A. Consider the two blocks in the figure, with known masses. Initially they are

Question

A. Consider the two blocks in the figure, with known masses. Initially they are resting on a surface when a known horizontal force F pointing to the right is applied to block of mass m1. Find the acceleration of the blocks and the contact forces F12 and F21, as a function of m1, m2 and g.

B. Now consider the that the known horizontal force F is applied to block of mass m2, and is pointing to the left. Find the acceleration of the blocks and the contact forces F12, and F21, as a function of m1, m2, and g. Why are the contact forces found in part A different than in part B

Explanation / Answer

A) Applying Newton's 2nd law to m1 in the horizontal direction, F12-F21=m1a ........(1)

Applying Newton's 2nd law to m2 in the horizontal direction, F21=m2a ...................(2)

Since the blocks are in contact,the accelerations are the same.Eliminating a from this pair of equations,

F21=m2F12/(m1+m2)

substituting this value in equation (2),

therefore, a=F12/(m1+m2)

B) Applying Newton's 2nd law to m1 in the horizontal direction, -F21=-m1a .......(3)

Applying Newton's 2nd law to m2 in the horizontal direction, F21-F12=-m2a .......(4)

Since the blocks are in contact,the accelerations are same.Note that I have explicitly put in my expectation that the acceleration will be to the left.Eliminating a from this pairs of equations,

F21=m1F12/(m1+m2)

substituting this value in equation (3)

a=F12/(m1+m2)

Consider m2<<m1, for the top situation then contact force is then very small,not much force is required to accelerate the small mass m2.For the bottom situation F21 equivalent to F12,most of the mass is in m1,so most of the force required to accelerate.

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