measured counterclockwise The forces in the ropes are N [125 ] and N [55 ] respectively. 19

20 8. (a) (b) 9. For example: 20

21 Student Book page 138 Concept Check Inertia is the property of an object that resists acceleration and is directly proportional to mass. Since the mass of the astronaut is the same everywhere in the universe, the inertia of the astronaut on Earth s surface, in orbit around Earth, and in outer space will be the same. Inertia cannot be zero. That would mean zero mass. Student Book page 139 Concept Check Interstellar space is located far from any celestial body, so the gravitational force acting on Voyager 1 is negligible. As a result, the probe will continue moving at a speed of 17 km/s in a straight line as predicted by Newton s first law. Student Book page 141 Concept Check (a) A steel barrier usually separates the cab of a truck from the load to protect the driver in the cab. According to Newton s first law, if the truck stops suddenly, the load will tend to continue moving at its original velocity. The steel barrier will prevent this motion and cause the load to maintain the same velocity as the truck. (b) Trucks carrying tall loads navigate corners slowly to prevent the top of the load from shifting or the truck from rolling over. Turning a corner slowly ensures that the force of friction between the load and the bed of the truck, and between the tie-down straps and the load will be large enough to prevent the load from shifting since without a large net force, the load will continue moving in its original direction. (c) Customers who order take-out drinks are provided with lids to prevent spillage when they walk. Without lids, the liquid in the cup will tend to keep moving at constant velocity and some may spill out of the cup. 3.2 Check and Reflect Student Book page 142 Knowledge 1. Newton s first law states that the motion of an object does not change unless it experiences an external non-zero net force. 2. An example that illustrates the property of inertia for a stationary object is removing a table cloth out from under stationary dishes. An example that illustrates the property of inertia for a moving object is a person carrying a cup of coffee on a tray and suddenly falling ahead due to his/her shoes getting stuck on the carpet. The person s upper torso as well as the cup of coffee continue to move ahead. 3. (a) A car that attempts to go around an icy curve too quickly slides off the curve because the car has a tendency to keep moving in a straight line at constant speed. 21

22 The direction of motion is along the tangent of the curve at the instant the car starts to slide. (b) Before the ball leaves the lacrosse stick, it has the same velocity as the stick. After leaving the stick, the ball maintains this velocity unless acted upon by an external non-zero net force. Air resistance acts on the ball causing it to slow down gradually as it moves through the air. The force of gravity acts downward on the ball, and causes the ball to accelerate downward. It is the downward force of gravity that causes the ball to have a parabolic trajectory through the air. 4. (a) According to Newton s first law, every object will continue being either in a state of rest or in motion at constant speed along a straight line unless acted upon by an external non-zero net force. Initially, both the beaker and the newspaper are at rest. If the newspaper is pulled slowly, the force of friction between the two contact surfaces will cause the beaker to move with the paper. The force of friction provides the net force to give the beaker the small acceleration needed to keep it moving with the paper. However, if the newspaper is pulled quickly, the force of friction is not large enough to give the beaker the large acceleration needed to move it along with the paper. The result is that the paper is removed quickly without the beaker moving significantly. (b) When released, the velocity of the ball has two components, x and y. In the x direction, the ball continues to move at the same speed as the snowmobile (ignoring air resistance). In the y direction, the rider provides an initial upward velocity to the ball. The ball slows down as it moves upward until it comes to rest at the top of its parabolic arc, then the ball speeds up on the way down. When the ball falls back down to the same height as the launch, its y velocity will be the same magnitude but opposite in direction to its y velocity at the launch. The ball returns to the hand of the rider because the x velocity of the ball is the same as that of the snowmobile, and both objects travel for the same length of time. If the snowmobile stops, the ball continues moving forward after it leaves the rider s hand. So the ball travels some horizontal distance before returning to the height of the launch. The snowmobile only travels the stopping distance. The result is that the ball lands ahead of the snowmobile. Applications 5. According to Newton s first law, an object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force. Before the students work with the air table, ensure it is level so there is no component of the force of gravity acting on an object on the air table, parallel to its surface. Caution students against holding onto conducting parts of the table when activating the spark timer. Place the newsprint and carbon paper flat on the surface and ensure that there are no ripples on either the carbon paper or the newsprint. One way to prevent ripples is to store both the newsprint and carbon paper under a flat pane of glass on a flat surface when not in use. To test the first part of the law, have students place the air puck in the centre of the table and observe its motion. According to Newton s first law, it should remain stationary in the centre. To test the second part of the law, use the launcher provided with the air table to propel a puck across the table and observe its motion and the spacing of the dots produced by the equal-interval sparker. If the velocity is constant, 22

23 the track should be a straight line and the spacing of the dots should be constant. To see what happens if there is a non-zero net force acting on the puck, give the table a slight slope and repeat the above steps. This time, the stationary puck will accelerate down the air table and if launched across the slope, it will trace out a parabolic path. Students can experiment to see if these observations are consistent for pucks with different masses. 6. (a) When the player skates across the ice, the puck is moving with him. When he shoots it toward the net, it stays in line with him. If he s past the net when the puck should be going in, the puck will be past the net the same distance he is. (b) This drill could be done in two steps. First, players line up at pylon P 1 and while at rest, aim at pylon P 2 in the centre of the net and try to score. Then, players skate along the blue line with the puck and shoot the puck when they reach pylon P 1. This time they try to aim at P 2, on another trial they aim at P 3, and so on. Players can even try skating slowly on one trial, and faster on another to determine where they must aim to compensate for their speed. Extensions 7. Students will find web site links at that describe both high-back and backless booster seats. These sites also provide additional information, links, flyers, and artwork. Another link provides a good source of information on the seat belt test. In Canada, most child-booster seat laws state that children must ride in a booster seat until they are a minimum of 145 cm tall, or a minimum of 36 kg, or a minimum of 9 years old. Booster seats are required by law to comply with the same standards and guidelines as child safety seats. 8. Students will find a web site link at that provides a summary of how seat belts and the retractor mechanism work together to improve the safety of a motorist in a moving vehicle. Clear diagrams and a summary of the retractor mechanism and pretensioner are provided. 23

24 9. For example: Student Book page 146 Concept Check An applied force is a force that is provided by a person or machine. A net force is the vector sum of all the forces acting simultaneously on an object. Suppose an object, say A, is in contact with a surface and experiences an applied force. If the surface is frictionless, the free-body diagram for A will have three forces (see below). Since N is equal in magnitude but opposite in direction to g, in the vertical direction, F net = 0 N. In the v horizontal direction, net = h app. Student Book page 148 Concept Check (a) If the mass and net force both decrease by a factor of 4, the acceleration will not change. a = F net m 24

25 1 a = 4 F net 1 4 m = F net m (b) If the mass and net force both increase by a factor of 4, the acceleration will not change. a = F net m = 4F net 4m = F net m (c) If the mass increases by a factor of 4, but the net force decreases by the same factor, the acceleration will change by a factor of a = F net m 1 = 4 F net 4m = 1 16 F net m (d) If the mass decreases by a factor of 4, and the net force is zero, the acceleration will be zero. a = F net m 0 = 1 4 m = 0 Example 3.5 Practice Problems 1. Given Student Book page 149 net = 12 N [left] m = 6.0 kg 25

26 acceleration of cart ( a ) The cart is not accelerating up or down. So in the vertical direction, F = 0 N. netv In the horizontal direction, the acceleration of the cart is in the direction of the net force. So use the scalar form of Newton s second law. F net h = ma F net a = h m = 12 N 6.0 kg m 12 kg 2 = s 6.0 kg = 2.0 m/s 2 a = 2.0 m/s 2 [left] The acceleration of the cart is 2.0 m/s 2 [left]. 2. Given net = 34 N [forward] a = 1.8 m/s 2 [forward] mass of curling stone (m) The curling stone is not accelerating up or down. So in the vertical direction, F = 0 N. netv In the horizontal direction, the acceleration of the curling stone is in the direction of the net force. So use the scalar form of Newton s second law. F net h = ma F net m = h a = 34 N 1.8 m/s 2 m 34 kg 2 = s m 1.8 s 2 = 19 kg The mass of the curling stone is 19 kg. 26

