(C) So F Is 12% of the Weight of the Motorcycle

Solutions

F F 4-1. (a) a  net  . m m m F 15 N kg· 2 (b) a    1.5 s  1.5 m . m 10 kg kg s2

F 4-2. (a) From a  net  F = ma. m net 5 m 6 m 6 (b) Fnet  ma  (3.2 10 kg) 3.5  1.110 kg   1.1  10 N.  s2  s2

F F 4-3. (a) From a  net  m  net . m a  m  F 5.0 N kg  2 (b) m  net   2.0 s   2.0 kg. a 2.5 m  m  s2  s2 

F F  f 4-4. (a) a  net  . m m kg  m F  f 5.0 N 1.0 N  2  (b) a    8.0 s  8.0 m . m 0.50 kg kg s2

F v v  v  v v  v  4-5. (a) F = ? From a  net and a   f 0  F  ma  m  m f 0 . m t t t t v  v  20.8 m 12.5 m (b) F  m f 0  210 kg s s  250 kg·m  250 N. t 6.9 s s2 F 250 N   0.12, (c) so F is 12% of the weight of the motorcycle. mg 210 kg 9.8 m  s2 

4-6. (a) anew =? The important thing here is that Fnet is unchanged. F m a From a  net  F  ma  (1.5m)a  a  a  0 . m net 0 new new 1.5m 0 1.5 F net 6000 N  m  m 4000 kg kg· 2 (b) a0     s m anew     1.0   1.0 2 . 1.5 1.5 1.5  kg  s

© Paul G. Hewitt and Phillip R. Wolf 4-1 F v v  v  mv 4-7. (a) F  ? From a  net  F  ma  m  m f 0  . m t t t (b) Since we'll want the answer in newtons, we'll need the mass to be in kilograms. 1 kg 56 g  32 m mv  1000 g  s  F    20 kg  m  20 N. t 0.090s s2

R 4-8. (a) m

mg F mg  R R (b) From a  net and calling downward the positive direction a   g  . m m m kg  m R 680 N  2  (c) a  g   9.8 m   3.0 s  3.0 m . m s2 100 kg kg s2

4-9. (a) f m F

F F  f (b) a  net  . m m kg  m F  f 415 N 130 N  2  (c) a    3.8 s  3.8 m . m 75 kg kg s2

 FA  F F  2  F 4-10. (a) a  A . a  B   A  a . So both cars have the same acceleration. A B m A mA mB A mA 2  m F 4000 N kg  2 (b) a    2 s  2 m . m 2000 kg kg s2

4-11. (a) anew =? The important thing here is that F is unchanged. F m a From a   F  ma  (0.6m)a  a  a  0 . m 0 new new 0.6m 0 0.6 2.0 m (b) a0 s2 m anew    3.3 . 0.6 0.6 s2

F F (ma ) m 4-12. (a) a  . a   m  a . m m M M M M m

m  2 kg m m (b) a5 kg  a2 kg  2  0.8 . M  5 kg s2 s2

4-14. (a) F (b) a = ? F pushes upward. W pulls down. So, calling upward the positive F F  W direction, a  net  . m m W F  W  F  W  From W  mg  m  so a     g. g W  W  W g 

 F  W   57 N  38 N m m (c) a    g  9.8 2  4.9 2 .  W   38 N   s  s

F 3mg 4-15. (a) a  net   3g  3 9.8 m  29.4 m  29 m . m m  s2  s2 s2 F 4mg (b) a = ? If F  4W  4mg then a  net   4g  4 9.8 m  39 m . m m  s2  s2

F 3mg  mg 2mg 4-16. (a) Calling upward the positive direction, a  net    2g  2 9.8 m  19.6 m . m m m  s2  s2 (b) The friend forgot to include the weight of the bucket as part of the net force.

4-17. (a) Fnet  ? P pulls up. The bucket's weight W pulls down. So Fnet  P  W  P  mg upward. F P  mg (b) a  net  . m m 95 N  8.5 kg 9.8 m P  mg  2  (c) a   s  1.4 m . m 8.5 kg s2

R 4-18. (a) (b) a = ? Let’s call downward the positive direction. The weight (W=mg) acts F mg  R R downward. R acts upward. So F  mg  R and a  net   g  . net m m m

W (c) If R > mg then the net force acts upward, thus the acceleration is upward. Since the parachutist is still falling, the velocity is still downward. R m 1000 N m (d) a  g   9.8   2.7 . The negative sign indicates that the acceleration is m s2 80 kg s2 upward, opposite the direction of motion, so the parachutist is slowing down.

