(1) F  =  m a  =  60 kg×1.15 m/s²  =  69 N
(3) F  =  9 ×10^-3 ×10⁴×9.8 m/s²  =  880 N
(5) (a) F_g,E  =  mg  =  66 kg×9.8 m/s²  =  650 N
(b) F_g,M  =  66 kg×1.7 m/s²  =  110 N
(c) F_g,Mars  =  66 kg×3.7 m/s²  =  240 N
(7) a  =  v/t  =  (90,000/3600/8)m/s²  =  3.13 m/s²
F  =  m a  =  1100 kg ×3.13 m/s²  =  3400 N (direction opposite the velocity)
(9) The tension force is T  =  m_min (a + g), so m_min  =  T/(a + g)  =  22 N/[(4.5 + 9.8)m/s²]  =  1.5 kg
(17) a_max  =  T_max/m - g  =  (21750/2100 - 9.8)m/s²  =  0.56 m/s²
(21) v  =  (2 g h)^1/2  =  9.39 m/s
a  =  v²/(2d)  =  9.39²/(2·0.7)m/s²  =  63 m/s²
F_avg  =  m a  =  45 kg×63 m/s²  =  2800 N
This force adds to his weight of 440 N to give a total of 3300 N
(23) The ideal (frictionless) pulley simply changes the direction of the tension force (equal to the weight of the hanging mass). In all cases, the normal force of the table is n  =  70 N - W
(a) 70 N - 30 N  =  40 N, (b) 70 - 60  =  10, (c) zero, since 90  >  70
(27) (a) F_1x  =  -10.2 N,  F_1y  =  0, F_2x  =  0, F_2y  =  -16.0 N
F_x  =  F_1x + F_2x  =  -10.2 N, F_y  =  F_1y + F_2y  =  -16.0 N
F  =  (F_x² + F_y²)^1/2  =  19 N
tan^-1(F_y/F_x)  =  tan^-1(-16/-10.2)  =  57.5^°
but in 3rd quadrant, so angle of force is 180 + 57.5  =  237^°
a  =  F/m  =  19 N/27 kg  =  0.70 m/s², same direction as the force.
(b) F_1x  =  10.2 N cos(-30^°)  =  8.83 N
F_1y  =  10.2 N sin(-30^°)  =  -5.10 N
F_2x  =  0, F_2y  =  16 N, F_x  =  F_1x + F_2x  =  8.83 N
F_y  =  F_1y + F_2y  =  10.9 N, F  =  (8.83²+10.9²)^1/2 N  =  14. N
tan^-1(10.9/8.83)  =  51^°, a  =  14/27 m/s²  =  0.52 m/s², same direction as the force.
(31) (a) F  =  (m_c + m_h)(a + g)  =  (6500 + 1200)(0.6 + 9.8) N  =  8.01×10⁴ N
(b) T  =  m_c(a + g)  =  1200(0.6 + 9.8) N  =  1.25×10⁴ N
(33) For one car, T₁  =  a×m, for two cars T₂  =  a×(2m),
(where a is the acceleration of the system; i.e., any of the cars).
So the tension is twice as great in the coupling between the locomotive and the first car as compared to that between the first and second car. Note: You should draw a free body diagram for each car to better understand this result.
(35) All vectors (F's and a) are 1-D, direction in algebraic sign.
Newton's 2nd law applied to each block gives:
F + F₁₂  =  m₁ a,  F₂₁ + F₂₃  =  m₂ a, F₃₂  =  m₃ a
where the subscript ij means ``on i by j''. Newton's 3rd law gives: F₂₁  =  -F₁₂, F₂₃  =  -F₃₂
The last 2nd law expression thus becomes: F₂₃  =  -m₃ a, and the 2nd expression becomes: -F₁₂ + F₂₃  =  m₂ a
Thus F₁₂  =  -(m₂ + m₃) a, which combines with the first to give
F - (m₂ + m₃) a  =  m₁ a. Rearranging yields F  =  (m₁ + m₂ + m₃) a
which is the (b) result.
Likewise, the earlier expressions give for the net force on individual blocks: (c) F₃  =  F₃₂  =  m₃ a, F₂  =  F₂₁ + F₂₃  =  (m₂ + m₃) a - m₃ a  =  m₂ a
F₁  =  F + F₁₂  =  (m₁ + m₂ + m₃) a - (m₂ +m₃) a  =  m₁ a
(d) The force between blocks are:
F₁₂  =  -(m₂ + m₃) a
and F₂₃  =  -m₃ a
(e) a  =  96 N/(3×12 kg)  =  2.67 m/s²
F₁  =  12×2.67  =  32 N,  F₃  =  F₂  =  F₁  =  32 N
|F₁₂|  =  2×12×2.67 N  =  64 N, F₂₃  =  32 N
(41) m_s  =  f_max/(mg)  =  8 N/(2 kg·9.8 m/s²)  =  0.41
(only if the sides and top of drawer are not touching, the usual cause for spilled contents-this problem is not realistic!)
(43) T  =  2×9.8 N  for ideal pulley  =  m_s m g, m_s  =  0.3
m  =  2 kg×9.8/(0.3×9.8)  =  6.7 kg
(49) a  =  m g sinq, a_fr  =  m g sinq - m_k m g cosq
v²  =  2 a x, so v µ a^1/2, v_fr  =  v/2 ® a_fr  =  a/4
Thus sinq/4  =  sinq - m_k cosq
or m_k  =  3 tanq/4  =  0.4
(67) (a) a  =  F/m  =  f_s,max/m  =  m_s m g/m
Thus m_s,min  =  a/g
(b) m_s,min  =  4/9.8  =  0.41, so the chair would slide since 0.25  <  0.41.

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File translated from T_EX by T_TH, version 1.95.
On 10 Sep 1999, 11:20.