27 Student Book page 150 Example 3.6 Practice Problem 1. Given m s = 255 kg m p = 98 kg m b = 97 kg A = 1220 N [forward] B = 1200 N [forward] f = 430 N [backward] average acceleration of bobsled-pilot-brakeman combination ( a ) The bobsled, pilot, and brakeman are a system because they move together as a unit. Find the total mass of the system. m T = m s + m p + m b = 255 kg + 98 kg + 97 kg = 450 kg Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on the system in both the horizontal and vertical directions. 27

28 horizontal direction net h = A + B + f F net h = F A + F B + F f = 1220 N N + ( 430 N) = 1220 N N 430 N = 1990 N vertical direction net = v N + g F net = 0 v Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h mt = 1990 N 450 kg kg m = s 450 kg = 4.4 m/s 2 a = 4.4 m/s 2 [forward] The bobsled-pilot-brakeman combination will have an average acceleration of 4.4 m/s 2 [forward]. Example 3.7 Practice Problems 1. Given Student Book page 152 T = N [up] m T = kg g = 9.81 m/s 2 [down] acceleration of person ( a ) Draw a free-body diagram for the person-elevator system. Calculations in the vertical direction are not required in this problem. 28

29 The system is not accelerating left or right. So in the horizontal direction, F = 0 N. neth Since the person is standing on the elevator floor, both the person and the elevator will have the same vertical acceleration. For the vertical direction, write an equation to find the net force on the system. net v = T + g Apply Newton s second law. m a = T + g ma = F T + F g ma = F T + mg a = F T m + g = N kg + ( 9.81 m/s2 ) 3 kg m = s 9.81 m/s kg a = 1.14 m/s 2 = 1.14 m/s 2 [down] 29

30 The acceleration of the person is 1.14 m/s 2 [down]. 2. Given a = 1.50 m/s 2 [up] T = N [up] g = 9.81 m/s 2 [down] mass of engine (m) Draw a free-body diagram for the engine. The engine is not accelerating left or right. So in the horizontal direction, F net = 0 N. h For the vertical direction, write an equation to find the net force on the engine. net v = T + g Apply Newton s second law. m a = T + g ma = F T + F g ma = F T + mg m(a g) = F T m = F T a g = N m/s ( 9.81 m/s ) 3 m kg = s m s 2 = 252 kg The mass of the engine is 252 kg. 30

31 Student Book page 153 Example 3.8 Practice Problems 1. Given m = 55 kg e = 825 N [up] g = 9.81 m/s 2 [down] acceleration of bungee jumper ( a ) Draw a free-body diagram for the bungee jumper. The bungee jumper is not accelerating left or right. So in the horizontal direction, F net = 0 N. h For the vertical direction, write an equation to find the net force on the bungee jumper. = e + g net v 31

32 Apply Newton s second law. m a = e + g ma = F e + F g ma = F e + mg a = F e m + g = 825 N 55 kg + ( 9.81 m/s2 ) kg m = s 9.81 m/s 2 55 kg = 5.2 m/s 2 a = 5.2 m/s 2 [up] The acceleration of the bungee jumper is 5.2 m/s 2 [up]. 2. (a), (b) During the entire jump, the force of gravity is acting on the bungee jumper. Ignoring air resistance, her acceleration is initially 9.81 m/s 2 [down] and decreases to zero when she reaches the bottom of the jump. Then she begins to accelerate up. So at the lowest part of the jump, she is accelerating up even though she is momentarily at rest. Example 3.9 Practice Problems 1. Given m A = 5.0 kg m B = 5.0 kg T = 60 N [along rope] g = 9.81 m/s 2 [down] acceleration of both buckets ( a ) The two buckets move together as a unit with the same acceleration. So calculate the acceleration of m B. Draw a free-body diagram for m B. 32

33 The bucket is not accelerating left or right. So in the horizontal direction, F net = 0 N. h For the vertical direction, write an equation to find the net force on m B. net v = T + g Apply Newton s second law. m B a = T + g m B a = F T + F g m B a = F T + m B g a = F T m + g B = 60 N 5.0 kg + ( 9.81 m/s2 ) kg m 60 2 = s 9.81 m/s kg = 2.2 m/s 2 a = 2.2 m/s 2 [up] Both buckets will have an acceleration of 2.2 m/s 2 [up]. 2. (a) Given m B = 10 kg a = m/s 2 [right] from Example 3.9 app = 55 N [right] F f = 14.7 N tension in rope connecting the blocks Draw a free-body diagram for the 10-kg block. 33

34 The block is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the block in both the horizontal and vertical directions. horizontal direction vertical direction net h = app + T + f net = v N + g Apply Newton s second law. F net = 0 v m B a = app + T + f Calculations in the vertical direction are not m B a = F app + F T + F f required in this problem. F T = m B a F app F f = (10 kg)(0.363 m/s 2 ) 55 N ( 14.7 N) = 37 N The tension in the rope connecting the two blocks is 37 N. 34

35 (b) For the rope between the hand and the block, the only force that creates a tension is the applied force. So the tension in this rope is 55 N. Student Book page 155 Example 3.10 Practice Problems 1. From Example 3.10, net = (m B m A )g and a = F net m. But m T = m A + m B, so T substitute m T and net into the equation for a. a = m B m A m A + m g B Substitute the appropriate values for m A and m B to determine the acceleration in each situation. (a) m A = 1 3 m B a = = m B 1 3 m B 1 3 m B + m B g g m B m B 1 2 = 3 g 4 3 = 2 4 g = 1 2 g The system will have an acceleration of magnitude 1 2 g. Since m A < m B, m A will move up and m B will move down. 35

36 (b) m A = 2m B a = m B 2m B 2m B + m g B m = B 1 2 m B g = 1 3 g The system will have an acceleration of magnitude 1 3 g. Since m A > m B, m A will move down and m B will move up. (c) Since m A = m B, the acceleration of the system will be zero. 2. Given m A = 25 kg a = 1.64 m/s 2 [up] from Example 3.10 g = 9.81 m/s 2 [down] tension in rope Draw a free-body diagram for the 25-kg block. The block is not accelerating left or right. So in the horizontal direction, F net = 0 N. h Write equations to find the net force on the block in the vertical direction. vertical direction net v = T + g Apply Newton s second law. m A a = T + g m A a = F T + F g F T = m A a F g = m A a m A g = (25 kg)(1.64 m/s 2 ) { (25 kg)(9.81 m/s 2 )} = N The tension in the rope is N. 36

37 3. Student Book page 157 Example 3.11 Practice Problems 1. From Example 3.11, net = (m B m C )g and a = F net m. But m T = m A + m B + m C, so T substitute m T and net into the equation for a. a = m B m C m A + m B + m g C Substitute the appropriate values for m A, m B, and m C to determine the acceleration. a = 12 kg 6.0 kg 20 kg + 12 kg kg (9.81 m/s2 ) 6.0 kg = (9.81 m/s 2 ) 38.0 kg = 1.5 m/s 2 a = 1.5 m/s 2 [left] The tire will have an acceleration of 1.5 m/s 2 [left]. 2. From Example 3.11 Practice Problem 1, a = m B m C m A + m B + m g. C Substitute the appropriate values for m A, m B, and m C to determine the acceleration. a = 8.0 kg 6.0 kg 15 kg kg kg (9.81 m/s2 ) 2.0 kg = (9.81 m/s 2 ) 29.0 kg = 0.68 m/s 2 a A = 0.68 m/s 2 [left] a B = 0.68 m/s 2 [down] a = 0.68 m/s 2 [up] C 37

38 The tire will have an acceleration of 0.68 m/s 2 [left], pail B an acceleration of 0.68 m/s 2 [down], and pail C an acceleration of 0.68 m/s 2 [up]. Student Book page Check and Reflect Knowledge 1. Newton s second law states that when a non-zero net force acts on an object, the object will accelerate in the direction of the net force. The magnitude of the acceleration will be directly proportional to the net force and inversely proportional to the mass of the object. 2. (a) and (b) A vehicle with a more powerful engine can exert a greater force on the axles, which would give the car of similar mass a greater acceleration. Applications 5. (a) Given dolphin = 320 N [up] a = 2.6 m/s 2 [up] mass of dolphin (m) Draw a free-body diagram for the dolphin. 38