(b) a  ? The weight mg acts downward. R acts upward. So, calling downward the positive F mg  R mg  1 mg direction, F  mg  R and a  net   5  4 g. net m m m 5 (c) a  4 g  4 9.8 m  7.8 m . 5 5  s2  s2 F 4-20. (a) F  ? From a  net  F  ma  mg. m (b) F  mg  1.1 kg 9.8 m  11kg·m  11N.  s2  s2

4-21. (a) F3m =? It requires three times the force to give three times the mass the same acceleration.

Fm F3m  3m Formally, from a    F3m    Fm  3Fm. m 3m  m  a (b) truck =? Logically, the same force acting on three times the mass should cause one-third the acar F atruck m 1 acceleration. Formally, from a   F  macar  (3m)atruck    3 . m acar 3m

F v v  v  v  v  4-22. (a) F  ? From a   F  ma  m  m f 0  m 0 . m t t t  m m  v  v0  3 12 s  32 s m 4 (b) F  m  3.0 10 kg   10,000 kg· 2  1.0  10 N. The negative sign t  6.0 s  s indicates that a braking force of 1.0  104 N acts opposite the direction of motion.

4-23. (a) and (b) The direction N of the braking force is opposite the direction of F the truck’s motion. mg F v v  v  0  v mv (c) F  ? From a   F  ma  m  m f 0  m  . m t t t t v  v  v  0 vt (d) x  vt  0 f t  t  . 2 2 2 3 m mv 2.0 10 kg18  (e) F   s  4.5 103 kg·m  4.5  103 N. t 8.0 s s2 (That is, 4.5  103 N acting opposite the direction of motion.)

© Paul G. Hewitt and Phillip R. Wolf 4-4 vt 18 m (8.0 s) (f) x   s  72 m. 2 2 F v v  v  0  v mv 4-24. (a) F  ? From a  net  F  ma  m  m f 0  m  . m net t t t t m mv 55 kg28  (b) F   s  7700 kg  m  7700 N. That is, 7700 N t 0.20s s2 acting opposite the direction of motion of the car.

F v v  v  0  v mv 4-25. (a) F  ? From a  net  F  ma  m  m f 0  m  . m net t t t t 1000 m 1 h 55 kg 86 km   mv  h 1 km 3600 s  (b) F    410 N. That is, 410 N t 3.2s acting opposite the direction of motion of the car.

F v v  v  v  0 mv 4-26. (a) F  ? From a  net  F  ma  m  m f 0  m  . m t t t t m mv (26,500 kg)72  2 (b) F   s  1.0 106 kg  m  1.0  106 N. t 1.9s s

F v v  v  v  0 mv 4-27. (a) F  ? From a  net  F  ma  m  m f 0  m  . m t t t t 31 6 m mv (9.1110 kg)3.0 10 s  (b) F    1.8 1017 kg  m  1.8  1017 N. t 1.5 107 s s2

F v v  v  v  0 mv 4-28. (a) F  ? From a  net  F  ma  m  m f 0  m  . m t t t t m mv 0.045 kg72  (b) F   s  650 kg  m  650 N. t 0.0050 s s2 (c) During the impact, the gravitational force on the ball is balanced out by an upward force on the ball from the tee. Also, compared to the impact force, the gravitational force on the ball  mg  0.045 kg 9.8 m  0.44 N is much, much less than the force on the ball from the club.  s2 

4-29. (a) R = ? Since the skier is moving at a constant velocity, the net force on the skier is zero. In the horizontal direction Fnet  T  R  0  R  T.

(b) Fup= ? Since the skier maintains a constant vertical position in the water, the net force on the skier is zero. In the vertical direction Fnet  Fup  mg  0  Fup  mg.

© Paul G. Hewitt and Phillip R. Wolf 4-5 4-30. (a) f  ? The firefighter’s weight acts downward. Friction acts upward. If we call down the positive F mg  f direction, then from a  net   f  mg  ma  m(g  a)  m(g  0.25g)  0.75mg. m m (b) f  ? To slide at a constant speed the net force would have to be zero, with f = mg.

4-31. (a) Fnet  ? If we call R the air resistance force and downward the positive direction, then Fnet  mg  R. At the moment Suzy steps from the basket R = 0 so Fnet  mg. mg mg (b) Fnet  mg  R  mg  2  2 .

(c) Fnet  ? At her terminal speed a = 0 so the net force on her is zero as well. (d) Whatever her terminal speed happens to be, when she gets there Fnet = 0 and a = 0 so the net force on her is zero.

