4 NEWTON’S LAWS OF MOTION 4.9 Frames of reference
the upward acceleration a of the platform is
2 T − W
a =.
m
Since T = F and m = W/g, we obtain
W (a/g + 1)
F =.
2Finally, given that W = 34 N and a = 3.2 m/s^2 , we have
34 (3.2/9.81 + 1)
F = = 22.55 N.
2
Worked example 4.4: Suspended block
Question: Consider the diagram. The mass of block A is 75 kg and the mass
of block B is 15 kg. The coefficient of static friction between the two blocks is
μ = 0.45. The horizontal surface is frictionless. What minimum force F must be
exerted on block A in order to prevent block B from falling?
FAnswer: Suppose that block A exerts a rightward force R on block B. By New-
ton’s third law, block B exerts an equal and opposite force on block A. Applying
Newton’s second law of motion to the rightward acceleration a of block B, we
obtain
R
a = ,
mB
where mB is the mass of block B. The normal reaction at the interface between
the two blocks is R. Hence, the maximum frictional force that block A can ex-
ert on block B is μ R. In order to prevent block B from falling, this maximum
(^) B
A