39 The dolphin is not accelerating left or right. So in the horizontal direction, F net = 0 N. h Write equations to find the net force on the dolphin in the vertical direction. vertical direction net v = dolphin + g Apply Newton s second law. m a = dolphin + g ma = F dolphin + F g ma = F dolphin + mg m(a g) = F dolphin F m = dolphin a g 320 N = 2.6 m/s 2 ( 9.81 m/s 2 ) m 320 kg 2 = s m s 2 = 26 kg The dolphin has a mass of 26 kg. (b) From part (a), ma = F dolphin + mg. Solve for the acceleration. F a = dolphin m + g 39

40 m 320 kg 2 a = s + ( 9.81 m/s 2 ) 25.8 kg 2 = 15 m/s 2 a = 15 m/s 2 [up] The dolphin would have an acceleration of 15 m/s 2 [up]. 6. Given m = 80 kg wind = 205 N [horizontally] F f = 196 N acceleration of hut ( a ) Draw a free-body diagram for the hut. The hut is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the hut in both the horizontal and vertical directions. horizontal direction vertical direction net h = wind + f netv F net h = F wind + F f netv = 205 N + ( 196 N) = 9 N = N F = 0 + g Calculations in the vertical direction are not required in this problem. 40

41 Apply Newton s second law to the horizontal direction. F net h = m a F a = net h m = 9 N 80 kg kg m 9 2 = s 80 kg = 0.11 m/s 2 a = 0.11 m/s 2 [horizontally] The ice hut will have an acceleration of 0.11 m/s 2 [horizontally]. 7. (a) Given g = 20 N [down] 1 = 36 N [45 ] 2 = 60 N [125 ] net force on object ( net ) Draw a free-body diagram for the object. Resolve all forces into x and y components. 41

42 Vector x component y component (36 N)(cos 45 ) (36 N)(sin 45 ) 1 (60 N)(cos 55 ) (60 N)(sin 55 ) 2 Add the x and y components of all force vectors separately. x direction net x = 1x + 2 x F net x = F 1x + F 2 x = (36 N)(cos 45 ) + { (60 N)(cos 55 )} = 8.96 N y direction net y = 1y + 2 y F net y = F 1y + F 2 y = (36 N)(sin 45 ) + (60 N)(sin 55 ) = 74.6 N Use the Pythagorean theorem to find the magnitude of F net. F net = 2 + F 2 net x net y = ( 8.96 N) 2 + (74.6 N) 2 = 75 N Use the tangent function to find the direction of F net. tan = opposite adjacent = 74.6 N 8.96 N = = tan 1 (8.328) = 83 42

43 From the vector addition diagram, this angle is between net and the negative x-axis. So the direction of net measured counterclockwise from the positive x- axis is = 97. net = 75 N [97 ] The net force on the object is 75 N [97 ]. (b) Find the mass of the object. F g = mg m = F g g 20 N = 9.81 m/s 2 m 20 kg 2 = s m 9.81 s 2 = 2.04 kg The acceleration of the object is in the direction of the net force. So use the scalar form of Newton s second law. F net = ma a = F net m = 75.1 N 2.04 kg m 75.1 kg 2 = s 2.04 kg = 37 m/s 2 a = 37 m/s 2 [97 ] The object will have an acceleration of 37 m/s 2 [97 ]. 8. Given m A = 25 kg m B = 15 kg = 30 N [right] app acceleration of the boxes ( a ) Both boxes are a system because they move together as a unit. Find the total mass of the system. 43

44 m T = m A + m B = 25 kg + 15 kg = 40 kg Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on the system in both the horizontal and vertical directions. 44

45 horizontal direction vertical direction net h = app net = v N + g F net h = F app F net = 0 v = 30 N Apply Newton s second law to the Calculations in the vertical direction are not required horizontal direction. in this problem. F net h = m T a Fnet a = h mt = 30 N 40 kg kg m 30 2 = s 40 kg = 0.75 m/s 2 a = 0.75 m/s 2 [right] The boxes will have an acceleration of 0.75 m/s 2 [right]. 9. (a) Given m A = 4.0 kg m B = 2.0 kg F f = 11.8 N g = 9.81 m/s 2 [down] acceleration of the system ( a ) The string has a negligible mass. So the tension in the string is the same on both sides of the pulley. The string does not stretch. So the magnitude of a A is equal to the magnitude of a B. Find the total mass of both objects. m T = m A + m B = 4.0 kg kg = 6.0 kg Choose an equivalent system in terms of m T to analyze the motion. 45

46 B is equal to the gravitational force acting on m B. Apply Newton s second law to find the net force acting on m T. net = B + f F net = F B + F f = m B g + ( 11.8 N) = m B g 11.8 N = (2.0 kg)(9.81 m/s 2 ) 11.8 N = 7.82 N Use the scalar form of Newton s second law to calculate the magnitude of the acceleration. F net = m T a a = F net m T = 7.82 N 6.0 kg kg m = s 6.0 kg = 1.3 m/s 2 a A = 1.3 m/s 2 [toward pulley] a B = 1.3 m/s 2 [down] The 4.0-kg block will have an acceleration of 1.3 m/s 2 [toward pulley] and the 2.0-kg block will have an acceleration of 1.3 m/s 2 [down]. (b) Given m B = 2.0 kg a B = 1.30 m/s 2 [down] from part (a) g = 9.81 m/s 2 [down] tension in string Draw a free-body diagram for the 2.0-kg block. The block is not accelerating left or right. So in the horizontal direction, F net = 0 N. h Write equations to find the net force on the block in the vertical direction. 46

47 vertical direction net v = T + g Apply Newton s second law. m B a B = T + g m B a B = F T + F g F T = m B a B F g = m B a B m B g = m B (a B g) = (2.0 kg)(1.30 m/s m/s 2 ) = 17 N The tension in the string is 17 N. Extension 10. Concept Check Student Book page 160 Forces always exist in pairs. Thus, it is not possible to have just an action force or just a reaction force. The reaction force is always equal in magnitude and opposite in direction to the action force. 47

48 Student Book page 161 Concept Check Since the action-reaction forces act on different objects, both objects may accelerate provided there is a non-zero net force acting on each one. For example, if two people are sitting on skateboards and each person applies a force on the other with their feet, both people will accelerate away from each other. Each person exerts an action force on the other, and each person experiences a reaction force exerted by the other person. Student Book page 164 Example 3.12 Practice Problems 1. Given m A = 8.0 kg m B = 10 kg m C = 5.0 kg a = 1.5 m/s 2 [right] force exerted by box B on box A ( B on A ) Since all three boxes have the same acceleration, the boxes are a system. Find the total mass of the system. m T = m A + m B + m C = 8.0 kg + 10 kg kg = 23 kg Draw a free-body diagram for the system. 48

49 The system is not accelerating up or down. So in the vertical direction, F net = 0 N. v For the horizontal direction, write an equation to find the net force on the system. net h = app Apply Newton s second law. m T a = app m T a = F app F app = (23 kg)(1.5 m/s 2 ) = 34.5 N app = 34.5 N [right] 49

50 Draw a free-body diagram for box A. Box A is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on box A in both the horizontal and vertical directions. horizontal direction vertical direction net h = app + B on A netv F net h = F app + F B on A netv Apply Newton s second law. m A a = F app + F B on A F B on A = m A a F app = (8.0 kg)(1.5 m/s 2 ) 34.5 N = 23 N = N F = 0 + g = 23 N [left] The force exerted by box B on box A is 23 N [left]. B on A Calculations in the vertical direction are not required in this problem. 50

51 2. Given m A = 5.0 kg m B = 5.0 kg T = 60 N [along rope] a = 2.19 m/s 2 [up] from Example 3.9 Practice Problem 1 g = 9.81 m/s 2 [down] applied force ( app ) The two buckets move together as a unit with the same acceleration. So the buckets are a system. Find the total mass of the system. m T = m A + m B = 5.0 kg kg = 10.0 kg Draw a free-body diagram for the system. The system is not accelerating left or right. So in the horizontal direction, F net = 0 N. h For the vertical direction, write an equation to find the net force on the system. net v = app + g Apply Newton s second law. m T a = app + g m T a = F app + F g m T a = F app + m T g F app = m T (a g) 51

