How To Solve The Equation For The Force Of Gravity

CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION PROBLEMS 1. REASONING AND SOLUTION According to Newton s second law, the acceleration is a = ΣF/m. Since the pilot and the plane have the same acceleration, we can write ΣF ΣF or ( F) m ΣF PILOT m = Σ = m m PILOT PILOT PLANE PLANE Therefore, we find 4 3.7 10 N ( Σ F ) = PILOT ( 78kg) 4 = 93N 3.1 10 kg. REASONING AND SOLUTION The acceleration of the bicycle/rider combination is a = ΣF m = 9.78 N = 0.103 m/s 13.1kg + 81.7 kg 3. SSM REASONING According to Newton's second law of motion, the net force applied to the fist is equal to the mass of the fist multiplied by its acceleration. The data in the problem gives the final velocity of the fist and the time it takes to acquire that velocity. The average acceleration can be obtained directly from these data using the definition of average acceleration given in Equation.4. SOLUTION The magnitude of the average net force applied to the fist is, therefore, v 8.0 m/s 0m/s F = ma= m ( 0.70 kg = ) = 37 N t 0.15 s 4. REASONING AND SOLUTION Using Equation.4 and assuming that t 0 = 0 s, we have for the required time that Since ΣF = ma, it follows that t = v v 0 a

114 FORCES AND NEWTON'S LAWS OF MOTION ( ) [( ) ( 0 m/s )] ( ) 4.0 10 3 m/s t = v v 0 ΣF / m = m v v 5.0 kg 0 = ΣF 4.9 10 5 = 4.1 10 s N 5. SSM REASONING The net force acting on the ball can be calculated using Newton's second law. Before we can use Newton's second law, however, we must use Equation.9 from the equations of kinematics to determine the acceleration of the ball. SOLUTION According to Equation.9, the acceleration of the ball is given by a = v v 0 x Thus, the magnitude of the net force on the ball is given by v v0 (45 m/s) (0 m/s) F = ma = m = (0.058 kg) = 130 N x (0.44 m) 6. REASONING AND SOLUTION The acceleration is obtained from where v 0 = 0 m/s. So Newton s second law gives x = v 0 t + 1 at a = x/t ΣF = ma = m x t = ( 7 kg) ( 18 m) ( 0.95 s) = 900 N 7. REASONING According to Newton s second law, Equation 4.1, the average net forceσ F is equal to the product of the object s mass m and the average acceleration a. The average acceleration is equal to the change in velocity divided by the elapsed time (Equation.4), where the change in velocity is the final velocity v minus the initial velocity v 0. SOLUTION The average net force exerted on the car and riders is ( ) v v0 3 45m/s 0m/s F = ma = m = 5.5 10 kg = 3.5 10 4 N t t 0 7.0s

Chapter 4 Problems 115 8. REASONING AND SOLUTION From Equation.9, v = v 0 + ax Since the arrow starts from rest, v 0 = 0 m/s. In both cases, x is the same so v 1 v = a 1 x a x = a 1 a or v 1 v = a 1 a Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and or v 1 v = F 1 F = ( 5.0 m/s ) = 35.4 m/s F v = v F 1 = v 1 F 1 1 F 1 9. SSM WWW REASONING Let due east be chosen as the positive direction. Then, when both forces point due east, Newton's second law gives where gives where F + F = ma 1443 A B 1 (1) ΣF a 1 = 0.50 m/s. When F A points due east and F B points due west, Newton's second law F F = ma 1443 A B () ΣF a = 0.40 m/s. These two equations can be used to find the magnitude of each force. SOLUTION a. Adding Equations 1 and gives F A ( ) ( ) ( 1 + a 8.0kg 0.50m/s + 0.40m/s ) m a = = = 3.6 N b. Subtracting Equation from Equation 1 gives

116 FORCES AND NEWTON'S LAWS OF MOTION ( ) ( )( 1 a 8.0kg 0.50m/s 0.40m/s ) m a FB = = = 0.40N 10. REASONING AND SOLUTION F E = F cos θ = (70 N) cos 38 = 570 N = F sin θ = (70 N) sin 38 = 440 N 11. REASONING According to Newton s second law, the acceleration has the same direction as the net force and a magnitude given by a = ΣF/m. SOLUTION Since the two forces are perpendicular, the magnitude of the net force is given by the Pythagorean theorem as ΣF = 40.0 N second law, the magnitude of the acceleration is ( ) + ( 60.0 N). Thus, according to Newton s a = ΣF m = ( 40.0 N) + ( 60.0 N) 4.00 kg = 18.0 m/s The direction of the acceleration vector is given by θ = tan 1 60.0 N = 56.3 above the + x axis 40.0 N 1. REASONING AND SOLUTION The free body diagram is shown at the right, where F 1 = 45.0 N F = 5.0 N θ = 65.0 When F 1 is replaced by its x and y components, we obtain the free body diagram in the second drawing. F W θ F 1

Chapter 4 Problems 117 If we choose the right as the positive direction, then F 1 cos θ F = ma x F F 1 cos θ or a x = F 1 cosθ F m ( a x = 45.0 N )cos65.0 5.0 N 5.00 kg ( ) = 1.0 m/s W F 1 sin θ The minus sign indicates that the acceleration is to the left. 13. SSM REASONING According to Newton's second law ( F = ma ), the acceleration of the object is given by a = F / m, where F is the net force that acts on the object. We must first find the net force that acts on the object, and then determine the acceleration using Newton's second law. SOLUTION The following table gives the x and y components of the two forces that act on the object. The third row of that table gives the components of the net force. Force x-component y-component F 1 40.0 N 0 N F (60.0 N) cos 45.0 = 4.4 N (60.0 N) sin 45.0 = 4.4 N F = F 1 + F 8.4 N 4.4 N The magnitude of F is given by the Pythagorean theorem as F = (8.4 N) + (4.4) = 9.7 N The angle θ that F makes with the +x axis is θ = tan 1 4.4 N 8.4 N = 7. Σ F θ 8.4 N 4.4 N According to Newton's second law, the magnitude of the acceleration of the object is