4-32. (a) Fthrust

mgmoon

(b) Fthrust =? The net force on the landing craft has to be upward and large enough to decelerate the craft from initial speed v to zero in a distance d. Taking downward to be the positive direction (and the magnitude of Fthrust to be positive), F mg  F a  net  moon thrust  F  mg  ma  m(g  a). What is a? m m thrust moon moon v2  v2 02  v2 v2 From v2  v2  2ad  a  f 0   . f 0 2y 2y 2y   v2    v2  So Fthrust  m gmoon      m gmoon   .   2y    2y (c) 2 2 2  9.8 m m   v   g v  2 15  F  m g   m Earth   12,000 kg s  s   28,000 N.  moon 2y  6 2y  6 2(160 m)      

F F 4-33. (a) v  ? From v  v  at with v  0 and a  net  v  t. f f 0 0 m f m kg m F 32 N  2 (b) v  t  (2.1s)  0.76 s ·s  0.76 m . f m 88 kg kg s

F F 4-34. (a) v =? From v  v  at (with v  0) and a  net  v  t. f f 0 0 m f m

© Paul G. Hewitt and Phillip R. Wolf 4-7 4-35. (a) The key concept to get is that the scale reads the normal force mRamon action on Ramon. That is, Fscale-on-Ramon  N. Since Ramon is at rest, the net force on him is zero, and N  mg. The scale reads mg. (b) Since the velocity of the elevator is constant the net force on mg N Ramon is zero, the scale reads mg. (c) N = ? Let’s call upward the positive direction. F N  mg From a  net =  N  ma  mg = m(a + g). m m (d) For parts (a) and (b) N  mg  70.0 kg 9.8 m  686 N.  s2  For part (c) N  m(a  g)  70.0 kg 2.0 m  9.8 m  826 N.  s2 s2 

4-36. (a) T = ? Since the elevator moves at a constant speed Fnet  T  Mg  0  T  Mg. T (b) T  Mg  (1960 kg) 9.8 m  19,200 N. (Note that the distance does not C  s2  Elevatorh figure into our solution.) a p Mgte 4-37. (a) The scale reads F  N. Calling upward the positive r scale-on-Maria m F N  W 1.2W  W 0.2mg Maria 2 direction, a  net     0.2g. m m m m w e (b) Scale reading  N  1.2W  1.2(550 N)  660 N. f mg N (c) Scale reading = ? Calling downward the positive direction, o F mg  N W a  net   N  mg  ma  m(g  0.2g)  (0.8g)  0.8W  0.8(550 N)  4u40 N. m m g e s 4-38. (a) N = ? During takeoff a normal force pushes upward on the astronaut, while gravity pulls down. The apparent weight of the astronaut is the magnitude of the normal force. We’ll take upward to be the positive direction.

Fnet N  mg v vf  v0  v  v From a    N  ma  mg  m(g  a). a    , so N  m g +  . m m t t t  t   v  (b) The scale reads the same as the normal force exerted on the astronaut  m g +  .  t   m   v m 162 s m (c) N  m  g    75 kg 9.8 2    1600 kg· 2  1600 N.  t   s 14s  s

(d) Before takeoff a = 0 and N = W  mg  75 kg 9.8 m  740 kg·m  740 N. Comparing the two,  s2  s2 the reading during blast-off is (1600 N)/(740 N) = 2.2 times the astronaut’s weight.

F F v v  v v F Ft m  ? From a  net  m  . Since a   f 0   m   . 4-39. (a) m a t t t v v t 

© Paul G. Hewitt and Phillip R. Wolf 4-8 (b) d = ? Since friction is negligible, after Tammy stops pushing the net force will be zero and the ice block will move in a straight line at constant speed. d From v   d  vt  2.4 m (3.0 s)  7.2 m. t s Fnet F 4-40. (a) a   . msystem m + M (b) With less force acting on the sail, the acceleration of the boat and occupant will be smaller. (c) If the boat were traveling at the same speed as the wind, the air behind the sail would be moving at the same speed as the sail and so would exert no force on it. The acceleration of the boat at that point would be zero.

4-41. (a) a = ? The force of the road on the truck’s tires acts to accelerate the system (truck + car). F F a  net  . msystem (m  M) (b) T = ? The tension in the chain provides the net force on the car. Fnet on the car T  F  From acar    T  ma  m  . mcar m  m  M 

 F   4800 N  m m (c) a      1.2 2 . T  ma  1100 kg 1.2 2  1300 N.  m  M   1100 kg  2900 kg s  s 

4-42. (a) The net force on the system is the weight mg of the hanging mass. The total mass of the system Fnet mg m is m + M . So a    g. msystem (m  M ) (m + M) m 2.20 kg (b) a  g  9.8 m  3.5 m . (m  M ) (2.20 kg  4.0 kg)  s2  s2 (c) This is the same expression above, except now M is the hanging mass and m is the mass on the M 4.0 kg table. a  g  9.8 m  6.3 m . (m  M ) (2.20 kg  4.0 kg) s2  s2

W 96.0 N m 4-43. gmoon  ? From W  mgmoon  gmoon    1.6 . m 60.0 kg s2

F 10.0 N 4-44. a  net   5.0 m . Note that the time does not enter into the calculation. m 2.0 kg s2

v v  v 0 m  9.0 m 4-45. a   f 0  s s  45 m . F  ma  (100 kg) 45 m  4500 N, about 1000 lbs. t t 0.2 s s2  s2 

m F v v  v  0  3.0  4-46. F  ? From a  net  F  ma  m  m f 0  5.0 kg s m t t 2.0 s  7.5 kg·m  7.5 N. s2 The magnitude of the stopping force is 7.5 N.