52 F app = (10.0 kg){2.19 m/s 2 ( 9.81 m/s 2 )} = (10.0 kg){2.19 m/s m/s 2 } = N app = N [up] An applied force of N [up] would cause both buckets to have an acceleration of 2.2 m/s 2 [up]. Student Book page 165 Example 3.13 Practice Problem 1. (a) Given m A = 150 kg m B = 250 kg app = 2600 N [forward] F f = 2400 N acceleration of the logs ( a ) Both logs are a system because they move together as a unit. Find the total mass of the system. m T = m A + m B = 150 kg kg = 400 kg Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on the system in both the horizontal and vertical directions. 52

53 horizontal direction net h = app + f netv F net h = F app + F f netv vertical direction = N F = 0 + g = 2600 N + ( 2400 N) = 2600 N 2400 N = 200 N Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h mt = 200 N 400 kg kg m = s 400 kg = m/s 2 a = m/s 2 [forward] The logs will have an acceleration of m/s 2 [forward]. (b) Given m A = 150 kg m B = 250 kg app = 2600 N [forward] f on A = 900 N [backward] a = m/s 2 [forward] from part (a) force exerted by log B on log A ( B on A ) Draw a free-body diagram for log A. Calculations in the vertical direction are not required in this problem. 53

54 Log A is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on log A in both the horizontal and vertical directions. horizontal direction vertical direction + B on A + f on A = N net h = app netv F net h = F app + F B on A + F f on A netv Apply Newton s second law. m A a = F app + F B on A + F f on A F B on A = m A a F app F f on A F = 0 + g = (150 kg)(0.500 m/s 2 ) 2600 N ( 900 N) = N B on A = N [backward] Log B exerts a force of N [backward] on log A. Calculations in the vertical direction are not required in this problem. 54

55 Student Book page Check and Reflect Knowledge 1. Newton s third law states that forces exist only in pairs. If one object exerts a force on a second object, the second object exerts a force back on the first object that is equal in magnitude and opposite in direction. 2. (a) The swimmer at the edge of the pool exerts an action force directed toward the wall of the pool. The wall, in turn, exerts a reaction force directed toward the swimmer. This reaction force causes the swimmer to accelerate away from the wall. (b) The person in the canoe exerts an action force on the package directed toward the shore. The package, in turn, exerts a reaction force directed on the person. Friction between the person and the canoe causes the canoe to move away from shore. 3. In order for a car to accelerate, the tires must exert a force of friction directed backward on the ground. The ground will, in turn, exert a reaction force directed forward on the tires causing the car to accelerate forward. If the ground is icy, the tires will not be able to exert a great enough force of friction on the ice, no matter how powerful the car engine. If the driver presses the accelerator too hard, the tires will spin, causing the ice beneath the tires to melt. This watery layer will make it even more difficult for the tires to exert any frictional force on the icy surface. In general, it is better to use lower gears and to gently press the accelerator so that the tires can exert the maximum force of friction on snow and ice. 4. (a) The action force is the water pushing against the centreboard with a magnitude of 600 N. The reaction force is the centreboard pushing against the water with a magnitude of 600 N. (b) There are three reaction pairs: force Earth exerts on the object force the object exerts on Earth; downward force of magnitude 30 N the object exerts on the spring upward force of magnitude 30 N the spring exerts on the object; and the force the spring exerts on the support force the support exerts on the spring. 55

56 Applications 5. The normal force N and the gravitational force g acting on an object are not actionreaction forces, because both of these forces act on the same object. The normal force is exerted by the contact surface on the object and is directed perpendicular to the contact surface. The reaction force to N is the force exerted by the object on the contact surface, and is directed toward the contact surface. The gravitational force is the force exerted by Earth on the object, and is directed toward Earth s centre. The reaction force to g is the force exerted by the object on Earth, and is directed toward the object

57 7. The hand exerts a force of 10 N [right] on spring A. Since the spring stays in place, the net horizontal force acting on it is zero. Consequently, spring B must be pulling spring A also by a force of 10 N, but toward the left. The two forces stretch spring A appropriately by the tension force of 10 N. Spring A pulls back on spring B to the right also by 10 N, according to Newton s third law. Spring B does not move either, thus the wall must be balancing out this force by pulling on spring B to the left, also with 10 N. Spring B is also stretched by 10 N. The reaction force to this latter force is pulling back on the wall to the right with the same 10 N magnitude. 8. (a) Given m X = 10 kg m Y = 5.0 kg app = 36 N [right] force exerted by block X on block Y ( X on Y ) force exerted by block Y on block X ( Y on X ) Blocks X and Y are a system because they move together as a unit. Find the total mass of the system. m T = m X + m Y = 10 kg kg = 15 kg Draw a free-body diagram for the system. 57

58 The system is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction vertical direction net h = app netv F net h = F app netv = 36 N Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h mt = 36 N 15 kg kg m 36 2 = s 15 kg = 2.40 m/s 2 a = 2.40 m/s 2 [right] = N F = 0 + g Calculations in the vertical direction are not required in this problem. 58

59 Draw a free-body diagram for block Y. Block Y is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on block Y in both the horizontal and vertical directions. horizontal direction vertical direction net h = X on Y netv F net h = F X on Y netv = N F = 0 Apply Newton s second law. m Y a = F Calculations in the vertical direction are not required X on Y F X on Y = (5.0 kg)(2.40 m/s 2 in this problem. ) = 12 N X on Y = 12 N [right] Apply Newton s third law. Y on X = X on Y = 12 N [left] The force exerted by block X on block Y is 12 N [right] and block Y exerts a force of 12 N [left] on block X. + g (b) Given m X = 10 kg m Y = 5.0 kg f on X = 8.0 N [left] f on Y = 4.0 N [left] app = 36 N [right] force exerted by block X on block Y ( X on Y ) force exerted by block Y on block X ( Y on X ) 59

60 Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction vertical direction net h = app + f netv F net h = F app + F f netv = N F = 0 = F app + F f on X + F f on Y Calculations in the vertical direction are not required = 36 N + ( 8.0 N) + ( 4.0 N) in this problem. = 36 N 8.0 N 4.0 N = 24 N Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h mt = 24 N 15 kg + g 60

61 kg m 24 2 a = s 15 kg = 1.60 m/s 2 a = 1.60 m/s 2 [right] Draw a free-body diagram for block Y. Block Y is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on block Y in both the horizontal and vertical directions. horizontal direction vertical direction net h = X on Y + f on Y netv F net h = F X on Y + F f on Y netv Apply Newton s second law. m Y a = F X on Y + F f on Y F X on Y = m Y a F f on Y = (5.0 kg)(1.60 m/s 2 ) ( 4.0 N) = 12 N X on Y = 12 N [right] Apply Newton s third law. = N F = 0 + g Calculations in the vertical direction are not required in this problem. Y on X = X on Y = 12 N [left] The force exerted by block X on block Y is 12 N [right] and block Y exerts a force of 12 N [left] on block X. 61

62 9. The box will begin to move in a direction opposite to the water flow out of the holes. Since the holes are diagonally opposite each other, the box will rotate. This motion can be explained using Newton s third law. The side of the box opposite each hole exerts a force on the water to accelerate it out of the hole. From Newton s third law, the water exerts a reaction force on the side of the box. This net force accelerates the box in a direction opposite to the direction in which the water is flowing. Example 3.14 Practice Problem 1. Student Book page 172 Student Book page 175 Example 3.15 Practice Problems 1. Given m A = 7.0 g = kg m B = 7.3 g = kg = 30.0 g = 9.81 m/s 2 force of static friction ( ) fstatic Since the two coins are stacked, the coins are a system. Find the total mass of the system. m T = m A + m B = kg kg = kg 62

