5. Since the sled accelerates at a constant rate ax ˆi in the positive x direction, ax can be
found using the kinematic equations (Table 2-1). Solving v2 x  v1 x  ax t for the case
where it starts from rest, we have ax  v2 x t .
The velocity component is v2 x = (1600 km/h) (1000 m/km)/(3600 s/h) = 444 m/s, so
F net  Fxnet  m ax  500 kg
444 m s
 1.2  105 N.
1.8 s
23. (a) The links are numbered from bottom to top. The forces on the bottom link are the
force of gravity mg , downward, and the force F21 y of link 2, upward. Take the positive
direction to be upward. Then Newton’s second law for this link is F21  mg  ma . Thus
F21  m(a  g)  (0.100 kg) 2.50 m / s2 ˆj + 9.8 m / s 2 ˆj = 1.23 N ˆj and so the magnitude of
the force on link 1 from link 2 is 1.23 N.
(b) The forces on the second link are the force of gravity mg , downward, the force F12 of
link 1, downward, and the force F32 of link 3, upward. According to Newton’s third law
F12 has the same magnitude but opposite direction as F21 . Newton’s second law for the
second link is
F32  F1 2  mg  ma 


F32  m(a  g)  F1 2 = (0.100 kg) 2.50 m / s2 ˆj + 9.8 m / s 2 ˆj + 1.23 N ˆj = 2.46 N ˆj


and so the magnitude of the force on link 2 from link 3 is 2.46 N.
(c) Newton’s second law for link 3 where, according to Newton’s third law, F23 y has the
same magnitude but opposite direction from F32 y is
F4 3  F23  mg  ma 


F4 3  m(a  g)  F23 = (0.100 N) 2.50 m / s2 ˆj + 9.8 m / s 2ˆj + 2.46 N ˆj = 3.69 N ˆj
so the magnitude of the force on link 3 from link 4 is 3.69 N.
(d) Newton’s second law for link 4 where according to Newton’s third law F34 y has the
same magnitude but opposite direction as F43 y is



F54  F34  mg  ma 


F54  m(a  
g)  F34  (0.100 kg) 2.50 m / s 2 ˆj + 9.8 m / s2 ˆj + 3.69 N ˆj = 4.92 N ˆj
so the magnitude of the force on link 4 from link 5 is 4.92 N.
(e) Newton’s second law for the top link where F45 y and F54 y have equal magnitude but
opposite directions is

F  F4 5  mg  ma 


F  m(a  g)  F45  (0.100 kg) 2.50 m / s2 ˆj + 9.8 m / s 2ˆj + 4.92 N ˆj = 6.15 N ˆj
so the magnitude of the force on the top link is 6.15 N.
(f) Each link has the same mass and the same acceleration, so the same net force acts on
each of them: F net  ma  (0.100 kg)(2.50 m / s2 )ˆj = 0.25 N ˆj .
39. We use the notation gc as the acceleration due to gravity near the surface of Callisto,
m as the mass of the landing craft, ay as the acceleration of the landing craft, and F as the
rocket thrust. We take down to be the negative y direction. Thus, Newton’s second law in
component form takes the form Fy – mgc = may. If the thrust is F1 y (= 3260 N), then the
acceleration is zero, so F1 y – mgc = 0 N. If the thrust is F2 y (= 2200 N), then the
acceleration is a2 y (= – 0.39 m/s2), so F2 y = mg – mgc = ma2 y.
(a) Equation 3-7 gives the weight of the landing craft: mgc = F1 y = 3260 N.
(b) The second equation above gives the mass:
m
F2 y  mgc
a2 y

2200 N  3260 N
 2.7  10 3 kg .
 0.39 m/s 2
(c) The weight divided by the mass gives the acceleration due to gravity:
gc = (3260 N)/(2.7  103 kg) = 1.2 m/s2 so its magnitude is 1.2 m/s2.