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susus · Oct 28, 2010, 03:22 PM

Two blocks with masses m1 = 24.9 kg and m2 = 51.9 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.25 but the surface beneath m2 is frictionless.

What is the minimum force F required to hold m1 against m2?

ebaines · Oct 29, 2010, 06:19 AM

The static force between m1 and m2 is equal to μF, and this must equal the weight of m1 in order for m1 not to slide. Now, this force will cause the whole system to slide to the right, but that doesn't matter - as long as you have that force applied to block m1 it will not slide down.

harum · Oct 29, 2010, 02:21 PM

Originally Posted by ebaines

... this force will cause the whole system to slide to the right, but that doesn't matter ...

Nope. It does! To put it in layman terms, some of the applied force will be used to accelerate both masses. Here is the proof:

(1) m1*a = F - N
(2) m1*g = μ*N -- this is when F is minimum.
(3) m2*a = N

Here m1 and m2 are masses, a is acceleration of both of the masses, F is applied force, N is the support reactive force that accelerates mass m2 and creates friction between m1 and m2. Note that eqns (1) and (3) are for the horizontal projections of forces and (2) is for the vertical projection.

Plug N from (3) into (1) and (2), then solve (2) for "a" and plug it in (1). What I get is:

F = ( (m1+m2)*m1*g ) / (μ*m2)

As an extra check, imagine that m2 is infinity -- that is mass m1 is pushed against a wall -- then: F ~ (m1*g)/μ. Looks right!

susus · Nov 1, 2010, 10:41 AM

Thanks !