63 Draw a free-body diagram for the system. Since the system is not accelerating, net = 0 N both parallel and perpendicular to the incline. Write equations to find the net force on the system in both directions. direction net = N + g direction + net = g Fnet = 0 F net = F g + Calculations in the direction are not required in this problem. 0 = F g + fstatic F fstatic F fstatic F f = F static g Now, F g = m T g sin So, F f = ( m static T g sin ) = m T g sin = ( kg)(9.81 m/s 2 )(sin 30.0 ) = N F f static indicates that = N [uphill] f static prevents the system from sliding downhill. The positive value for the direction of f static is uphill. fstatic The force of static friction acting on the loonie-toonie combination is N [uphill]. 2. Given m = 7.00 g = kg F = N f static g = 9.81 m/s 2 maximum angle of the incline ( ) 63

64 Draw a free-body diagram for the loonie. Since the loonie is not accelerating, net = 0 N both parallel and perpendicular to the incline. f prevents the loonie from sliding downhill. So static f is directed static uphill. Write equations to find the net force on the loonie in both directions. direction direction net = N + g net = g + fstatic Fnet = 0 F net = F g + Calculations in the direction are not required in this problem. 0 = F g + F fstatic F fstatic F f = F static g Now, F g = mg sin So, F f = ( mg sin ) static = mg sin F f sin = static mg N = 3 2 ( kg)(9.81 m/s ) = = sin 1 (0.6407) = 39.8 The maximum angle that the book can be inclined at is

65 Student Book page 178 Concept Check The angle between the normal force and the force of friction is always 90º, whether or not the object is on a horizontal surface. The free-body diagram below shows the forces acting on an object being pulled at constant velocity along a horizontal, rough surface. The free-body diagram below shows the same object being pulled up a rough incline at constant velocity with app acting parallel to the incline. Example 3.16 Practice Problems 1. Given m = 3.5 kg = 15.0 g = 9.81 m/s 2 acceleration of block ( a ) Student Book page

66 Draw a free-body diagram for the block. Since the block is accelerating downhill, net 0 N parallel to the incline, but net = 0 N perpendicular to the incline. Write equations to find the net force on the block in both directions. direction net = N + g direction net = g Fnet = 0 F net = F g Calculations in the ma = F g direction are not required Now, F g = mg sin in this problem. So, m a = m g sin = g sin a = (9.81 m/s 2 )(sin 15.0 ) = 2.5 m/s 2 The positive value for a indicates that the direction of a is downhill. a = 2.5 m/s 2 [downhill] The acceleration of the block will be 2.5 m/s 2 [downhill]. 66

67 2. Given m = 55.0 kg = 35.0 a = 4.41 m/s 2 g = 9.81 m/s 2 force of kinetic friction ( ) fkinetic Draw a free-body diagram for the skier. Since the skier is accelerating downhill, net 0 N parallel to the incline, but net = 0 N perpendicular to the incline. Write equations to find the net force on the skier in both directions. direction net = N + g direction + net = g Fnet = 0 F net = F g + fkinetic F fkinetic Calculations in the ma = F g + F fkinetic direction are not required Now, F g = mg sin in this problem. So, ma = mg sin + F fkinetic F f kinetic = m(a g sin ) = (55.0 kg){4.41 m/s 2 (9.81 m/s 2 ) (sin 35.0 )} = 66.9 N The negative value for F f indicates that the direction of kinetic fkinetic is uphill. 67

68 f kinetic = 66.9 N [uphill] The force of kinetic friction on the skier is 66.9 N [uphill]. Student Book page 180 Concept Check If the magnitude of the force of kinetic friction were greater than the maximum magnitude of the force of static friction, an object would be unable to move. Student Book page 185 Example 3.17 Practice Problems 1. Given app = 24 N [forward] g = 9.81 m/s 2 [down] s = 0.15 from Table 3.4 (steel on greased steel) mass of block (m) Draw a free-body diagram for the block. 68

69 Since the block is not accelerating, net = 0 N in both the horizontal and vertical directions. Write equations to find the net force on the block in both directions. 69

70 horizontal direction vertical direction net h = app + f static net = v N + g F net h = F app + F f F static net = F v N + F g 0 = F app + F f 0 = F static N + ( mg) = F app + ( s F N ) = F N mg = F app s F N F N = mg F app = s F N Substitute F N = mg into equation for F app. F app = s mg F m = app sg 24 N = (0.15) 9.81 m s 2 m 24 kg s 2 = m s 2 = 16 kg The mass of the steel block is 16 kg. 2. Given app = 125 N [forward] g = 9.81 m/s 2 [down] s = 0.20 from Table 3.4 (waxed hickory skis on wet snow) m = 78 kg from Example 3.17 determine if sled will move Draw a free-body diagram for the sled. 70

71 Since the sled is not accelerating in the vertical direction, F net = 0 N. h Write equations to find the net force on the sled in both directions. horizontal direction vertical direction net h = app + f static net = v N + g F net h = F app + F f F static net = F v N + F g = F app + F f 0 = F static N + ( mg) = F app + ( s F N ) = F N mg = F app s F N F N = mg Substitute F N = mg into equation for F net. h F net h = F app s mg = 125 N (0.20)(78 kg)(9.81 m/s 2 ) = 28 N 71

72 Since net is negative, F h f > F static app. The sled will not move on a wet snowy surface if an applied force of 125 N acts on it. Student Book page 186 Example 3.18 Practice Problems 1. Draw a free-body diagram for the toboggan. From Example 3.18, s depends on the angle of the hill, not the mass of the toboggan. Substitute = 30.0 into equation for s. s = tan = tan 30.0 = 0.58 The coefficient of static friction for the toboggan on the snow is Given m = 80 kg = 25.0 g = 9.81 m/s 2 (a) coefficient of static friction ( s ) (b) maximum force of static friction ( ) fstatic (a) Draw a free-body diagram for the skier. 72

73 When the angle of the incline is just enough for the skier to start moving, the surface of the incline is exerting the maximum magnitude of the force of static friction on the skier. Just before the skier begins to slide, net = 0 N in both the parallel and perpendicular directions to the incline. Write equations to find the net force on the skier in both directions. direction = N net + g direction net = g + fstatic Fnet = F N + F g F net = F g + F fstatic 0 = F N + F g 0 = F g + F fstatic F N = F g F f = F static g Now, F g = mg cos F g = mg sin So, F N = ( mg cos ) F f = ( mg sin ) static = mg cos = mg sin s F N = mg sin Substitute F N = mg cos into the last equation for the direction. 73

74 s ( mg cos ) = mg sin s cos = sin s = sin cos = tan = tan 25.0 = (b) Substitute s into the equation F f = static s F N. F f static = s mg cos = (0.466)(80 kg)(9.81 m/s 2 )(cos 25.0 ) = N f static = N [uphill] (a) The coefficient of static friction for the skier on the snow is (b) The maximum force of static friction on the skier is N [uphill]. Student Book page 188 Example 3.19 Practice Problems 1. Given app = 450 N [forward] g = 9.81 m/s 2 [down] m = 1000 kg a = 0 m/s 2 coefficient of kinetic friction ( k ) Draw a free-body diagram for the crate. 74

75 Since the crate is moving at constant speed, net = 0 N in both the horizontal and vertical directions. Write equations to find the net force on the crate in both directions. 75

76 horizontal direction vertical direction net h = app + f kinetic net = v N + g F net h = F app + F f F kinetic net = F v N + F g 0 = 450 N + ( k F N ) 0 = F N + ( mg) = 450 N k F N = F N mg k F N = 450 N F N = mg Substitute F N = mg into the last equation for the horizontal direction. k mg = 450 N k = 450 N mg 450 N = (1000 kg) 9.81 m s 2 m 450 kg 2 s = m 1000 kg 9.81 s 2 = The coefficient of kinetic friction for the crate on the floor is Given m = 1640 kg = 15.0 a = 0 m/s 2 g = 9.81 m/s 2 k = 0.5 from Table 3.4 (rubber tires on wet concrete) force of kinetic friction ( ) fkinetic Draw a free-body diagram for the lift truck. 76

77 Since the lift truck is moving at constant speed downhill, net = 0 N both parallel and perpendicular to the incline. Write equations to find the net force on the lift truck in both directions. direction net = N + g direction + net = g Fnet = 0 F net = F g + fkinetic F fkinetic Calculations in the 0 = F g + F fkinetic direction are not required F f = F kinetic g in this problem. Now, F g = mg sin 77