F m F 40 N m 4-48. a = ? From aM   F  Ma  8.0 kg 5.0  40 N. Now am    20 . M  s2  m 2.0 kg s2

 M  8.0 kg m m  Or, from F  MaM  mam  am  aM  5.0 2  20 2 .  m  2.0 kg s s 

F F 10.0 N m  ? From a  net  m    5.0 kg. 4-49. m a 2.0 m s2

4-50. m F v v  v  0  22.0  F  ? From a  net  F  ma  m  m f 0  2000 kg s  11,000 N. m t t 4.0 s The magnitude of the braking force is 11,000 N.

F F  f 4-51. f  ? From a  net   f  F  ma  12.0 N  (5.0 kg)(0.8 m )  8.0 N. m m s2 R 4-52. The weight (W=mg) acts downward, which we’ll call the positive F mg  R direction. R acts upward. So F  mg  R and a  net  . net m m  R  mg  ma  m(g  a)  50 kg 9.8 m  -6.2 m  800 kg·m  800 N.  s2  s2  s2 mg (Can you see that we’d get the same answer if we chose upward to be the positive direction?)

4-53. (a) Since the block is on a horizontal surface and not accelerating up of down so Fy =0  N  mg. (b) Since it takes a force P to get the block to just start to move, right before that instant the friction force equals the pull so Fx =0  f  P. f P (c)    . s N mg P 13 N     0.60. (d) s mg 2.2 kg 9.8 m  s2  f 4-54. (a)   . From F =0  N  mg. When the couch is just s N y f P about to start sliding F =0  f  P. So    . x s N mg

© Paul G. Hewitt and Phillip R. Wolf 4-10 (b) Now  P  m  M N  (m  M )g and P  f   N  (m  M )g P. new new s new  mg m P 180 N     0.57. (c) s mg 32 kg 9.8 m  s2  m  M (32 kg  72 kg) (d) = P  P  (180 N)  585 N. new m 32 kg

f 4-55. (a)   .The mass M starts to move when the tension in the string exceeds the maximum s N friction force = sN = sMg. The tension necessary to get M to move is between mg and m (m – 1 gram)  g or approximately mg. From f = T   Mg  mg    . s s M m 119 g (b)     0.46. s M 256 g N 4-56. (a) Note that the static friction force from the board on the brick provides the net force on the brick. Fnet fs m (b) a   . The block’s maximum acceleration fs m m occurs when the static friction force has its maximum  mg value   N   mg  a  s   g. s s max m s mg m m (c) amax  sg  0.60 9.8  5.9 .  s2  s2

4-57. (a)   ? The force diagram shows the four forces on the s fs book. When the book is at rest but on the verge of slipping, fs is at its maximum value, sN . F  0  f  mg  0  f  mg From y and from m P N Fx  0  P  N  0  P  N (Remember that, by definition, the normal force acts perpendicular to the surface, in this case horizontally.) f (maximum) mg mg   s  . s N P 1.2 kg 9.8 m mg  2  (b)   = s  0.34. s P 35 N

Direction of motion 4-58. (a) k  ? Since each block moves at a constant velocity, N F= 0. Pull and f have equal magnitudes, Pull acting

f m Pull © Paul G. Hewitt and Phillip R. Wolf 4-11

mg forward and f acting backward. Likewise, N = mg. f P So    . k N mg P 13 N     0.60. (b) brick-floor mg (2.20 kg) 9.8 m  s2  P 2.2 N P 5.5 N pine-floor    0.40. steel-floor    0.47. mg (0.56 kg) 9.8 m mg (1.20 kg) 9.8 m  s2   s2  (c) Their results would have been the same. The normal force would have been doubled, so the friction force and so the force required to pull the blocks across the floor at constant speed would have doubled too, but the ratio f/N would have remained the same.

f f 4-59. (a)   ? Since the box moves along a horizontal surface, N = W. So    . k k N W f 22 N (b)     0.39. k W 56 N

4-60. (a) k  ? Since the crate moves at a constant velocity, Direction of F= 0. P and f have equal magnitudes, P acting forward motion N and f acting backward. Likewise, N = mg. f P So    . k N mg f m P P 87 N So     0.28. (b) k mg 32 kg 9.8 m  s2  mg

4-61. (a) P = ? Since the box is moving in a straight line at constant speed the net force on it is zero. Then Fy  0  N  mg and Fx  0  P  f  k N  kmg. m m (b) T  kmg  0.31(25 kg) 9.8  76 kg·  76 N.  s2  s2

mg (b) Since Stan is moving in a straight line at constant speed the net force on him is zero. From the force diagram Fy  0  N  mg and Fx  0  T  f  k N  kmg. m m (c) P  kmg  0.60(82 kg) 9.8  480 kg·  480 N.Note that neither the mass of the truck nor  s2  s2 Stan’s speed enter into the solution. (d) Tension would remain the same. As in (c) above, Stan’s speed does not enter into the solution.