118 FORCES AND NEWTON'S LAWS OF MOTION a = F m = 9.7 N = 30.9 m/s 3.00 kg Since Newton's second law is a vector equation, we know that the direction of the right hand side must be equal to the direction of the left hand side. In other words, the direction of the acceleration a is the same as the direction of the net force F. Therefore, the direction of the acceleration of the object is 7. above the + x axis. 14. REASONING Equations 3.5a x = v 0 x t + 1 a x t and 3.5b y = v t + 1 a 0 y yt give the displacements of an object under the influence of constant accelerations a x and a y. We can add these displacements as vectors to find the magnitude and direction of the resultant displacement. To use Equations 3.5a and 3.5b, however, we must have values for a x and a y. We can obtain these values from Newton s second law, provided that we combine the given forces to calculate the x and y components of the net force acting on the duck, and it is here that our solution begins. SOLUTION Let the directions due east and due north, respectively, be the +x and +y directions. Then, the components of the net force are ΣF x = 0.10 N + ( 0.0 N)cos 5 = 0.31 N ΣF y = ( 0.0 N)sin 5 = 0.1576 N According to Newton s second law, the components of the acceleration are a x = ΣF x m = 0.31 N.5 kg = 0.0894 m/s a y = ΣF y m = 0.1576 N.5 kg = 0.06304 m/s From Equations 3.5a and 3.5b, we now obtain the displacements in the x and y directions: x = v 0 x t + 1 a x t = ( 0.11 m/s ) 3.0 s y = v 0 y t + 1 a y t = ( 0 m/s ) 3.0 s The magnitude of the resultant displacement is ( )+ 1 ( 0.0894 m/s ) 3.0 s ( )+ 1 ( 0.06304 m/s ) 3.0 s ( ) = 0.7316 m ( ) = 0.837 m

Chapter 4 Problems 119 r = x + y = ( 0.7316 m) + ( 0.837 m) = 0.78 m The direction of the resultant displacement is θ = tan 1 0.837 m = 0.7316 m 1 south of east 15. REASONING AND SOLUTION Combining Newton's second law with v = v 0 + at, we have ΣF = ma = m v v 0 = 0.38 kg t ( ) (.0 m/s ) +.1 m/s 3.3 10 3 s ( ) = 470 N The answer is negative, indicating that the force is directed away from the cushion. 16. REASONING For both the tug and the asteroid, Equation.8 x = v 0 t + 1 at applies with v 0 = 0 m/s, since both are initially at rest. In applying this equation, we must be careful and use the proper acceleration for each object. Newton s second law indicates that the acceleration is given by a = ΣF/m. In this expression, we note that the magnitudes of the net forces acting on the tug and the asteroid are the same, according to Newton s action-reaction law. The masses of the tug and the asteroid are different, however. Thus, the distance traveled for either object is given by, where we use for ΣF only the magnitude of the pulling force x = v 0 t + 1 at = 1 SOLUTION Let L be the initial distance between the tug and the asteroid. When the two objects meet, the distances that each has traveled must add up to equal L. Therefore, ΣF m t Solving for the time t gives L = x T + x A = 1 a T t + 1 a A t L = 1 ΣF t + 1 ΣF t = 1 ΣF 1 + 1 m T m T m A m A t

10 FORCES AND NEWTON'S LAWS OF MOTION t = L ΣF 1 + 1 m T = ( 450 m ) = 64 s 1 ( 490 N) 3500 kg + 1 600 kg m A 17. SSM WWW REASONING We first determine the acceleration of the boat. Then, using Newton's second law, we can find the net force F that acts on the boat. Since two of the three forces are known, we can solve for the unknown force F W once the net force F is known. SOLUTION Let the direction due east be the positive x direction and the direction due north be the positive y direction. The x and y components of the initial velocity of the boat are then v 0 x = (.00 m/s) cos 15.0 = 1.93 m/s v 0 y = (.00 m/s) sin 15.0 = 0.518 m/s Thirty seconds later, the x and y velocity components of the boat are v x = (4.00 m/s) cos 35.0 = 3.8 m/s v y = ( 4.00 m/s) sin 35.0 =.9 m/s Therefore, according to Equations 3.3a and 3.3b, the x and y components of the acceleration of the boat are a x = v x v 0x 3.8 m/s 1.93 m/s = = 4.50 10 m/s t 30.0 s a y = v y v 0y t =.9 m/s 0.518 m/s 30.0 s = 5.91 10 m/s Thus, the x and y components of the net force that act on the boat are F x = ma x = (35 kg) (4.50 10 m/s ) = 14.6 N F y = ma y = (35 kg) (5.91 10 m/s ) =19. N The following table gives the x and y components of the net force F and the two known forces that act on the boat. The fourth row of that table gives the components of the unknown force F W.

Chapter 4 Problems 11 Force x-component y-component F F 1 14.6 N (31.0 N) cos 15.0 = 9.9 N 19. N (31.0 N) sin 15.0 = 8.0 N F W = F (3.0 N ) cos 15.0 =. N (3.0 N) sin 15.0 = 5.95 N F F 1 F 14.6 N 9.9 N +. N = 6.9 N 19. N 8.0 N + 5.95 N = 17.1 N The magnitude of F W is given by the Pythagorean theorem as F W = (6.9 N) + (17.1N) = 18.4 N The angle θ that F W makes with the x axis is θ = tan 1 17.1 N 6.9 N = 68 Therefore, the direction of F W is 68, north of east. θ 6.9 N 17.1 N 18. REASONING AND SOLUTION Newton's law of gravitation gives 11 3 3 ( 6.67 10 N m /kg ) ( 1.50 10 kg) ( 870.0 10 kg) Gmm 1 F = = = r ( 0.100 m ) 1 8.70 10 N 19. SSM REASONING AND SOLUTION a. According to Equation 4.5, the weight of the space traveler of mass m = 115 kg on earth is W = mg = (115 kg) (9.80 m/s ) = 1.13 10 3 N b. In interplanetary space where there are no nearby planetary objects, the gravitational force exerted on the space traveler is zero and g = 0 m/s. Therefore, the weight is W = 0 N. Since the mass of an object is an intrinsic property of the object and is independent of its location in the universe, the mass of the space traveler is still m =115 kg.

1 FORCES AND NEWTON'S LAWS OF MOTION 0. REASONING AND SOLUTION The forces that act on the rock are shown at the right. Newton's second law (with the direction of motion as positive) is R Solving for the acceleration a gives ΣF = mg R = ma ( 45kg)( 9.80 m/s ) ( 50 N) mg R a = = = m 45kg 4. m/s 1. REASONING The magnitude of the gravitational force that each part exerts on the other is given by Newton s law of gravitation as F = Gmm 1 / r. To use this expression, we need the masses m 1 and m of the parts, whereas the problem statement gives the weights W 1 and W. However, mg the weight is related to the mass by W = mg, so that for each part we know that m = W/g. SOLUTION The gravitational force that each part exerts on the other is Gmm F = = r ( / ) ( / ) GW g W g 1 1 r 11 ( 6.67 10 N m /kg ) ( 11000 N) ( 3400 N) ( 9.80 m/s ) ( 1 m) = = 7 1.8 10 N. REASONING Each particle experiences two gravitational forces, one due to each of the remaining particles. To get the net gravitational force, we must add the two contributions, taking into account the directions. The magnitude of the gravitational force that any one particle exerts on another is given by Newton s law of gravitation as F = Gmm 1 / r. Thus, for particle A, we need to apply this law to its interaction with particle B and with particle C. For particle B, we need to apply the law to its interaction with particle A and with particle C. Lastly, for particle C, we must apply the law to its interaction with particle A and with particle B. In considering the directions, we remember that the gravitational force between two particles is always a force of attraction. SOLUTION We begin by calculating the magnitude of the gravitational force for each pair of particles:

Chapter 4 Problems 13 F AB = Gm A m B r = F BC = Gm B m C r = F AC = Gm A m C r = ( 6.67 10 11 N m / kg ) 363 kg ( )( 517 kg) ( 0.500 m) = 5.007 10 5 N ( 6.67 10 11 N m / kg ) 517 kg ( )( 154 kg) ( 0.500 m) = 8.497 10 5 N ( 6.67 10 11 N m / kg ) 363 kg ( )( 154 kg) ( 0.500 m ) = 6.69 10 6 N In using these magnitudes we take the direction to the right as positive. a. Both particles B and C attract particle A to the right, the net force being F A = F AB + F AC = 5.007 10 5 N + 6.69 10 6 N = 5.67 10 5 N, right b. Particle C attracts particle B to the right, while particle A attracts particle B to the left, the net force being F B = F BC F AB = 8.497 10 5 N 5.007 10 5 N = 3.49 10 5 N, right c. Both particles A and B attract particle C to the left, the net force being F C = F AC + F BC = 6.69 10 6 N + 8.497 10 5 N = 9.16 10 5 N, left 3. SSM REASONING AND SOLUTION a. Combining Equations 4.4 and 4.5, we see that the acceleration due to gravity on the surface of Saturn can be calculated as follows: g Saturn = G M Saturn r Saturn ( ( ) 5.67 10 6 kg) = 6.67 10 11 N m /kg b. The ratio of the person s weight on Saturn to that on earth is (6.00 10 7 = 10.5 m/s m) WSaturn mgsaturn gsaturn 10.5 m/s = = = = W mg g 9.80 m/s earth earth earth 1.07

14 FORCES AND NEWTON'S LAWS OF MOTION 4. REASONING AND SOLUTION The magnitude of the net force acting on the moon is found by the Pythagorean theorem to be F SM Sun F Moon F EM F = F SM + F EM Earth Newton's law of gravitation applied to the sun-moon (the units have been suppressed) F SM = G m S m M r SM =(6.67 x 10-11 ) (1.99 x 1030 )(7.35 x 10 ) (1.50 x 10 11 m) = 4.34 x 10 0 N. A similar application to the earth-moon gives F EM = G m r m E M EM = (6.67 x 10-11 ) (5.98 x 104 )(7.35 x 10 ) (3.85 x 10 8 m) = 1.98 x 10 0 N. The net force on the moon is then F = ( 4.34 10 0 N) + ( 1.98 10 0 N) = 4.77 10 0 N 5. REASONING According to Equation 4.4, the weights of an object of mass m on the surfaces of planet A (mass = M A, radius = R ) and planet B (mass = M B, radius = R ) are W A = GM A m R and W B = GM B m R The difference between these weights is given in the problem. SOLUTION The difference in weights is W A W B = GM A m R Rearranging this result, we find GM B m R = Gm R ( ) M A M B

Chapter 4 Problems 15 ( M A M B = W A W B )R Gm = ( ) ( 360 N) 1.33 10 7 m (6.67 10 11 N m /kg ) 5450 kg ( ) = 1.76 10 4 kg 6. REASONING The acceleration due to gravity at the surface of a planet is given by Equations 4.4 and 4.5 as gplanet = GM / r, where M is the mass of the planet and r is its radius. By applying this relation to the unknown planet and to the earth, and noting the ratios of the masses and radii, we will be able to find the acceleration due to gravity on the surface of the planet. SOLUTION Taking the ratio of g Planet to g Earth, and noting that M Planet /M Earth = 0.10 and r Earth /r Planet = 1/0.50, we have that GMPlanet gplanet rplanet MPlanet rearth 1 = = ( 0.10) 0.40 g GM = = M r 0.50 Earth Earth Earth Planet rearth Therefore, the acceleration due to gravity on the planet is Planet ( ) Earth ( ) ( ) g = 0.40 g = 0.40 9.80m/s = 3.9m/s 7. SSM REASONING AND SOLUTION According to Equations 4.4 and 4.5, the weight of an object of mass m at a distance r from the center of the earth is mg = GM E m r In a circular orbit that is 3.59 10 7 m above the surface of the earth (radius = 6.38 10 6 m, mass = 5.98 10 4 kg), the total distance from the center of the earth is r = 3.59 10 7 m + 6.38 10 6 m. Thus the acceleration g due to gravity is g = GM E = (6.67 10 11 N m /kg )(5.98 10 4 kg) (3.59 10 7 m + 6.38 10 6 m) = 0.3 m/s r

16 FORCES AND NEWTON'S LAWS OF MOTION 8. REASONING AND SOLUTION The figure at the right shows the three spheres with sphere 3 being the sphere of unknown mass. Sphere 3 feels a force F 31 due to the presence of sphere 1, and a force F 3 due to the presence of sphere. The net force on sphere 3 is the resultant of F 31 and F 3. Note that since the spheres form an equilateral triangle, each interior angle is 60. Therefore, both F 31 and F 3 make a 30 angle with the vertical line as shown. 3 F 31 F 3 1.0 m 30 1 Furthermore, F 31 and F 3 have the same magnitude given by F = GMm 3 r where M is the mass of either sphere 1 or and m 3 is the mass of sphere 3. The components of the two forces are shown in the following drawings: F 31 30.0 F cos θ F cos θ 30.0 F 3 F sin θ F sin θ Clearly, the horizontal components of the two forces add to zero. Thus, the net force on sphere 3 is the resultant of the vertical components of F 31 and F 3 : F 3 = Fcos θ = GMm 3 r cos θ. The acceleration of sphere 3 is given by Newton's second law: ( ) ( ) cos 30.0 a 3 = F 3 = GM (6.67 10 11 N m /kg ).80 kg m 3 r cosθ = 1.0 m =.5 10 10 m/s