78 So, F f = mg sin kinetic = (1640 kg)(9.81 m/s 2 )(sin 15.0 ) = N The negative value for F f indicates that the direction of kinetic f is uphill. kinetic f kinetic = N [uphill] The force of kinetic friction on the lift truck is N [uphill]. Student Book page 189 Example 3.20 Practice Problem 1. (a) Given m A = 15 kg m B = 15 kg k = 0.50 g = 9.81 m/s 2 a = 0 m/s 2 applied force on bundles ( app ) Calculate the angle of the roof. tan = 1.0 m 2.0 m = = tan 1 (0.5000) = 26.6 Both bundles are a system because they move together as a unit. Find the total mass of the system. m T = m A + m B = 15 kg + 15 kg = 30 kg Draw a free-body diagram for the system. 78

79 Since the system is moving at constant speed up the roof, net = 0 N both parallel and perpendicular to the roof. Write equations to find the net force on the system in both directions. direction net = N + g Fnet = F N + Fg direction + g + fkinetic net = app F net = F app + F g + F fkinetic 0 = F N + Fg 0 = F app + F g + F fkinetic F N = F g F app = F g F fkinetic Now, F g = m T g cos F g = m T g sin and F f = kinetic k F N So, F N = ( m T g cos ) F app = ( m T g sin ) ( k F N ) = m T g cos = m T g sin + k F N Substitute F N = m T g cos into the equation for F app. F app = m T g sin + k m T g cos = m T g(sin + k cos ) = (30 kg)(9.81 m/s 2 ){sin (0.50)(cos 26.6 )} = N The positive value for F app indicates that the direction of app app = N [up roof] is up the roof. The roofer must apply a force of N [up roof] to drag the bundles at constant speed. (b) Given m A = 15 kg m B = 15 kg k = 0.50 g = 9.81 m/s 2 a = 0 m/s 2 79

80 force exerted by bundle A on bundle B ( A on B ) Draw a free-body diagram for bundle B. Since bundle B is moving at constant speed up the roof, net = 0 N both parallel and perpendicular to the roof. Write equations to find the net force on bundle B in both directions. direction direction net = N + g net = A on B + g + fkinetic Fnet = F N + Fg F net = F A on B + F g + F fkinetic 0 = F N + Fg 0 = F A on B + F g + F fkinetic F N = F g F A on B = F g F fkinetic Now, F g = m B g cos F g = m B g sin and F f = kinetic k F N So, F N = ( m B g cos ) F A on B = ( m B g sin ) ( k F N ) = m B g cos = m B g sin + k F N Substitute F N = m B g cos into equation for F A on B. F A on B = m B g sin + k m B g cos = m B g(sin + k cos ) = (15 kg)(9.81 m/s 2 ){sin (0.50)(cos 26.6 )} = N The positive value for F A on B indicates that the direction of A on B is up the roof. A on B = N [up roof] Bundle A exerts a force of N [up roof] on bundle B. 80

81 (c) Given m A = 15 kg m B = 15 kg k = 0.50 g = 9.81 m/s 2 a = 2.0 m/s 2 [up roof] applied force on bundles ( app ) Since the system is accelerating up the roof, net 0 N parallel to the roof, but net = 0 N perpendicular to the roof. Write equations to find the net force on the system in both directions. direction direction net Fnet = N + g = F N + Fg net = app + g + fkinetic F net = F app + F g + F fkinetic 0 = F N + Fg m T a = F app + F g + F fkinetic F N = F g F app = m T a F g F fkinetic Now, F g = m T g cos F g = m T g sin and F f = kinetic k F N So, F N = ( m T g cos ) F app = m T a ( m T g sin ) ( k F N ) = m T g cos = m T a + m T g sin + k F N Substitute F N = m T g cos into equation for F app. F app = m T a + m T g sin + k m T g cos = m T a + m T g(sin + k cos ) = (30 kg)(2.0 m/s 2 ) + (30 kg)(9.81 m/s 2 ){sin (0.50)(cos 26.6 )} = N The positive value for F app indicates that the direction of app is up the roof. app = N [up roof] The roofer must apply a force of N [up roof] to accelerate the bundles. 3.5 Check and Reflect Student Book page 190 Knowledge 1. Friction is the force that opposes the motion of an object in contact with another material, or the direction the object would be moving if there were no friction. 2. Friction can be ignored if an object is in a vacuum, or on an air track or air table. 3. Static friction is the force that opposes the moving component of an applied force on a stationary object. The magnitude of static friction varies from zero to a maximum value, trying to exactly balance the applied force. 81

82 Kinetic friction is the force that opposes the motion of an object. The magnitude of kinetic friction is usually less than the maximum value of static friction. Applications 4. Given g = 15 N [down] s = 0.06 from Table 3.4 (waxed hickory skis on dry snow) s = 0.20 from Table 3.4 (waxed hickory skis on wet snow) difference in the maximum force of static friction on wet and dry snow ( ) Draw a free-body diagram for the skis on each type of surface. f static Since the skis are not accelerating, net directions. Calculate F on each surface. fstatic = 0 N both in the horizontal and vertical 82

83 F f static = s F N = s mg Dry Snow Wet Snow F f static = s mg F f = static s mg = (0.06)(15 N) = (0.20)(15 N) = 0.90 N = 3.0 N f static = 0.90 N [backward] f = 3.0 N [backward] static Calculate the difference between the two values of F fstatic f static = 3.0 N 0.90 N = 2 N = 2 N [backward] F f. static The difference in the maximum force of static friction on wet and dry snow is 2 N [backward]. 5. Given app = 31 N [forward] g = 9.81 m/s 2 [down] m = 8.0 kg coefficient of static friction ( s ) Draw a free-body diagram for the steel slider. 83

84 Since the slider is not accelerating, net = 0 N in both the horizontal and vertical directions. Write equations to find the net force on the slider in both directions. horizontal direction net h = app + fstatic F net h = F app + fstatic F v vertical direction net = v N + g F net = F N + F g 0 = 31 N + ( s F N ) 0 = F N + ( mg) = 31 N s F N = F N mg s F N = 31 N F N = mg Substitute F N = mg into the last equation for the horizontal direction. s mg = 31 N s = 31 N mg 84

85 s = 31 N (8.0 kg) 9.81 m s 2 m 31 kg 2 s = m 8.0 kg 9.81 s 2 = 0.40 The coefficient of static friction for the steel slider on the steel rail is Given = 2350 N [down] g = 9.81 m/s 2 [down] g k = 0.7 from Table 3.4 (rubber tires on dry concrete) force of kinetic friction ( ) fkinetic Draw a free-body diagram for the biker-bike system. Since the system is skidding, net N in the vertical direction. Calculate F. fkinetic 0 N in the horizontal direction, but net = 0 85

86 F f kinetic = k F N = k mg = (0.7)(2350 N) = N f kinetic = N [backward] The force of kinetic friction on the biker-bike system is N [backward]. 7. Given m = 15 kg s = 0.45 g = 9.81 m/s 2 maximum angle of the incline ( ) Draw a free-body diagram for the box. Since the box is not accelerating, net = 0 N both parallel and perpendicular to the incline. f static prevents the box from sliding downhill. So f is directed uphill. static Write equations to find the net force on the box in both directions. direction net = N Fnet + g = F N + Fg 0 = F N + Fg F N = F g Now, F g = mg cos direction + net = g F net = F g + 0 = F g + fstatic F fstatic F fstatic F f = F static g F g = mg sin and F f = static s F N So, F N = ( mg cos ) s F N = ( mg sin ) = mg cos = mg sin 86

87 Substitute F N = mg cos into the last equation for the direction. s ( mg cos ) = mg sin s cos = sin s = sin cos s = tan 0.45 = tan = tan 1 (0.45) = 24 The maximum angle of the incline before the box starts to move is Given s = 0.10 m = 85 kg = 8.0 g = 9.81 m/s 2 determine if wheelchair will move Draw a free-body diagram for the wheelchair. Since the wheelchair is not accelerating, net = 0 N both parallel and perpendicular to the incline. f static prevents the wheelchair from sliding downhill. So f is directed uphill. static Write equations to find the net force on the wheelchair in both directions. direction net = N Fnet + g = F N + Fg 0 = F N + Fg direction net F net = g + = F g + 0 = F g + fstatic F fstatic F fstatic F N = F g Now, F g = mg cos F f = F static g F g = mg sin and F f = static s F N 87