Direction of F P  f P   N P   mg 4-63. (a) P  ? From a  net   k  k motion N m m m m  ma  P  kmg

 P  ma  kmg  m(a  k g). f m P m m (b) P  m(a  kg)  18 kg 0.50  0.34 9.8  69 N.  s2  s2 

mg Direction of F P  f P   N P   mg motion 4-64. (a)   ? From a  net   k  k N k m m m m P  ma   = . k mg f m P 143 N  42 kg 1.3 m P  ma  s2  (b) k    0.21. mg m 42 kg 9.8 2  s  mg 4-65. (a) Friction is the net force acting on the calculator, opposite the direction of the calculator’s motion. f f N (b)    . What is f ? k N mg Direction of motion From the diagram,  f  ma  f  ma. What is a? m v 2 f From v 2  v 2  2ad  2ax  a  0 f 0 2x  v 2   0 f ma  2x  v 2 mg Therefore      0 . k mg mg g 2gx 2 2 1.8 m v0  s  (c) k    0.37. 2gx 2 9.8 m (0.45 m)  s2 

© Paul G. Hewitt and Phillip R. Wolf 4-13 4-66. (a) Friction is the net force acting on the car, opposite the direction of the car’s motion. f f N (b) k   . What is f ? N mg Direction of From the diagram,  f  ma  f  ma. What is a? motion v 2 f m From v 2  v 2  2ad  2ax  a  0 f 0 2x  v 2   0 f ma  2x  v 2 Therefore      0 . mg k mg mg g 2gx 1 hr 1000 m 2 2 62 km   v0  h 3600 s 1 km  (c) k    0.54. 2gx 2 9.8 m (28 m)  s2 

Direction of 4-67. (a) m = ? From motion F N a  net  ma  F  P  f  P   N  P   mg m net k k P m a   g  P  m  .  k  f m P a  k g P 14 N m    2.2 kg. (b) a  kg 0.40 m  (0.60) 9.8 m s2  s2  mg

4-68. (a) Direction of motion N

mg F P  f P   N P   mg (b) P  ? From a  net   k  k m m m m  ma  P  kmg  P  ma  kmg  m(a  kg). What is a? v2 From v 2  v 2  2ad with v  v  a  . f 0 f 2d 2 So P  m(a   g) becomes P  m v   g k 2d k  (c) 2  2.5 m  v2   s  m  P  m  kg  (49 kg)  0.52 9.8  280 N. 2d   2(5.0 m)  s2   

(b) k  ? Calling the crate’s direction of motion positive: f m P

Fnet P  f P  k N P  kmg P  ma From a      k  . m m m m mg Note that the acceleration a will be a negative number. mg

75 N  22 kg 0.80 m F  ma  s2  (c) k    0.43. mg 22 kg 9.8 m  s2 

4-70. (a) The kinetic friction force provides the N net force acting on the car. This force Direction of acts opposite the direction of motion motion of the car. The sign of this force will be negative. fk (b) v0  ? In the y-direction Fy  0  N  mg. In the x-direction, F  ma. So 2 F 2  f   mg 2 2 net vf nve0t k k and from vf  v0  2ax  a  2 . 2 2xvf  v0  2 mg So Fnet  ma becomes  kmg  m   v0  vf  2k xg.  2x 

2 m 2 m (c) v0  vf  2k xg  6.7  2(0.70)(26 m) 9.8  20 m/s.(This is 45 mi/h.)  s   s2 

4-71. (a) amax= ? The truck is speeding up. If the crate is not N N slipping that means that acrate= atruck and that the Direction of Direction of horizontal force accelerating the crate is static friction. motion motion (Note that here the static friction force acts in the direction of motion.) At the maximum non-slipping m f f m forward acceleration of the crate, F f  N  mg a  net  s, max  s  s   g. max m m m m s mg mg (b) amax= ? Again, if the crate is not slipping that means for 4-58(a) for 4-58(b) that acrate= atruck and that the horizontal force acting to slow down the crate is static friction. At the maximum non-slipping deceleration of the crate, F  f  N  mg a  net  s, max  s  s   g. m m m m s (c) At the instant the truck suddenly stops, the crate will be moving at speed v—the speed of the truck. As it slides along the truck bed the crate will slow somewhat. If the crate is immediately behind the cabin when the truck stops the crate will hardly slow at all, so the maximum possible crate impact speed is v.