Chapter 4 Problems 17 9. REASONING According to Equation 4.4, the weights of an object of mass m on the surface of a planet (mass = M, radius = R ) and at a height H above the surface are W = GMm 14 R4 3 On surface GMm and W H = ( R + H) 144443 At height H above surface The fact that W is one percent less than W H tells us that the W H /W = 0.9900, which is the starting point for our solution. SOLUTION The ratio W H /W is W H W = Solving for H/R gives GMm ( R + H) GMm R 1 + H R = 1 0.9900 R = ( R + H) = 1 1+ H / R or ( ) = 0.9900 H R = 0.0050 30. REASONING The gravitational force that the sun exerts on a person standing on the earth is given by Equation 4.3 as Fsun = GMsunm/ rsun-earth, where M sun is the mass of the sun, m is the mass of the person, and r sun-earth is the distance from the sun to the earth. Likewise, the gravitational force that the moon exerts on a person standing on the earth is given by Fmoon = GMmoonm/ rmoon-earth, where M moon is the mass of the moon and r moon-earth is the distance from the moon to the earth. These relations will allow us to determine whether the sun or the moon exerts the greater gravitational force on the person. SOLUTION Taking the ratio of F sun to F moon, and using the mass and distance data found on the inside of the front cover of the text, we have that GMsunm Fsun rsun-earth Msun rmoon-earth = = F GM m M r moon moon moon sun-earth rmoon-earth 30 8 1.99 10 kg 3.85 10 m = 11 = 7.35 10 kg 1.50 10 m 178

18 FORCES AND NEWTON'S LAWS OF MOTION Therefore, the sun exerts the greater gravitational force. 31. SSM WWW REASONING AND SOLUTION There are two forces that act on the balloon; they are, the combined weight of the balloon and its load, Mg, and the upward buoyant force F B. If we take upward as the positive direction, then, initially when the balloon is motionless, Newton's second law gives F B Mg = 0. If an amount of mass m is dropped overboard so that the balloon has an upward acceleration, Newton's second law for this situation is F B (M m)g = (M m)a But F B = mg, so that Mg ( M m)g = mg = ( M m)a Solving for the mass m that should be dropped overboard, we obtain m = Ma g + a = (310 kg )(0.15 m/s ) 9.80 m/s = 4.7 kg +0.15 m/s 3. REASONING AND SOLUTION The acceleration due to gravity at the surface of the neutron star is 11 30 Gm (6.67 10 N m /kg )(.0 10 kg) a = = = 5.3 10 m/s 3 r (5.0 10 m) 1 Since the gravitational force is assumed to be constant, the acceleration will be constant and the speed of the object can be calculated from v = v0 + ay, with v 0 = 0m/s since the object falls from rest. Solving for v yields v = ay = ( 5.3 10 1 m/s )( 0.010 m) = 3.3 10 5 m/s 33. REASONING We place the third particle (mass = m 3 ) as shown in the following drawing: D L m m 3 m

Chapter 4 Problems 19 The magnitude of the gravitational force that one particle exerts on another is given by Newton s law of gravitation as F = Gm 1 m /r. Before the third particle is in place, this law indicates that the force on each particle has a magnitude F before = Gmm/L. After the third particle is in place, each of the first two particles experiences a greater net force, because the third particle also exerts a gravitational force on them. SOLUTION For the particle of mass m, we have Gmm Gmm 3 + after D L Lm3 Gmm before md F F For the particle of mass m, we have = = + 1 L Gmm Gmm 3 + after ( L D) L Lm3 F = = + 1 F Gm m before m( L D) L Since F after /F before = for both particles, we have L m 3 md +1 = L m 3 m( L D) +1 or D = ( L D) Expanding and rearranging this result gives the quadratic formula: D + LD L = 0, which can be solved for D using ( ) ( ) ( ) L± L 41 L D= = 0.414L or.414l 1 ( ) The negative solution is discarded because the third particle lies on the +x axis between m and m. Thus, D = 0.414L. 34. REASONING In each case the object is in equilibrium. According to Equation 4.9b, ΣF y = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of the object(s) and the normal force exerted on them.

130 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION a. There are three vertical forces acting on the crate: an upward normal force + that the floor exerts, the weight m 1 g of the crate, and the weight m g of the person standing on the crate. Since the weights act downward, they are assigned negative numbers. Setting the sum of these forces equal to zero gives F + ( mg) + ( mg) = 0 14444444443 N 1 ΣF y The magnitude of the normal force is = m 1 g + m g = (35 kg + 65 kg)(9.80 m/s ) = 980N b. There are only two vertical forces acting on the person: an upward normal force + that the crate exerts and the weight m g of the person. Setting the sum of these forces equal to zero gives The magnitude of the normal force is F + ( mg) = 0 144443 N ΣF y = m g = (65 kg)(9.80 m/s ) = 640N 35. SSM REASONING AND SOLUTION According to Equation 3.3b, the acceleration of the astronaut is a y = (v y v 0y )/ t = v y /t. The apparent weight and the true weight of the astronaut are related according to Equation 4.6. Direct substitution gives vy { FN = { mg + may = m ( g + ay) = m g + t Apparent weight True weight 45 m/s = + = 15 s (57 kg) 9.80 m/s 7.3 10 N 36. REASONING AND SOLUTION a. The apparent weight of the person is = mg + ma = (95.0 kg)(9.80 m/s + 1.80 m/s ) = 1.10 X 10 3 N

Chapter 4 Problems 131 b. = (95.0 kg)(9.80 m/s ) = 931 N c. = (95.0 kg)(9.80 m/s 1.30 m/s ) = 808 N 37. REASONING AND SOLUTION The block will move only if the applied force is greater than the maximum static frictional force acting on the block. That is, if F > µ s = µ s mg = (0.650)(45.0 N) = 9. N The applied force is given to be F = 36.0 N which is greater than the maximum static frictional force, so the block will move. The block's acceleration is found from Newton's second law. a = ΣF m = F f k m = F µ k mg = 3.7 m/s m 38. REASONING It is the static friction force that accelerates the cup when the plane accelerates. The maximum magnitude that this force can have will determine the maximum acceleration, according to Newton s second law. SOLUTION According to Newton s second law, we have ΣF = f s MAX = µ s = µ s mg = ma In this result, we have used the fact that the magnitude of the normal force is = mg, since the plane is flying horizontally and the normal force acting on the cup balances the cup s weight. Solving for the acceleration gives a = µ s g = ( 0.30) ( 9.80 m/s )=.9 m/s 39. SSM REASONING In order to start the crate moving, an external agent must supply a force that is at least as large as the maximum value f s MAX = µs, where µ s is the coefficient of static friction (see Equation 4.7). Once the crate is moving, the magnitude of the frictional force is very nearly constant at the value f k = µ k, where µ k is the coefficient of kinetic friction (see

