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2 Particle Dynamics
2.4 (a) The force of contact F c between the blocks is equal to the force exerted on
m 2 :
F c = m 2 a
(1)
where the acceleration of the whole system is
a =
F
m 1 + m 2
(2)
∴ F c =
m 2 F
m 1 + m 2
(b) Here the contact force F
c is given by
F
c = m 1 a =
m 1 F
m 1 + m 2
Notice that F
c = F c simply because m 1 = m 2 .
2.5 (a) When the box is dragged, the horizontal component of F is F cos θ and the
vertical component (upward) is F sin θ as in Fig. 2.19. The reaction force
N on the box by the floor will be
Fig. 2.19
N = mg − F sin θ
(1)
(b) The equation of motion will be
ma = F cos θ − μN = F cos θ − μ(mg − F sin θ)
∴ a =
F
m
(cos θ + μ sin θ) − μg
(2)
(c) When the box is pushed the horizontal component of F will be F cos θ and
the vertical component F sin θ (downwards), Fig. 2.20. The reaction force
exerted by the floor on the box will be
N
= mg + F sin θ
(3)
which is seen to be greater than N (the previous case).
2 Particle Dynamics
2.4 (a) The force of contact F c between the blocks is equal to the force exerted on
m 2 :
F c = m 2 a
(1)
where the acceleration of the whole system is
a =
F
m 1 + m 2
(2)
∴ F c =
m 2 F
m 1 + m 2
(b) Here the contact force F
c is given by
F
c = m 1 a =
m 1 F
m 1 + m 2
Notice that F
c = F c simply because m 1 = m 2 .
2.5 (a) When the box is dragged, the horizontal component of F is F cos θ and the
vertical component (upward) is F sin θ as in Fig. 2.19. The reaction force
N on the box by the floor will be
Fig. 2.19
N = mg − F sin θ
(1)
(b) The equation of motion will be
ma = F cos θ − μN = F cos θ − μ(mg − F sin θ)
∴ a =
F
m
(cos θ + μ sin θ) − μg
(2)
(c) When the box is pushed the horizontal component of F will be F cos θ and
the vertical component F sin θ (downwards), Fig. 2.20. The reaction force
exerted by the floor on the box will be
N
= mg + F sin θ
(3)
which is seen to be greater than N (the previous case).