4-73. s = ? The sandwich just starts to move when P just exceeds the maximum value of the static P 1.4 N P  f   N   mg      0.42. friction force. s, max s s s mg (0.34 kg) 9.8 m  s2 

4-74. P  ? Since the sled moves at a constant Direction of N velocity F= 0. P and f have equal motion magnitudes, P acting forward and f acting backward. Likewise, N = mg. f m P So P  f  k N  kW  0.050(2000 N)  100 N.

f mg 4-75. (a)   .The mass M starts to move when the tension in the string exceeds the maximum s N friction force = sN = sMg. The tension necessary to get M to move is between the weight of the 500- grams and the 600-grams. So m m m g   Mg  m g  500-g    600-g 500-g s 600-g M s M 0.50 kg 0.60 kg      0.42    0.50 1.2 kg s 1.2 kg s

f f N 4-76. k   . What is f ? N mg Direction of From the diagram,  f  ma  f  ma. What is a? motion

1 2 2(d  v0t) f m From d  v0t  at  a  2 t 2 24.0 m  2.9 m (1.6 s)  s  0.50 m . (1.6 s)2 s2 mg  0.50 m f ma a  2  Therefore      s  0.051. k mg mg g 9.8 m s2

© Paul G. Hewitt and Phillip R. Wolf 4-16 Fnet  fk k N kmg N 4-77. k  ? a  = = = =  kg. m m m m Direction of 2 2 2 v0 motion From vf  v0  2ad  a  . 2x f m 2 2 2 8.5 m v0 v0  s   kg   k    0.080. 2x 2gx m 2 9.8 2 (46 m)  s  mg

4-78. P =? Let’s draw a force diagram for the calendar. If the calendar is just about to slip (but still at rest) then f is at its maximum value, f

sN . From Fy  0  f  mg  0  f  mg and from F  0  P  N  0  P  N. x m P N fs mg From fs, max  sN  N  = s s (0.13kg) 9.8 m mg  s2  mg so P    2.5 N.  0.51 s N 4-79. Direction of motion v v F  f  N  mg t  ? From a   t  . a  x   k  k   g. t a m m m m k f m  6.3 m v vf  v0  s  So t     7.3 s. a kg (0.088) 9.8 m  s2  mg

4-80. a = ? The net external force on the system of the two masses is the difference between the gravitational force pulling down on the hanging mass and the friction force pulling back on the sliding mass.

Fnet on system mg  f mg  k N mg  kmg asystem     msystem 2m 2m 2m

 1 k   1 0.2    g    g  0.4g.  2   2 

4-81. (a) If each of the scales pulls at an angle, each can exert a vertical force component of 5 N for a total vertical force of 10 N. F = 10 N (b) For each scale we want an overall force of 10 N, but a F = 5 N. vertical component of 5 N. From the diagram,  y 5 N sin   0.5   sin1(0.5)  30º.Each scale 10 N 10 N 10 N needs to be oriented 30º above the horizontal as shown. The overall force diagram would look like this  30º 30º

© Paul G. Hewitt and Phillip R. Wolf 4-17 4-82. (a) T = ? Draw a force diagram. Since Gracie is a T T body in equilibrium, the tensions in the two ropes have to balance out her weight. m mg Gracie F  0  2T  mg  0  T = . y 2 (b) T = ? Draw a force diagram. Since Gracie is a body in equilibrium, the vertical components of mg the tension in the two ropes have to balance out (5-27(b)) her weight: Fy  0  2Ty  mg  0  2T sin  mg  0   mg T T T  T = . T y y 2sin   T T x x (c) For each of the vertical ropes, Gracie 55 kg 9.8 m mg  2  T   s  270 N.For each of 2 2 the ropes at 53º, 55 kg 9.8 m mg  2  T   s  340 N. 2sin 2sin53º mg

4-83. (a) P = ? At constant speed Fnet = 0, so R  P BOAT R  2P cos  R  0  P  . P 2cos R 150 N (b) P    81 N. 2 cos 2 cos22º

4-84. (a) f = ? Draw a force diagram for the mower. Since the mower N N moves at a constant velocity, the forces acting on the mower must  sum to zero. We can resolve F = Fcos f f  x Bronco’s push F into horizontal and vertical components and mg mg F = Fsin F redraw the diagram. Since the F y horizontal forces have to sum to zero: Fx  0  Fx  f  0  F cos  f  0  f = Fcos. (b) If the mower were accelerating we couldn’t assume that the horizontal forces would sum to zero. (c) f  F cos  210 N cos55º  120 N.