13 FORCES AND NEWTON'S LAWS OF MOTION Equation 4.8). In both cases described in the problem statement, there are only two vertical forces that act on the crate; they are the upward normal force, and the downward pull of gravity (the weight) mg. Furthermore, the crate has no vertical acceleration in either case. Therefore, if we take upward as the positive direction, Newton's second law in the vertical direction gives mg = 0, and we see that, in both cases, the magnitude of the normal force is = mg. SOLUTION a. Therefore, the applied force needed to start the crate moving is f s MAX = µs mg = (0.760)(60.0 kg)(9.80 m/s ) = 447 N b. When the crate moves in a straight line at constant speed, its velocity does not change, and it has zero acceleration. Thus, Newton's second law in the horizontal direction becomes P f k = 0, where P is the required pushing force. Thus, the applied force required to keep the crate sliding across the dock at a constant speed is P = f k = µ k mg = (0.410)(60.0 kg)(9.80 m/s ) = 41 N 40. REASONING AND SOLUTION a. f k = µ k mg = (0.61)(9 kg)(9.80 m/s ) = 550 N b. From kinematics v 0 = at = f k t = 550 N ( 1. s)= 7. m/s m 9 kg 41. REASONING In each of the three cases under consideration the kinetic frictional force is given by f k = µ k. However, the normal force varies from case to case. To determine the normal force, we use Equation 4.6 ( = mg + ma) and thereby take into account the acceleration of the elevator. The normal force is greatest when the elevator accelerates upward (a positive) and smallest when the elevator accelerates downward (a negative). SOLUTION a. When the elevator is stationary, its acceleration is a = 0 m/s. Using Equation 4.6, we can express the kinetic frictional force as

Chapter 4 Problems 133 f k = µ k = µ k ( mg + ma) = µ k m( g + a) = ( 0.360 )( 6.00 kg) 9.80 m/s [( )+( 0 m/s )] = 1. N b. When the elevator accelerates upward, a = +1.0 m/s. Then, f k = µ k = µ k ( mg + ma) = µ k m( g + a) = ( 0.360 )( 6.00 kg) 9.80 m/s [( )+( 1.0 m/s )] = 3.8 N c. When the elevator accelerates downward, a = 1.0 m/s. Then, f k = µ k = µ k ( mg + ma) = µ k m( g + a) = ( 0.360 )( 6.00 kg) 9.80 m/s [( )+( 1.0 m/s )] = 18.6 N 4. REASONING The initial and final velocities of the car are known, so we may use Equation.9 ( v = v 0 + ax) of the equations of kinematics to determine the stopping displacement x, x x x provided the acceleration a x can be found. The acceleration is related to the net force ΣF x acting on the car by Equation 4., Fx = max, where m is its mass. The only force acting on the car in the x-direction is the static frictional force that acts on the tires. The minimum stopping distance occurs when the static frictional force is a maximum, so the net force can be expressed with the aid MAX of Equation 4.7 as Fx = fs = µ sfn, where µ s is the coefficient of static friction and is the magnitude of the normal force that acts on the tires. The minus sign indicates that the force points to the left, along the x axis. By combining these relations, will be able to find the displacement of the car. SOLUTION Solving Equation.9 for x and substituting Newton s second law ( a = F m) into this equation yields vx v0x vx v0x x = = a F / m x ( Σ ) x x x / The net force acting on the car is equal to the maximum static frictional force. The maximum static MAX frictional force is given by f = µ F. With this substitution, the displacement becomes: s s N v v v v x = = / / x 0x x 0x ( ΣF m) ( µ F m) x s N

134 FORCES AND NEWTON'S LAWS OF MOTION There are two forces acting on the car in the vertical, or y, direction, the downward-acting weight mg and the upward-acting normal force +. Since the car is traveling on a horizontal road and does not leave it, the net force in the vertical direction must be zero, so that mg = 0, or = mg. Thus, the displacement can be written as (notice that the mass m algebraically cancels out of this expression) v v v v x = = µ F / m µ g x 0x x 0x ( ) s N s The stopping displacements with and without antilock brakes are x ( 0m/s) ( 17.9m/s ) vx v0x with µ sg 0.764 9.80m/s = = = ( ) ( ) 1.4m x ( 0m/s) ( 17.9m/s) vx v0 x without µ s g 0.615 9.80m/s = = = ( )( ) 6.6m The difference between these two displacements is 6.6 m 1.4 m = 5.m. 43. SSM REASONING If we assume that kinetic friction is the only horizontal force that acts on the skater, then, since kinetic friction is a resistive force, it acts opposite to the direction of motion and the skater slows down. According to Newton's second law ( F = ma ), the magnitude of the deceleration is a = f k / m. The magnitude of the frictional force that acts on the skater is, according to Equation 4.8, f k = µ k where µ k is the coefficient of kinetic friction between the ice and the skate blades. There are only two vertical forces that act on the skater; they are the upward normal force and the downward pull of gravity (the weight) mg. Since the skater has no vertical acceleration, Newton's second law in the vertical direction gives (if we take upward as the positive direction) mg = 0. Therefore, the magnitude of the normal force is = mg and the magnitude of the deceleration is given by a = f k m = µ k m = µ k mg m = µ k g SOLUTION a. Direct substitution into the previous expression gives

Chapter 4 Problems 135 a = µ k g = (0.100)(9.80 m/s ) = 0.980 m/s Since the skater is slowing down, the direction of the acceleration must be opposite to the direction of motion. b. The displacement through which the skater will slide before he comes to rest can be obtained from Equation.9 (v v 0 = ax). Since the skater comes to rest, v = 0 m/s. If we take the direction of motion of the skater as the positive direction, then, solving for x, we obtain x = v 0 a = (7.60 m/s) ( 0.980 m/s ) = 9.5 m 44. REASONING The free-body diagrams for the large cube (mass = M) and the small cube (mass = m) are shown in the following drawings. In the case of the large cube, we have omitted the weight and the normal force from the surface, since the play no role in the solution (although they do balance). f s MAX P mg In these diagrams, note that the two blocks exert a normal force on each other; the large block exerts the force on the smaller block, while the smaller block exerts the force on the larger block. In accord with Newton s third law these forces have opposite directions and equal magnitudes. Under the influence of the forces shown, the two blocks have the same acceleration a. We begin our solution by applying Newton s second law to each one. SOLUTION According to Newton s second law, we have ΣF = P = Ma 144 4443 Large block = ma 14 43 Small block Substituting = ma into the large-block expression and solving for P gives P = (M + m) a For the smaller block to remain in place against the larger block, the static frictional force must balance the weight of the smaller block, so that f s MAX = mg. But f s MAX is given by