88 So, F N = ( mg cos ) s F N = ( mg sin ) = mg cos = mg sin Substitute F N = mg cos into the last equation for the direction. s ( mg cos ) = mg sin s cos = sin s = sin cos s = tan 0.10 = tan = tan 1 (0.10) = 5.7 In order for the wheelchair to remain stationary, 5.7. The given value of is greater than 5.7. An angle of 8.0 will not prevent the wheelchair from moving. 9. Given = 10.0 s = 0.30 g = 9.81 m/s 2 maximum acceleration uphill ( a ) Draw a free-body diagram for the crate. Since the crate is not accelerating, net the incline. = 0 N both parallel and perpendicular to Write equations to find the net force on the crate in both directions. direction direction net Fnet = N + g = F N + Fg net = g + fstatic F net = F g + F fstatic 0 = F N + Fg ma = F g + F fstatic 88

89 F N = F g Now, F g = mg cos F g = mg sin and F f = static s F N So, F N = ( mg cos ) ma = mg sin + s F N = mg cos Substitute F N = mg cos into the last equation for the direction. m a = m g sin + s m g cos a = g( s cos sin ) = (9.81 m/s 2 ){(0.30)(cos 10.0 ) sin 10.0 } = 1.2 m/s 2 a = 1.2 m/s 2 [uphill] The maximum acceleration uphill is 1.2 m/s 2 [uphill]. 10. Given m = 400 kg g = 9.81 m/s 2 [down] v = 4.0 m/s [N] i k = coasting distance of sled (d) Draw a free-body diagram for the sled. 89

90 Since the sled is coasting, net the vertical direction. 0 N in the horizontal direction, but net Write equations to find the net force on the sled in both directions. horizontal direction vertical direction net h = f kinetic net = v N + g F net h = F f kinetic F net = F v N + F g ma = k F N 0 = F N + ( mg) = F N mg F N = mg Substitute F N = mg into the last equation for the horizontal direction. m a = k m g a = k g = (0.0500)(9.81 m/s 2 ) = 0.49 m/s 2 a = 0.49 m/s 2 [S] = 0 N in 90

91 Since the sled coasts to a stop, calculate the coasting distance of the sled. (v f ) 2 = (v i ) 2 + 2ad 0 = (v i ) 2 + 2ad 2ad = (v i ) 2 d = (v i) 2 2a (4.0 m/s) 2 = 2( 0.49 m/s 2 ) = 16 m The sled will coast for 16 m before it stops. Extensions 11. Given = 120 N [12.0 ] app m = 35 kg k = 0.30 g = 9.81 m/s 2 [down] v f = 1.2 m/s [0 ] elapsed time ( t) Draw a free-body diagram for the crate. Resolve app into x and y components. Vector x component y component (120 N)(cos 12.0 ) (120 N)(sin 12.0 ) app 91

92 Since the crate is accelerating in the x direction, F net x 0 N but F net y = 0 N. Write equations to find the net force on the crate in the x and y directions. x direction y direction net x = appx + f kinetic net y = N + app y + g F net x = F appx + F f F kinetic net y = F N + F app y + F g ma = (120 N)(cos 12.0 ) + ( k F N ) 0 = F N + (120 N)(sin 12.0 ) + ( mg) = (120 N)(cos 12.0 ) k F N F N = mg (120 N)(sin 12.0 ) = (35 kg)(9.81 m/s 2 ) (120 N)(sin 12.0 ) = 318 N Substitute F N = 318 N into the last equation for the x direction. ma = (120 N)(cos 12.0 ) (0.30)(318 N) = 21.9 N a = 21.9 N 35 kg = m/s 2 The crate is accelerating along the 0 direction. a = m/s 2 [0 ] Calculate the elapsed time. a = v t v f v i = t m/s m/s 0 = t t = m 1.2 s m s = 1.9 s The elapsed time will be 1.9 s

93 13. A tire is made up of natural and synthetic rubber, synthetic fabrics and chemicals, and possibly metal, and is designed to provide traction, cushion road shock, and carry a load under any condition. Traction is the force of friction, exerted by the tire tread on a road surface that provides grip. Traction is determined by measuring the force needed to start dragging a tire over a road surface. This force depends on the temperature of the road surface and the tire, and the nature of the road surface, e.g., dry concrete, dry asphalt, packed snow, black ice, or a surface covered with a layer of water. Coefficients of static and kinetic friction are measured for different tire tread F f patterns and surfaces using the equation = F. N The tread is the groove design on the operating surface of the tire and is made from a mixture of many different types of natural and synthetic rubbers. The tread pattern is designed to dissipate heat, maintain contact with the road, and improve traction. Some common tire types are slicks, snow tires, ice tires, all-season tires, and rain tires. Slicks are racing tires and have no treads in order to get the maximum amount of rubber in contact with the road. This enables the tires to dissipate heat during drag racing. Snow tires are designed to bite into the snow to provide better traction. Ice tires now have a winter-rubber compound, siping technologies, and an aggressive V-shaped tread design. Siping is the process of cutting small slits perpendicular to the face of the tire, which in turn creates thousands of tiny edges that improve 93

94 traction and braking effectiveness. The National Safety Council performed tests on siped tires and found that there was a 22% increase in traction on snow and ice. All-season tires are tires on passenger cars that are designed for use on wet and dry surfaces, and provide traction on snow and ice. The tire must have a row of fairly large grooves that start at the edge of the tire and extend toward the centre. At least 25% of the surface must be grooved so that the tire can bite through snow and provide traction. Rain tires are used during wet conditions. They have grooves at the centre and sides that carry water away from the centre of the tire, enabling a good contact area, called the patch, with the road and minimizing hydroplaning. Hydroplaning is the skimming effect caused by tires losing contact with a surface covered with water. If the water cannot squirt out from under the tire fast enough, the tire will float on the water surface and the driver may lose control of the vehicle. All tires have rolling friction, the resistance of a tire as it rolls rather than slides over a road surface. The coefficient of rolling friction ( rf ) can be used to calculate the amount of rolling resistance caused by the tire. The table below lists typical coefficients of rolling friction for different tires. Tire Type Coefficient of Rolling Friction ( rf ) Low-rolling resistance tire Car tire Truck tire Train wheel The force required to roll a car on tires is calculated using the equation F frolling = rf F N. Below is a sample calculation to determine the force of rolling friction and the power used to move a car on a level road having a mass of 1814 kg and a coefficient of rolling friction of The weight of the car is given by g = m g. g = (1814 kg)(9.81 m/s 2 ) = N [down] Since the car is not accelerating in the vertical direction, g + N = 0. So N = mg [up]. The force of rolling friction acts parallel to the surface and is directed backward. F frolling = rf F N = (0.015)( N) = 267 N The power required to move the car on level ground at a particular speed is given by the equation P = F frolling v, where v is the average speed. If the car has a speed of 100 km/h (27.8 m/s), the power to overcome rolling friction is 7.41 kw or about 10 hp. 94

95 For interesting Web sites on the topic of tires, follow the links at A possible Inquiry Lab to determine the coefficients of static and kinetic friction for a curling stone is shown below. Students will find that s is slightly greater than k, and that temperature has an effect on the coefficients. They will also find that not all curling stones have exactly the same coefficients on the same surface. That is why professional curlers always check their set of eight stones before a competition. Question What are the coefficients of static and kinetic friction on a curling stone on ice? How does the ice temperature affect these coefficients? Hypothesis State a hypothesis relating the value of s to k. State another hypothesis relating the ice temperature to the coefficients of friction. Variables Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials balance thermometer string spring scale of appropriate range Procedure 1. Determine the mass of the curling stone and calculate its weight in newtons. 2. Place the curling stone on the icy surface and allow its temperature to equilibrate to the ice. Measure the temperature of the ice surface. 3. Attach a durable string around the rim of the curling stone and make a loop at the end. Attach the string to the spring scale. 4. Pull gently, slowly increasing the force until the curling stone just begins to move. Measure the maximum value of the force of static friction in newtons just as the curling stone begins to move. Repeat the procedure several times and average the results. 5. Repeat step 4 for a curling stone being pulled at a slow constant velocity along the ice surface. 6. Have the icemaker change the temperature of the ice. Allow sufficient time for the ice to stabilize and repeat steps 2 to 5. Do this for several temperatures. 7. Average the forces of static and kinetic friction for each temperature. Determine the coefficients of static and kinetic friction. Analysis 1. Compare the coefficients of static and kinetic friction. 2. How did the ice temperature affect the coefficients of friction? 3. What would be the ideal ice temperature to make a curling stone slide the fastest? 4. Design and conduct an experiment to determine if all curling stones in a set of eight have the same coefficient of kinetic friction. Analyze your data and form a conclusion. 95