© Paul G. Hewitt and Phillip R. Wolf 4-18 y’ 4-85. (a)  = ? This is similar to sample problem 3. The W  W sin. N component parallel to the plane is P The component perpendicular to the plane is W  W cos. These two components are equal when 1 sin  cos  tan  1   tan (1)  45º.  Wsin (b)  = ? The perpendicular component would equal W Wcos x' when the box sits on level ground. That is, when  0º .  (Formally,W  W cos  W when cos 1   0º.) W (c)  = ? The parallel component would equal W when the slope is vertical. That is, when  90º . (Formally, W  W sin  W when sin 1   90º P .) (d) a = ? In each case the net force is the component of the weight parallel to the incline. In F W sin mgsin 45º the first case a  net    6.9 m . In the second case the component m m m s2 of force parallel to the incline is zero, so a  0. In the last case the parallel component is a  g  9.8 m . W=mg so s2

4-86. (a) T = ? Since the lantern is a body at rest the T sum of all of the forces on it must be zero in T T y   both the horizontal and the vertical   From F  0  T  mg  0 directions. y y T T T h x h mg  T cos  mg  0  T = .   cos mg mg (b) Th  ? From Fx  0  Th  Tx  0  Th  T sin  0  mg   T  T sin  sin  mg tan h   cos 

12 kg 9.8 m mg  s2  m (c) Th  mg tan  12 kg 9.8 tan 40º  99 N. T    150 N.  2  cos cos 40º s

© Paul G. Hewitt and Phillip R. Wolf 4-19 4-87. (a) T = ? T = ? Since the lamp is a body at 1 2 T rest the sum of all of the forces on it must   T 2y T 1 2 2 T T 2 be zero in both the horizontal and the T 1 1y  1    vertical directions. 1 2 T 1 2 T 1x 2x From Fx  0  T2x  T1x  0 cos1  T2 cos2  T1 cos1  T2  T1 cos2 

From Fy  0  T2y  T1y  mg  0 mg mg cos1  sin2   T2 sin2  T1 sin1  T1 sin2  T1 sin1  mg  T1 cos1  sin1  mg cos2   cos2    mg cos1  mg  cos1  mg   T1  ; T2  T1  = . cos2   sin    cos1tan2  sin1 2 cos2  sin2 + cos2tan1   cos1  sin1   cos2  

(Note the symmetry between the answers for T1 and T2. The answers are the same, with just the subscripts interchanged. It HAS to be this way. Deciding differently about how to label the angles can’t change the answer. The physics has to be the same!)

(b) 15 kg 9.8 m mg  s2  T1    84 N; cos1 tan2  sin1 cos 35º tan 55º  sin 35º 15 kg 9.8 m mg  s2  T2    120 N. cos2 tan1  sin2 cos55º tan 35º  sin 55º

N y’ 4-88. (a) v = ? The net force on Gracie will be the component of weight pointing parallel to the slope. So her acceleration will be F mgsin a  net   gsin . m m  2 2 mgsin From vf  v0  2ax with v0  0, a  gsin and x  L mgcos  x'  v  2gLsin. (b) mg v  2gL sin  2 9.8 m (4.2 m)sin 39º  7.2 m .  s2  s

© Paul G. Hewitt and Phillip R. Wolf 4-20 y' F N N 4-89. (a) a = net Three forces act on m . the descending saucer—gravity, f f the normal force, and friction. If mgsin we choose axes oriented parallel x' and perpendicular to the slope,  mgcos  then from mg mg Fy'  N  mg cos  N  mg cos. (mgsin   mg cos) F  mgsin  f  mgsin   mg cos  a  k  g(sin   cos). x' k m k

1 2 2L 2L (b) t = ? From x  v0t  2 at with v0  0 and x  L  t   . a g(sin  kcos) m m a  g(sin  k cos)  9.8 sin11º (0.081)cos11º 1.1 . (c) s2 s2 2L 2(55 m) t    10 s. a 1.1 m s2

4-90. (a) F = ? Since Sherry is moving up the slope at a constant y’ velocity, the sum of all of the N N forces acting on her (the pull, x' gravity, friction, and the F F normal force) must add to zero. As in inclined plane f k f sample problem (pages 34- mgsin k  35), we choose a set of axes parallel and perpendicular to   the slope and separate the mgcos gravitational force into mg mg components.

From Fy  0  N  mg cos  0  N  mg cos.

From Fx  0 and fk  k N  F  fk  mgsin  0  F  kmg cos  mgsin  mg(kcos  sin) (b) m F  mg(k cos  sin)  46 kg 9.8 (0.11cos20º  sin 20º )  200 N.  s2 

© Paul G. Hewitt and Phillip R. Wolf 4-21 y’ 4-91. (a) T = ? Since the log moves N N constant v x' up the slope at a constant velocity, T T the sum of all of the forces acting on it (the pull, gravity, friction, and the f normal force) must add to zero. As in k mgsin f  sample problem 3, we choose a set of k axes parallel and perpendicular to the   slope and separate the gravitational mgcos force into components. mg mg

From Fy  0  N  mg cos  0  N  mg cos.