136 FORCES AND NEWTON'S LAWS OF MOTION f s MAX = µ s, where, from the Newton s second law, we know that = ma. Thus, we have µ s ma = mg or a = g/µ s. Using this result in the expression for P gives P = ( M + m)a = ( M + m)g = µ s ( ) 0.71 ( 5kg + 4.0 kg) 9.80 m/s = 4.0 10 N 45. REASONING The free-body diagram for the box is shown in the following drawing on the left. On the right the same drawing is repeated, except that the pushing force P is resolved into its horizontal and vertical components. P P sin θ θ P cos θ f k f k mg mg Since the block is moving at a constant velocity, it has no acceleration, and Newton s second law indicates that the net vertical and net horizontal forces must separately be zero. SOLUTION Taking upward and to the right as the positive directions, we write the zero net vertical and horizontal forces as follows: mg Psin θ = 0 1444 4443 Vertical P cosθ f k = 0 144 443 Horizontal From the equation for the horizontal forces, we have P cos θ = f k. But the kinetic frictional force is f k = µ k. Furthermore, from the equation for the vertical forces, we have = mg + P sin θ. With these substitutions, we obtain Pcosθ = f k = µ k = µ k ( mg + P sinθ) Solving for P gives µ P = k mg cosθ µ k sin θ

Chapter 4 Problems 137 The necessary pushing force becomes infinitely large when the denominator in this expression is zero. Hence, we find that cosθ µ k sin θ = 0, which can be rearranged to show that sin θ cosθ = tanθ = 1 or θ = tan 1 1 = 68 µ k 0.41 46. REASONING AND SOLUTION a. In the horizontal direction the thrust, F, is balanced by the resistive force, f r, of the water. That is, or f r = F = F x = 0 5 7.40 10 N b. In the vertical direction, the weight, mg, is balanced by the buoyant force, F b. So gives F y = 0 F b = mg = (1.70 10 8 kg)(9.80 m/s ) = 9 1.67 10 N 47. SSM REASONING AND SOLUTION There are three vertical forces that act on the lantern. The two upward forces of tension exerted by the wires, and the downward pull of gravity (the weight). If we let upward be the positive direction, and let T represent the tension in one of the wires, then Equation 4.9b gives Solving for T gives F y = T mg = 0 T = mg = (1.0 kg )(9.80 m/s ) = 58.8 N 48. REASONING AND SOLUTION The free body diagram for the plane is shown below to the left. The figure at the right shows the forces resolved into components parallel to and perpendicular to the line of motion of the plane. L T L T R R W θ W sin θ W cos θ

138 FORCES AND NEWTON'S LAWS OF MOTION If the plane is to continue at constant velocity, the resultant force must still be zero after the fuel is jettisoned. Therefore (using the directions of T and L to define the positive directions), T R W(sin θ) = 0 (1) L W (cos θ) = 0 () From Example 13, before the fuel is jettisoned, the weight of the plane is 86 500 N, the thrust is 103 000 N, and the lift is 74 900 N. The force of air resistance is the same before and after the fuel is jettisoned and is given in Example 13 as R = 59 800 N. After the fuel is jettisoned, W = 86 500 N 800 N = 83 700 N From Equation (1) above, the thrust after the fuel is jettisoned is T = R + W (sin θ) = [(59 800 N) + (83 700 N)(sin 30.0 )] = 101 600 N From Equation (), the lift after the fuel is jettisoned is L = W (cos θ) = (83 700 N)(cos 30.0 ) = 7 500 N a. The pilot must, therefore, reduce the thrust by 103 000 N 101 600 N = 1400 N b. The pilot must reduce the lift by 74 900 N 7 500 N = 400 N

Chapter 4 Problems 139 49. SSM REASONING AND SOLUTION The figure at the right shows the forces that act on the wine bottle. Newton s second law applied in the horizontal and vertical directions gives F y = F 1 cos 45.0 + F cos 45.0 W = 0 (1) y F 1 45.0 45.0 F x W F x = F sin 45.0 F 1 sin 45.0 = 0 () From Equation (), we see that F 1 = F. According to Equation (1), we have Therefore, F 1 = W cos 45.0 = mg cos 45.0 F 1 = F = (1.40 kg) (9.80 m/s ) = 9.70 N cos 45.0 50. REASONING The drawing shows the I-beam and the three forces that act on it, its weight W and the tension T in each of the cables. Since the I-beam is moving upward at a constant velocity, its acceleration is zero and it is in vertical equilibrium. According to Equation 4.9b, F y = 0, the net force in the vertical (or y) direction must be zero. This relation will allow us to find the magnitude of the tension. T T SOLUTION Taking up as the +y direction, Equation 4.9b becomes 70.0 70.0 3 + Tsin70.0 + Tsin70.0 8.00 10 N = 0 1444444 4444444443 Σ F y Solving this equation for the tension gives T = 460N. W = 8.00 10 3 N

140 FORCES AND NEWTON'S LAWS OF MOTION 51. REASONING The book is kept from falling as long as the total static frictional force balances the weight of the book. The forces that act on the book are shown in the following free-body diagram, where P is the pressing force applied by each hand. f s MAX f s MAX P P W In this diagram, note that there are two pressing forces, one from each hand. Each hand also applies a static frictional force, and, therefore, two static frictional forces are shown. The maximum static frictional force is related in the usual way to a normal force, but in this problem the normal force is provided by the pressing force, so that = P. SOLUTION Since the frictional forces balance the weight, we have f MAX s = ( µ s )= ( µ s P)= W Solving for P, we find that P = W = 31 N µ s 0.40 ( ) = 39 N 5. REASONING AND SOLUTION Newton s second law applied in the vertical and horizontal directions gives L cos 1.0 W = 0 (1) L sin 1.0 R = 0 ()

Chapter 4 Problems 141 a. Equation (1) gives L = W 53 800 N = = 57 600 N cos1.0 cos 1.0 b. Equation () gives R = L sin 1.0 = ( 57 600 N)sin 1.0 = 0 600 N R 1.0 W L 53. SSM REASONING In order for the object to move with constant velocity, the net force on the object must be zero. Therefore, the north/south component of the third force must be equal in magnitude and opposite in direction to the 80.0 N force, while the east/west component of the third force must be equal in magnitude and opposite in direction to the 60.0 N force. Therefore, the third force has components: 80.0 N due south and 60.0 N due east. We can use the Pythagorean theorem and trigonometry to find the magnitude and direction of this third force. SOLUTION The magnitude of the third force is F 3 = (80.0 N) + (60.0 N) = 1.00 10 N The direction of F 3 is specified by the angle θ where θ = tan 1 80.0 N 60.0 N = 53.1, south of east 80.0 N θ 60.0 N N θ F 3 E

