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5.2. Visualize: 5.6. Model: An object’s acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F a and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an acceleration of 2.4 m/s2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F ma, 2F (2m)a a F/m 0.6 m/s2 5.10 Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the time, d t 2 , so d should be constant. We have t2 2.0 furlongs 2.0 s 2 x 4.0 s 2 Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs. Assess: A longer time should result in a longer distance traveled. 5.14 Solve: Newton’s second law tells us that F ma. Compute F for each case: (a) F (0.200 kg)(5 m/s2 ) 1N. (b) F (0.200 kg)(10 m/s2 ) 2 N . Assess: To double the acceleration we must double the force, as expected. 5.18. Visualize: Solve: The object will be in equilibrium if F3 has the same magnitude as F1 F2 but is in the opposite direction so that the sum of all the three forces is zero. 5.22. Visualize: Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible description is: “A rocket accelerates upward.” 5.26. Visualize: Assess: Since the velocity is constant, the acceleration is zero, and the net force is zero. 5.30. Visualize: Solve: According to Newton’s second law F ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object. 5.34. Model: Use the particle model for the object. Solve: (a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the acceleration of the mass is 10 m/s2 . According to Newton’s second law, F0 m0 (10 m/s2 ). The force then becomes 1 2 F0 . Newton’s second law gives 1 1 F0 m0a m0 10 m s 2 2 2 This means a is 5 m/s2 . (b) The force is F0 and the mass is now 1 2 m0 . Newton’s second law gives 1 F0 m0a m0 10 m s 2 2 This means a 20 m/s2 . (c) A similar procedure gives a 10 m/s2 . (d) A similar procedure gives a 2.5 m/s2 . 5.38. Visualize: Solve: (d) This is an object on a surface because FG n. It must be moving to the left because the kinetic friction is to the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic block to the left.” 5.42. Visualize: Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so the object is not on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there is no drag force.” 5.46. Visualize: The normal force is perpendicular to the hill. The frictional force is parallel to the hill. 5.50. Visualize: The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a force of the spring to the right which causes an acceleration. 6.2. Model: We can assume that the ring is a particle. Visualize: This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle . Solve: Because the ring is in equilibrium it must obey Fnet 0 N. This is a vector equation, so it has both xand y-components: Fnet x T3 cos T2 0 N T3 cos T2 Fnet y T1 T3 sin 0 N T3 sin T1 We have two equations in the two unknowns T3 and . Divide the y-equation by the x-equation: T3 sin T 80 N tan 1 1.6 tan 1 1.6 58 T3 cos T2 50 N Now we can use the x-equation to find T3 T2 50 N 94 N cos cos58 The tension in the third rope is 94 N directed 58 below the horizontal. 6.6. Visualize: Please refer to Figure EX6.6. Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the –y-axis is the same 20 by which the coordinates are tilted. Applying Newton’s second law, ax Fnet x m ay Fnet y m 5 N 1 N 3sin 20 N 1.49 m/s 2 2 kg 2.82 N 3cos 20 N 0 m/s 2 2 kg For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15 with the positive xaxis. The other 2-newton force makes an angle of 15 with the negative y-axis. The accelerations are ax ay Fnet x 2cos15 N 2sin15 N 3 N 0.28 m/s 2 m Fnet y m 2 kg 1.414 N 2sin15 N 2cos15 N 0 m/s 2 2 kg 6.10. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize: Solve: (a) Since the box is at rest, ax 0 m/s2 , and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero. (b) For this situation again, ax 0 m/s2 , so Fnet T 0 N. (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ax 5.0 m/s2 , Fnet T max 50 kg 5.0 m/s 2 250 N Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s2. 6.14. Model: Use the particle model for the woman. Solve: (a) The woman’s weight on the earth is wearth mgearth 55 kg 9.80 m/s 2 540 N (b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth. Her weight on the moon is wmoon mgmoon 55 kg 1.62 m/s 2 89 N Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense. 6.18. Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be subject to kinetic friction. Visualize: Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n FG mg. The maximum friction force is fsmax s mg 0.8120 kg 9.80 m/s2 940 N The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not be able to budge the mule. Assess: The farmer should have known better. 6.22. Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics. Visualize: Solve: We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain: a v 82 m/s 0 m/s 2.34 m/s 2 t 35 s Fnet Fx Fthrust fr ma Fthrust fr ma For rubber rolling on concrete, r 0.02 (Table 6.1), and since the runway is horizontal, n FG mg. Thus: Fthrust r FG ma r mg ma m r g a 75,000 kg 0.02 9.8 m/s2 2.34 m/s2 190,000 N Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that 190,000 N is enough to produce the required acceleration. 6.26. Model: We will represent the tennis ball as a particle. Visualize: The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion. Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is Fnet FG D 0 N Since only the vertical direction matters, one can write: Fy 0 N Fnet D FG 0 N When this condition is satisfied, the speed of the ball becomes the constant terminal speed v vterm . The magnitudes of the gravitational and drag forces acting on the ball are: FG mg m 9.80 m/s 2 1 2 2 2 2 Avterm 0.25 R 2 vterm 0.25 0.0325 m 26 m/s 0.56 N 4 The condition for dynamic equilibrium becomes: 0.56 N 57 g 9.80 m/s2 m 0.56 N 0 N m 9.80 m/s 2 D Assess: The value of the mass of the tennis ball obtained above seems reasonable. 6.30. Model: The plastic ball is represented as a particle in static equilibrium. Visualize: Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the string’s tension plus two long-range forces. The equilibrium condition is Fnet x Tx Felec x T sin Felec 0 N Fnet y Ty FG y T cos mg 0 N We can solve the y-equation to get T 0.001 kg 9.8 m/s mg cos cos 20 2 0.0104 N Substituting this value into the x-equation, Felec T sin 1.04 102 N sin 20 0.0036 N (b) The tension in the string is 0.0104 N. 6.34. Model: We can assume your body is a particle moving in a straight line under the influence of two forces: gravity and the support force of the scale. Visualize: Solve: The weight (see Equation 6.10) of an object moving in an elevator is a w w mg 1 a 1 g g mg When accelerating upward, the acceleration is 170 lb a 1 9.80 m/s 2 1.3 m/s 2 150 lb When braking, the acceleration is 120 lb a 1 9.80 m/s 2 2.0 m/s 2 150 lb Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire State Building must be. Also note that we did not have to convert the units of the weights from pounds to newtons because the weights appear as a ratio. 6.38. Model: We will represent the bullet as a particle. Visualize: Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant to the problem.) The kinematic equation is v02 400 m/s 6.67 105 m/s2 2x 2 0.12 m 2 v12 v02 2ax a Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood exerts two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the wood. Second, a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points to the left and thus has a negative x-component. The x-component of Newton’s second law is Fnet x fk ma fk ma 0.01 kg 6.67 105 m/s2 6670 N Notice how the signs worked together to give a positive value of the magnitude of the force. (b) The time to stop is found from v1 v0 at as follows: t v0 6.00 104 s 600 μs a (c) Using the above kinematic equation, we can find the velocity as a function of t. For example at t 60 μs, vx 400 m/s (6.667 105 m/s2 )(60 106 s) 360 m/s 6.42. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize: Solve: From the height of the slope and its angle, we can calculate its length: h h 50 m sin x1 x0 288 m x1 x0 sin sin10 Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force: Fnet y Fy n FG cos 0 N n FG cos mg cos (75 kg)(9.80 m/s2 )(cos10) 724 N One-dimensional kinematics gives us Sam’s acceleration: v12 v02 2ax x x0 ax v12 v02 40 m/s 0 m2/s 2 2.78 m/s2 2 x1 x2 2 288 m 2 Then, from Newton’s second law and the equation f k k n : Fnet x Fx FG sin Fthrust f k max k mg sin Fthrust ma n 75 kg 9.80 m/s2 sin10 200 N 75 kg 2.78 m/s2 724 N Assess: This coefficient seems a bit high for skis on snow, but not impossible. 0.165 6.46. Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics. Assume w sin fs , so it does not hang up at the top. Visualize: The block ends where it starts, so x2 x0 0 m. We expect v2 to be negative, because the block will be moving in the –x-direction, so we’ll want to take v2 as the final speed. Because of friction, we expect to find v2 v0 . Solve: (a) The friction force is opposite to v , so f k points down the slope during the first half of the motion and up the slope during the second half. FG and n are the only other forces. Newton’s second law for the upward motion is ax a0 Fnet x m a y 0 m/s 2 FG sin f k mg sin f k m m Fnet y m n FG cos n mg cos m m The friction model is f k k n. First solve the y-equation to give n mg cos . Use this in the friction model to get f k k mg cos . Now substitute this result for f k into the x-equation: a0 mg sin k mg cos g sin k cos 9.8 m/s 2 sin30 0.20cos30 6.60 m/s 2 m Kinematics now gives v12 v02 2a0 x1 x0 x1 2 2 v12 v02 0 m /s 10 m/s 7.6 m 2a0 2 6.60 m/s 2 2 The block’s height is then h x1 sin (7.6 m)sin30 3.8 m. (b) For the return trip, f k points up the slope, so the x-component of the second law is ax a1 Fnet x m FG sin f k mg sin f k m m Note the sign change. The y-equation and the friction model are unchanged, so we have a1 g sin k cos 3.20 m/s2 The kinematics for the return trip are v22 v12 2a1 x2 x1 v2 2a1x1 2 3.20 m/s2 7.6 m 7.0 m/s Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction. The final speed is v2 7.0 m/s. 6.50. Model: The antiques (mass m) in the back of your pickup (mass M ) will be treated as a particle. The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the road. Visualize: Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m M , Newton’s second law is ( Fnet ) x Fx N x FG PA ( f ) x 0 N 0 N f (m M )ax (m M )a x ( Fnet ) y Fy N y FG PA ( f ) y N (m M ) g 0 N 0 N y The model of static friction is f N , where is the coefficient of friction between the tires and the road. These equations can be combined to yield a g. Since constant-acceleration kinematics gives v12 v02 2a( x1 x0 ), we find v12 v02 v02 25 m/s min 0.58 2 x1 x0 2 g x1 x0 2 9.8 m/s 2 55 m 2 a The truck cannot stop if is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80 respectively (see Table 6.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for min . This value is smaller than the given coefficient of static friction (s 0.60) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m. Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck. 6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity. Visualize: Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot straight up, F y 1 1 2 4mg ma FG D mg Av 2 mg A 2vterm mg A 5mg 4 4 A Thus a 5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the drag force and 6.19 for the terminal velocity. (b) When the ball is shot straight down, F y ma D FG 1 2 A 2vterm mg 3mg 4 Thus a 3 g , this time directed upward. (c) The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the terminal velocity because the drag force decreases as the ball slows. 6.58. Model: The ball hanging from the ceiling of the truck by a string is represented as a particle. Visualize: Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between “at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball moves with the truck, so its acceleration is 5 m/s2 in the forward direction. (d) The free-body diagram shows that the horizontal component of T provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton’s second law for the ball is ax Fnet x m Tx T sin10 m m a y 0 m2 /s2 Fnet y m Ty FG y m T cos10 mg m We can solve the second equation for the magnitude of the tension: T 1.0 kg 9.8 m/s mg cos10 cos10 2 9.95 N Then the first equation gives the acceleration of the ball and truck: ax T sin10 9.95 N sin10 =1.73 m/s 2 m m The truck’s velocity cannot be determined. 6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force is balanced by the upward drag force. FG D mg bvterm 6 Rvterm vterm mg 6 R 4 (b) The mass of the spherical sand grain of density p 2400 kg/m3 is m R3 . 3 3 2 4 2 gR 2 2 2400 kg/m 9.80 m/s 5.0 10 m 1.3 m/s 9 9 3 Ns 1.0 10 m2 50 m 38 s. The time required for the sand grain to fall 50 m at this speed is t 1.3 m/s Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right. 2 Thus vterm 6.66. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does the auto slide before coming to rest? (b) (c) Newton’s second law is Fy ny FG y n mg may 0 N Fx 0.80n max The y-component equation gives n mg 1500 kg 9.8 m/s2 . Substituting this into the x-component equation yields 1500 kg ax 0.80 1500 kg 9.8 m/s2 ax 0.80 9.8 m/s2 7.8 m/s2 Using the constant-acceleration kinematic equation v12 v02 2ax, we find v02 40 m/s 102 m 2a 2 7.8 m/s 2 2 x 7.2. Visualize: Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational force on them. The interaction forces between the two are equal and opposite. (b) The system consists of the soccer ball and bowling ball, as indicated in the figure. (c) Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the forces on it is determined by Newton’s second law, a Fnet / m, which depends on the mass. Since the masses of the balls are different, their accelerations are different. 7.6. Visualize: Please refer to Figure EX7.6. Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with the surface, and a tension force due to the rope. (b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c) Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since then one component of a is zero, simplifying the mathematical expression of Newton’s second law. 7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. Visualize: The force applied on block 1 is FA on 1 12 N. The acceleration for all the blocks is the same and is denoted by a. Solve: (a) Newton’s second law for the three blocks along the x-direction is F on 1 x FA on 1 F2 on 1 m1a F on 2 x F1on 2 F3 on 2 m2a F on 3 x F2 on 3 m3a Adding these three equations and using Newton’s third law ( F2 on 1 F1 on 2 and F3 on 2 F2 on 3 ), we get FA on 1 m1 m2 m3 a 12 N 1 kg 2 kg 3 kg a a 2 m/s2 Using this value of a, the force equation on block 3 gives F2 on 3 m3a 3 kg 2 m/s2 6 N (b) Substituting into the force equation on block 1, 12 N F2 on 1 1 kg 2 m/s 2 F2 on 1 10 N Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg block exerts on the 1 kg block is reasonable. 7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as objects in the system. Visualize: Solve: The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts on the rear end of the rope. Newton’s second law along the x-axis for the ice block and the rope is F FR on I mI a 10 kg 2.0 m/s2 20 N F Fext FI on R mR a Fext FR on I mR a on I x on R x Fext FR on I mR a 20 N 0.500 kg 2.0 m/s2 21 N 7.18. Visualize: Please refer to Figure P7.18. Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block B gives nB FG B mB g. The static friction force on B is thus fs B snB smB g. fs B TA on B , and the same TB on A FG A mA g. Since TA on B TB on A , we have fs B mA g , so Applying Newton’s first law to the horizontal forces on B gives vertical forces on A gives s mB g mA g mA s mB 0.60 20 kg 12.0 kg analysis of the 7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight of this segment of rope is a downward force. It is balanced by the tension force at height y. Solve: The mass m of this segment of rope is the same fraction of the total mass M 2.2 kg as length y is a fraction of the total length L 3.0 m. That is, m/M y/L, from which we can write the mass of the rope segment m M y L This segment of rope is in static equilibrium, so the tension force pulling up on it is T FG mg Mg (2.2 kg)(9.8 m /s2 ) y y 7.19 y N L 3.0 m where y is in m. The tension increases linearly from 0 N at the bottom ( y 0 m) to 21.6 N at the top ( y 3 m). This is shown in the graph. 7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize: Solve: (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However, the very different masses of the two crafts means that the distances they each move will also be very different. The pictorial representation shows that they meet at time t1 when xM1 xm1. There’s only one force on each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair, FM on m Fm on M Ftractor beam 4.0 104 N The accelerations of the two craft are aM Fm on M M FM on m 4.0 104 N 4.0 104 N 2 0.020 m/s a 2.0 m/s 2 m 2.0 106 kg m 2.0 104 kg Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The accelerations are very different even though the forces are the same. Now we have a constant-acceleration problem in kinematics. At a later time t1 the positions of the crafts are xM1 xM0 vM0 t1 t0 12 aM t1 t0 12 aMt12 2 xm1 xm0 vm0 t1 t0 12 am t1 t0 xm0 12 amt12 2 The craft meet when xM1 xm1, so 1 2 aMt12 xm0 12 amt12 t1 2 10,000 m 2 xm0 2 xm0 99.5 s aM am aM am 2.02 m/s 2 Knowing t1 , we can now find the starship’s position as it meets the shuttlecraft: xM1 12 aMt12 99 m The starship moves 99 m as it pulls in the shuttlecraft from 10 km away. 7.30. Model: The two blocks form a system of interacting objects. Visualize: Please refer to Figure P7.30. Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newton’s first law applied to the y-direction on Block L yields F L y 0 nL FG L cos20 nL mL g cos20 Therefore fk L k L mL g cos20 0.201.0 kg 9.80 m/s2 cos20 1.84 N A similar analysis of the vertical forces on Block R gives f k R 1.84 N as well. Using Newton’s second law in the x-direction for Block L, FL x mLa TR on L fk L FG L sin 20 mLa TR on L 1.84 N mL g sin 20. For Block R, F R x mR a FG R sin 20 1.84 N TL on R mR a mR g sin 20 1.84 N TL on R These are two equations in the two unknowns a and TL on R TR on L T . Solving them, we obtain a 2.12 m/s2 and T 0.61 N. Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks, which is reasonable. 7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a massless rope and frictionless pulley. Visualize: Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a2 a a1, where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freebody diagrams, especially force f1 exerted on block 2 by block 1. We have quite a few pieces of information to include. First, Newton’s second law for blocks 1 and 2: F net on 1 x f1 T1 k n1 T1 m1a1 m1a F F net on 1 y n1 m1g 0 N n1 m1g T f f T T f n T m a F n n m g 0 N n n m g net on 2 x pull net on 2 y 1 2 2 2 1 pull 2 1 k 2 2 2 1 2 2 m2a 2 We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law: n1 n1 m1g f1 f1 k n1 k m1g T1 T2 T Knowing n1, we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns: k m1g T m1a Tpull T k m1g k m1 m2 g m2a Subtract the first equation from the second to get Tpull k 3m1 m2 g m1 m2 a a Tpull k 3m1 m2 g m1 m2 1.77 m/s 2 7.38. Model: Assume the particle model for m1 , m2 , and m3 , and the model of kinetic friction. Assume the ropes to be massless, and the pulleys to be frictionless and massless. Visualize: Solve: Newton’s second law for m1 is T1 ( FG )1 m1a1. Newton’s second law for m2 is F on m2 (F y n2 FG 2 0 N n2 2.0 kg 9.8 m/s2 19.6 N ) T2 f k2 T m2a2 T2 k n2 T1 (2.0 kg)a2 on m2 x Newton’s second law for m3 is F on m3 y T2 FG 3 m3a3 Since m1, m2 , and m3 move together, a1 a2 a3 a. The equations for the three masses thus become T1 FG 1 m1a 1.0 kg a T2 k n2 T1 m2a 2.0 kg a T2 FG 3 m3a 3.0 kg a Subtracting the third equation from the sum of the first two equations yields: FG 1 k n2 FG 3 6.0 kg a 1.0 kg 9.8 m/s2 0.3019.6 N 3.0 kg 9.8 m/s2 6.0 kg a a 2.3 m/s 2 7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable. Visualize: Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2). Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1 to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet 0 N for both systems. Newton’s second law for the cable car is F net on 1 x Newton’s second law for the counterweight is F F T1 FB m1g sin1 0 N net on 2 x m2 g sin 2 T2 0 N net on 1 y F net on 2 y n1 m1g cos1 0 N n2 m2 g cos 2 0 N From the x-equation for the counterweight, T2 m2 g sin 2 . By Newton’s third law, T1 T2 . Thus the x-equation for the cable car becomes FB m1g sin1 T1 m1g sin1 m2 g sin2 3770 N (b) If the brakes fail, then FB 0 N. The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is a1x a2 x a, where a will have a negative value. Using T1 T2 T , the two x-equations are Fnet on 1 x T m1g sin1 m1a1x m1a F net on 2 x m2 g sin 2 T m2a2x m2a Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T: m sin1 m2 sin 2 m1g sin1 m2 g sin m1 m2 a a 1 g 0.991 m/s 2 m1 m2 Now we have a problem in kinematics. The speed at the bottom is calculated as follows: v12 v02 2a x1 x0 2ax1 v1 2ax1 2 0.991 m/s2 400 m 28.2 m/s Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of 30 is reasonable. 7.46. Model: Use the particle model for the wedge and the block. Visualize: The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note: n1 on 2 n2 on 1. Solve: The y-component of Newton’s second law for block m2 is F on 2 y n1 on 2 cos FG 2 0 N n1 on 2 m2 g cos Combining this equation with the x-component of Newton’s second law yields: F on 2 x n1 on 2 sin m2a a n1 on 2 sin g tan m2 Now, Newton’s second law for the wedge is F on 1 x F n2 on 1 sin m1a F m1a n2 on 1 sin m1a m2a (m1 m2 )a (m1 m2 ) g tan