From Fx  0 and fk  k N  T  fk  mgsin  T  kmg cos  mgsin  0  Tconstant v  kmg cos  mgsin  mg(kcos  sin). (b) T = ? If instead the log has acceleration a the y’-equations from part (a) are still the same. The x’-equation becomes Fx  max and fk  k N  T  fk  mgsin  max  T  fk  mgsin  max

 T  kmg cos  mgsin  max  mg(kcos  sin )  ax .

(Note that this is just Tconstant v + ma—enough pull to keep the log moving at a constant speed, plus the additional pull needed to accelerate it). (c) T = ? For the log moving at constant speed up the ramp, the tension m T  mg(k cos  sin)  48 kg 9.8 (0.30 cos27º  sin 27º )  340 N. For the log  s2  accelerating up the ramp at 0.4 m/s2,  m m  T  mg(k cos  sin)  ax  48 kg 9.8 (0.30 cos27º  sin 27º )  0.4  360 N.  s2 s2 

4-92. (a) T = ? Draw a force diagram showing all four forces acting on the box. The box will start to N N T Tsin T move when the horizontal component of the  pulling force exceeds the maximum static  Tcos m m friction force. That is, whenT cos  fs,max . f f Assuming that Tammy doesn’t pull hard enough to lift the box from the ground, F  0  N  T sin  mg  0 y mg mg  N  mg  T sin. Since

fs,max  sN, the box will start to move when T cos  s (mg  T sin).

We want to solve for T : T cos  smg  sT sin  T cos  sT sin  smg  mg  T  s . cos  ssin 0.32(23 kg) 9.8 m  mg  2  (b) T  s  s  69 N. cos  s sin cos20º  0.32sin 20º

© Paul G. Hewitt and Phillip R. Wolf 4-22 4-93. T = ? P = ? Draw a force diagram. Since your T nephew is a body at rest the sum of all of the forces T T y on him must be zero in both the horizontal and the   vertical directions. Let T be the combined tension T P x P in the two ropes. From Fy  0  Ty  mg  0 mg  T cos  mg  0  T  . mg cos mg From Fx  0  P  Tx  0  P  T sin  0  P  T sin. Plugging in numbers, 31 kg 9.8 m mg  2  T   s  330 N and P  T sin  (330 N)sin 23º  130 N. cos cos23º 4-94. The tension is divided equally between the two ropes. Since T mg 330 N T  2T  T  combined    165 N. combined individual individual 2 2 cos 2 4-95. f = ? Since the wagon moves at N N constant speed P Psin T F  0  P cos  f  0 net   Pcos  f  P cos  (91 N)cos21º  85 N f m f m

mg mg 4-96. a = ? Two forces act on the box—gravity y’ and the normal force. The gravitational N N force can be broken into two components —one parallel to the slope (=mgsin) and one perpendicular to the slope (=mgcos). mgsin The perpendicular component is balanced  x'  out by the normal force, so the net force on the box is mgsin). So mgcos mg mg F mgsin a  net   gsin  9.8 m sin16º  2.7 m . m m  s2  s2 y’

© Paul G. Hewitt and Phillip R. Wolf 4-23 N N 4-97. (a) N = ? Since the object is moving at P P f constant speed Fnet = 0. From the force f mgsin diagram N  mg cos  x'   (2.4 kg) 9.8 m cos27º  21 N.  2  mgcos s mg mg m (b) f  k N  kmg cos  0.33(2.4 kg) 9.8 cos27º  6.9 N.  s2 

(c) From Fx  0  P  fk  mgsin  P  kmg cos  mgsin  mg(k cos  sin)  (2.4 kg) 9.8 m (0.33cos27º  sin 27º )  17.6 N, a bit less than 18 N.  s2  y’ N N 4-98.  = ? When the box is just about to slip fs has its maximum value and Fnet = 0 fs fs mgsin From Fy'  0  N  mg cos.  x'  From Fx'  0  fs, max  mgsin. f mgsin mgcos   max   tan   tan-1  . mg mg s N mg cos s

4-99. (a) F = ? The box will start to move when the horizontal component of the N N pushing force exceeds the maximum F F Fsin static friction force. That is, when  Fcos  F cos  f s,max . f m f m From Fy  0  N  F sin  mg  0  N  mg  T sin. The box will start to move when F  f ; x s, max mg mg that is, when F cos  s (mg  F sin).

We want to solve for F : F cos  smg  sF sin  F cos  sF sin  smg smg  F  . cos  ssin (b) Note that as  gets larger, less of F is pushing the box forward and more of F is pushing the box into the ground, increasing the normal force and the friction force. At some point (where cos  s sin) you reach an angle where the box will not move no matter how hard you push on it.

1 sin 1 -1  1  From cos  s sin    tan    tan   . s cos s  s 