14 FORCES AND NEWTON'S LAWS OF MOTION 54. REASONING The free-body diagram in the drawing at the right shows the forces that act on the clown (weight = W). In this drawing, note that P denotes the pulling force. Since the rope passes around three pulleys, forces of magnitude P are applied both to the clown s hands and his feet. The normal force due to the floor is, and the maximum static frictional P P f s MAX force is f s MAX. At the instant just before the clown s feet move, the net vertical and net horizontal forces are zero, according to Newton s second law, since there is no acceleration at this instant. W SOLUTION According to Newton s second law, with upward and to the right chosen as the positive directions, we have + P W = 0 144 443 Vertical forces MAX and f s P = 0 1 4 4 3 Horizontal forces From the horizontal-force equation we find P = f s MAX. But f s MAX = µ s. From the verticalforce equation, the normal force is = W P. With these substitutions, it follows that Solving for P gives ( ) P = f s MAX = µ s = µ s W P ( )( 890 N) P = µ s W = 0.53 1+ µ s 1 + 0.53 = 310 N 55. REASONING The drawing shows the bicycle (represented as a circle) moving down the hill. Since the bicycle is moving at a constant velocity, its acceleration is zero and it is in equilibrium. Choosing the x axis to be parallel to the hill, Equation 4.9a states that F x = 0, so the net force along the x axis is zero. This relation will allow us to find the value of the numerical constant c that appears in the expression for f air. x 8.0 +x mg f air SOLUTION Taking up the hill as the +x direction, the x-component of the weight is mg sin 8.0, and the force due to air resistance is +f air = +cv, where the plus sign indicates that this force points opposite to the motion of the bicycle, or up the hill. Equation 4.9a can be written as

Chapter 4 Problems 143 Solving this equation for the constant c gives 14444443 + cv mg sin8.0 = 0 ΣF x ( )( ) mg sin8.0 85kg 9.80m/s sin8.0 c = = = v 8.9m/s 13kg/s 56. REASONING AND SOLUTION If the +x axis is taken in the direction of motion, F x = 0 gives F N F f k mg sin θ = 0 f k where Then f k = µ k F µ k mg sin θ = 0 (1) W 5.0 F Also, F y = 0 gives so mg cos θ = 0 = mg cos θ () Substituting Equation () into Equation (1) and solving for F yields F = mg( sin θ + µ k cos θ ) F = (55.0 kg)(9.80 m/s )[sin 5.0 + (0.10)cos 5.0 ] = 86 N 57. SSM REASONING There are four forces that act on the chandelier; they are the forces of tension T in each of the three wires, and the downward force of gravity mg. Under the influence of these forces, the chandelier is at rest and, therefore, in equilibrium. Consequently, the sum of the x components as well as the sum of the y components of the forces must each be zero. The figure below shows the free-body diagram for the chandelier and the force components for a suitable system of x, y axes. Note that the free-body diagram only shows one of the forces of tension; the second and third tension forces are not shown. The triangle at the right shows the geometry of one of the cords, where l is the length of the cord, and d is the distance from the ceiling.

144 FORCES AND NEWTON'S LAWS OF MOTION y θ T θ θ x d l mg We can use the forces in the y direction to find the magnitude T of the tension in any one wire. SOLUTION diagram that Remembering that there are three tension forces, we see from the free-body 3T sin θ = mg or T = mg 3sinθ = mg 3(d / l) = mgl 3d Therefore, the magnitude of the tension in any one of the cords is T = (44 kg)(9.80 m/s )(.0 m) = 1.9 10 N 3(1.5 m) 58. REASONING AND SOLUTION Let the tension in wire 1 be T 1 and the tension in wire be T. The sum of the vertical forces acting on the point where the wires join must be zero. T 1 sin 43.0 + T sin 55.0 mg = 0 (1) Similarly, the horizontal forces must add to zero so T 1 cos 43.0 + T cos 55.0 = 0 () Solving Equation () for T 1, substituting into Equation (1), and rearranging yields T = 317 N Using this result in Equation () gives T 1 = 49 N

Chapter 4 Problems 145 59. REASONING The tree limb is in equilibrium, so we will apply Newton s second law to it. This law will allow us to determine the tension in both segments of the wire. 0.800 m 3.0 m SOLUTION Let T 1 be the tension in the shorter segment of the wire, T be the tension in the longer segment, and W be the weight of the limb (see the drawing). In the horizontal direction, F x = 0 (Equation 4.9a) gives T1 cosθ1+ T cosθ = 0 14444443 ΣF x 1 0.00m 1 0.00m where θ1 = tan = 14.0 and θ = tan = 3.58 0.800m 3.0m. In the vertical direction, F y = 0 (Equation 4.9b) gives + T sinθ + T sinθ W= 0 1444444443 1 1 ΣF y a. Solving Equation (1) for T in terms of T 1, and substituting the result into Equation () yields T 1 θ 1 0.00 m T W θ (1) () T 1 W 151N = = = sinθ + sinθ cos θ / cosθ sin14.0 + sin 3.58 cos14.0 / cos3.58 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 499 N b. Substituting T 1 = 499 N into Equation (1) and solving for T gives ( ) ( ) ( ) ( ) ( ) T = T cos θ / cosθ = 499 N cos14.0 / cos3.58 = 485N 1 1 60. REASONING AND SOLUTION a. If the block is not to slide down the wall, then the vertical forces acting on the block must sum to zero. F cos 40.0 mg + µ s = 0 Additionally, the horizontal forces must sum to zero. Eliminating gives F = F sin 40.0 = 0 mg cos40.0 + µ s sin 40.0 = 79.0 N

146 FORCES AND NEWTON'S LAWS OF MOTION b. The above analysis applies to the case where the block is starting to slide up the wall except, that the frictional force will be in the opposite direction. Hence, mg F = cos40.0 µ s sin 40.0 = 19 N 61. SSM REASONING The following figure shows the crate on the incline and the free body diagram for the crate. The diagram at the far right shows all the forces resolved into components that are parallel and perpendicular to the surface of the incline. We can analyze the motion of the crate using Newton's second law. The coefficient of friction can be determined from the resulting equations. P y x P fs MAX P sin θ f S MAX θ θ P cos θ mg θ mg sin θ mg cos θ SOLUTION Since the crate is at rest, it is in equilibrium and its acceleration is zero in all directions. If we take the direction down the incline as positive, Newton's second law indicates that MAX F x = P cosθ + mgsin θ f s = 0 According to Equation 4.7, f s MAX = µs. Therefore, we have Pcosθ+ mgsin θ µ s = 0 (1) The expression for the normal force can be found from analyzing the forces that are perpendicular to the incline. Taking up to be positive, we have F y = Psin θ+ mgcosθ = 0 or = mg cosθ Psin θ Equation (1) then becomes Pcosθ+ mgsin θ µ s (mgcosθ P sin θ) = 0 Solving for the coefficient of static friction, we find that

