Two blocks with masses m 1 = 15.9 kg and m 2 = 60.5 kg, shown in the figure, are

Question

Two blocks with masses m1 = 15.9 kg and m2 = 60.5 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.52 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?

Two blocks with masses m1 = 15.9 kg and m2 = 60.5 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.52 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?

Explanation / Answer

first consider the horizontal forces
the force F is applied to m1
m1 applies a force N, say, to m2
m2 applies a reaction force -N to m1

use newtons law
ma = sum of forces on m
to each mass m1 & m2
you get

m1a = F - N.......(1)
m2a = N.............(2)
you aren't interested in a, you just need the normal force N in terms of F, so you need to eliminate a from the two equations
rearrange (1)
a = (F-N)/m1
substitute this in (2) gives
N = m2(F-N)/m1
now solve this for N, after some algebra you get
N = m2F/(m1+m2)

now that you know the normal force you can consider the vertical forces on m1
the friction force upwards on m1 is
msN
substitute N gives
msm2F/(m1+m2)
the downward force on m1 due to gravity is
m1g
the minimum force required is when the gravity and friction forces are equal
msm2F/(m1+m2) = m1g
so
F = m1(m1+m2)g/(msm2)
F= {15.9(15.9+60.5)*9.81}/0.52*60.5

=378.8N

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