96 Student Book pages Chapter 3 Review Knowledge 1. Given A = 300 N [E] B = 350 N [E] m = 2000 kg f = 550 N [W] acceleration of truck ( a ) Draw a free-body diagram for the truck. The truck is not accelerating up or down. So in the vertical direction, F = 0 N. netv 96

97 Write equations to find the net force on the truck in both the horizontal and vertical directions. horizontal direction vertical direction net h = A + B + f F net h = F A + F B + F f = 300 N N + ( 550 N) = 300 N N 550 N = 100 N netv netv = N F = 0 + g Apply Newton s second law to the horizontal direction. F net h = m a F a = net h m = 100 N 2000 kg kg m = s 2000 kg Calculations in the vertical direction are not required in this problem. = m/s 2 a = m/s 2 [E] The truck will have an acceleration of m/s 2 [E]. 2. (a) A figure skater during a glide travels at constant velocity. Newton s first law states that an object will continue being at rest or moving at constant velocity unless acted upon by an external non-zero net force. Since the ice is smooth, it can be considered frictionless. The net force on the figure skater is zero and the skater s velocity remains constant. 97

98 (b) A hockey puck during a cross-ice pass travels at constant velocity. Since the ice can be considered frictionless, the net force on the hockey puck is zero and the velocity of the puck remains constant. 3. If a transport truck pulls a trailer with a force of 1850 N [E], according to Newton s third law, the trailer exerts a force of 1850 N [W] on the truck. This is an equal, but opposite force to the one the truck exerts on the trailer. 4. If the driver presses the accelerator too hard, the tires will spin, causing the snow beneath the tires to melt. This watery layer will make it even more difficult for the tires to exert any frictional force on the snow. Also, when the tires spin, kinetic friction is present, not static friction. Since k is usually less than s, the traction that the tires will have with the snow will be reduced. In general, it is better to use lower gears and to gently press the accelerator so that the tires can exert the maximum force of friction on the snow. Applications 5. (a) Given T 1 T2 1 = = 45.0 m = 3.0 kg g = 9.81 m/s 2 [down] force exerted by foot ( ) Draw a free-body diagram for the pulley. 98

99 Resolve all given forces into x and y components. Vector x component y component F T 1 T (cos 45.0 ) F 1 T (sin 45.0 ) 1 F T (cos 45.0 ) F 2 T 2 (sin 45.0 ) T 2 The rope has a negligible mass. So the tension in the rope is the same on both sides of the pulley. F T 1 = F T 2 The hanging object will provide the tension in the rope. F T 1 = mg = (3.0 kg)(9.81 m/s 2 ) = 29.4 kg m s 2 = 29.4 N Since the hanging object is not accelerating, the net force in both the x and y directions is zero. F net x = F net y = 0 N Add the x and y components of all force vectors in the vector addition diagram. 99

100 x direction y direction net x = T1 x + T2 x + net y = T 1y + T2 y F net x = F T1 x + F T2 x + F F net y = 0 0 = 2(29.4 N)(cos 45.0 ) + F Calculations in the vertical direction are F = 42 N not required in this problem. From the vector addition diagram, is directed along the rope connecting the foot to the pulley, which is also along the negative x-axis. = 42 N [along rope connecting foot to pulley] The foot exerts a force of 42 N [along rope connecting foot to pulley]. (b) If 1 and 2 decrease, the magnitude of will increase because the x components of T and 1 T will increase Given m s = 255 kg m p = 98 kg m b = 97 kg A = 1200 N [forward] f = 400 N [backward] a = 4.4 m/s 2 [forward] average force exerted by rider B ( B ) The bobsled, pilot, and brakeman are a system because they move together as a unit. Find the total mass of the system. m T = m s + m p + m b = 255 kg + 98 kg + 97 kg = 450 kg Draw a free-body diagram for the system. 100

101 The system is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction net h = A + B + f F net h = F A + F B + F f Apply Newton s second law to the horizontal direction. m T a = F A + F B + F f F B = m T a F A F f B vertical direction netv = N + g F = 0 netv = (450 kg)(4.4 m/s 2 ) 1200 N ( 400 N) = (450 kg)(4.4 m/s 2 ) 1200 N N = N = N [forward] Rider B exerts an average force of N [forward]. 7. Given m = kg = 20.0 T = N [forward] a = 0 m/s 2 g = 9.81 m/s 2 [down] magnitudes of L and R Draw a free-body diagram for the jet. Calculations in the vertical direction are not required in this problem. 101

102 Resolve all given forces into x and y components. Vector x component y component T N 0 mg(sin 20.0 ) mg(cos 20.0 ) g Since the jet is not accelerating, the net force in both the x and y directions is zero. F net x = F net y = 0 N Add the x and y components of all force vectors in the vector addition diagram. x direction y direction net x = T + R + g x net y = L + g y F net x = T + R + F g x F net y = L + F g y 0 = N + R + { mg(sin 20.0 )} 0 = L + { mg (cos 20.0 )} = N + R mg(sin 20.0 ) = L mg(cos 20.0 ) R = mg(sin 20.0 ) N L = mg(cos 20.0 ) = ( kg)(9.81 m/s 2 )(sin 20.0 ) N = ( kg) (9.81 m/s 2 )(cos 20.0 ) = N = N 102

103 The magnitudes of L and R are N and N respectively. 8. From the equation f = kinetic k F N, F fkinetic F N and F N = mg. So, F fkinetic mg The free-body diagram below represents the situation of the problem. F fkinetic F fkinetic (m + 2m)g 3mg Calculate F fkinetic. 3F fkinetic = 3 (2.5 N) = 7.5 N The new force of kinetic friction will be 7.5 N [backward]. 9. (a) Given m 1 = 1385 kg m 2 = 453 kg a = 0.75 m/s 2 [forward] tension in the hitch ( T ) Draw a free-body diagram for the trailer. 103

104 The trailer is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the trailer in both the horizontal and vertical directions. horizontal direction vertical direction net h = T netv F net h = F T netv = N F = 0 + g Calculations in the vertical direction are not required in this problem. Apply Newton s second law to the horizontal direction. m 2 a = F T F T = m 2 a = (453 kg)(0.75 m/s 2 ) = N The tension in the hitch is N. (b) Given m 1 = 1385 kg m 2 = 453 kg a = 0.75 m/s 2 [forward] F T = N from part (a) force of friction on truck ( f ) Draw a free-body diagram for the truck. 104

105 The truck is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the truck in both the horizontal and vertical directions. horizontal direction vertical direction net h = f + T netv F net h = F f + F T netv = N F = 0 Apply Newton s second law to the horizontal direction. m 1 a = F f + F T F f = m 1 a F T = (1385 kg)(0.75 m/s 2 ) ( N) = (1385 kg)(0.75 m/s 2 ) N = N f + g Calculations in the vertical direction are not required in this problem. = N [forward] The road exerts a force of friction of N [forward] on the truck. 105

106 (c) From the free-body diagram in part (a), T = 1 on 2. Apply Newton s third law. 2 on 1 = 1 on 2 = N [backward] The trailer exerts a force of N [backward] on the truck. 10. (a) Given m A = 50 kg m B = 80 kg f on A = 24.5 N [E] f on B = 39.2 N [W] A on B = 60 N [E] net force on each player ( net and A net ) B Draw a free-body diagram for each player. 106