Chapter 4 Problems 147 Pcosθ+ mg sin θ (535 N) cos 0.0 + (5 kg)(9.80 m/s ) sin 0.0 µ s = = = mgcos θ Psin θ (5 kg)(9.80 m/s ) cos 0.0 (535 N) sin 0.0 0.665 6. REASONING The weight of the part of the washcloth off the table is m off g. At the instant just before the washcloth begins to slide, this weight is supported by a force that has magnitude equal to f s MAX, which is the static frictional force that the table surface applies to the part of the washcloth on the table. This force is transmitted around the bend in the washcloth hanging over the edge by the tension forces between the molecules of the washcloth, in much the same way that a force applied to one end of a rope is transmitted along the rope as it passes around a pulley. SOLUTION Since the static frictional supports the weight of the washcloth off the table, we have f s MAX = m off g. The static frictional force is f s MAX = µ s. The normal force is applied by the table to the part of the washcloth on the table and has a magnitude equal to the weight of that part of the washcloth. This is so, because the table is assumed to be horizontal and the part of the washcloth on it does not accelerate in the vertical direction. Thus, we have MAX f s = µ s = µ s m on g = m off g The magnitude g of the acceleration due to gravity can be eliminated algebraically from this result, giving µ s m on = m off. Dividing both sides by m on + m off gives m m on off µ = or µ f = f mon m + off mon + moff s s on off where we have used f on and f off to denote the fractions of the washcloth on and off the table, respectively. Since f on + f off = 1, we can write the above equation on the left as µ s ( 1 f off )= f off or f off = µ s = 0.40 1 + µ s 1 + 0.40 = 0.9 63. REASONING AND SOLUTION Newton's second law, F = ma, implies that the acceleration a and the net force are in the same direction. This is 64 N of E. The magnitude of the net force is F = ma = (350 kg)(0.6 m/s ) = 0 N

148 FORCES AND NEWTON'S LAWS OF MOTION 64. REASONING Suppose the bobsled is moving along the +x direction. There are two forces acting on it that are parallel to its motion; a force +F x propelling it forward and a force of 450 N that is resisting its motion. The net force is the sum of these two forces. According to Newton s second law, Equation 4.a, the net force is equal to the mass of the bobsled times its acceleration. Since the mass and acceleration are known, we can use the second law to determine the magnitude of the propelling force. SOLUTION a. Newton s second law states that Solving this equation for F x gives + Fx 450N = ma 144443 x (4.a) Σ F x F x ( ) ( ) = ma + 450N= 70kg.4m/s + 450N=1100N x b. The magnitude of the net force that acts on the bobsled is ( ) ( ) Σ F = ma = 70kg.4m/s = 650N (4.a) x x 65. SSM REASONING If we assume that the acceleration is constant, we can use Equation.4 (v = v 0 + at ) to find the acceleration of the car. Once the acceleration is known, Newton's second law ( F = ma ) can be used to find the magnitude and direction of the net force that produces the deceleration of the car. SOLUTION The average acceleration of the car is, according to Equation.4, v v0 17.0m/s 7.0m/s a = = = 1.5m/s t 8.00s where the minus sign indicates that the direction of the acceleration is opposite to the direction of motion; therefore, the acceleration points due west. According to Newton's Second law, the net force on the car is F = ma = (1380 kg)( 1.5 m/s ) = 1730 N

Chapter 4 Problems 149 The magnitude of the net force is 1730N. From Newton's second law, we know that the direction of the force is the same as the direction of the acceleration, so the force also points due west. 66. REASONING AND SOLUTION From Newton's second law and the equation: v = v 0 + at, we have v v F = ma = m 0 t a. When the skier accelerates from rest (v 0 = 0 m/s) to a speed of 11 m/s in 8.0 s, the required net force is F m v- v0 (11 m/s) 0m/s = = (73 kg) = 1.0 10 N t 8.0 s b. When the skier lets go of the tow rope and glides to a halt (v = 0 m/s) in 1 s, the net force acting on the skier is v v0 0m/s (11m/s) F = m = (73kg) = 38N t 1 s The magnitude of the net force is 38 N. 67. REASONING AND SOLUTION Newton's second law applied to block 1 (4 N) gives Object 1 Object T = m 1 a 1 N 1 T T Similarly, for block (185 N) T m g = m a W 1 W If the string is not to break or go slack, both blocks must have accelerations of the same magnitude. Then a 1 = a and a = a. The above equations become T = m 1 a (1)

150 FORCES AND NEWTON'S LAWS OF MOTION T m g = m a () a. Substituting Equation (1) into Equation () and solving for a yields a = b. Using this value in Equation (1) gives m g m 1 + m =.99 m/s T = m 1 a = 19 N 68. REASONING Let s assume that the rocket is moving in the +y direction. There are three forces acting on it: the upward thrust of +T, the force f air of air resistance, and the weight mg of the rocket. The net force is the sum of these three forces. According to Newton s second law, Equation 4.b, the net force is equal to the mass of the rocket times its acceleration. We can use the second law to determine the acceleration. SOLUTION Newton s second law states that T mg f = ma 144443 air ΣF y y (4.b) Solving this equation for a y gives a y + T mg f = m air ( ) ( ) 6 5 6 7.5 10 N 4.5 10 kg 9.80m/s.4 10 N = = + 1.5m/s 5 4.5 10 kg 69. SSM WWW REASONING The speed of the skateboarder at the bottom of the ramp can be found by solving Equation.9 (v = v 0 + ax where x is the distance that the skater moves down the ramp) for v. The figure at the right shows the free-body diagram for the skateboarder. The net force F, which accelerates the skateboarder down the ramp, mgsin θ mgcos θ

Chapter 4 Problems 151 is the component of the weight that is parallel to the incline: F = mg sin θ. Therefore, we know from Newton's second law that the acceleration of the skateboarder down the ramp is a = F m = mg sin θ m = g sin θ SOLUTION Thus, the speed of the skateboarder at the bottom of the ramp is v = v 0 + ax = v0 + gx sinθ = (.6 m/s) + (9.80 m/s )(6.0 m) sin18 = 6.6 m/s 70. REASONING AND SOLUTION a. A free-body diagram for the crate gives Crate T B W C = m C a T B T B = W C + m C a T 1510 N = 1510 N+ ( 0.60 m/s ) = 1610 N 9.80 m/s B W C b. An analysis of the free-body diagram for the platform yields T A T B W W = m W a T T A = T B + W W + m W a 965 N = 1610 N+ ( 0.60 m/s ) + 965 N 9.80 m/s A Man T A W W T B = 640 N 71. SSM REASONING We can use the appropriate equation of kinematics to find the acceleration of the bullet. Then Newton's second law can be used to find the average net force on the bullet. SOLUTION According to Equation.4, the acceleration of the bullet is