A block m 1 rests on a frictionless surface. A second block m 2 sits on top of t

Question

A block m1 rests on a frictionless surface. A second block m2 sits on top of the first block. Here m1 = m2 = m. A horizontal force F is applied to the bottom block to pull it to the right as shown below. Assume that the blocks move together across the surface (block m2 does not slide off of block m1).

(a) What is the magnitude of the acceleration of the blocks? (Use the following as necessary: m, g, and F.)
a =  

(b) What are the magnitude and direction of the static friction force acting on m1 due to m2? (Use the following as necessary: m, g, and F.)

(c) What are the magnitude and direction of the static friction force acting on m2 due to m1? (Use the following as necessary: m, g, and F.)

magnitude     f = direction     A block m1 rests on a frictionless surface. A second block m2 sits on top of the first block. Here m1 = m2 = m. A horizontal force F is applied to the bottom block to pull it to the right as shown below. Assume that the blocks move together across the surface (block m2 does not slide off of block m1). (a) What is the magnitude of the acceleration of the blocks? (Use the following as necessary: m, g, and F.) a = (b) What are the magnitude and direction of the static friction force acting on m1 due to m2? (Use the following as necessary: m, g, and F.) (c) What are the magnitude and direction of the static friction force acting on m2 due to m1? (Use the following as necessary: m, g, and F.)

Explanation / Answer

(a) Since the surface is frictionless, the only horizontal force acting on the system is F and the blocks move together, therefore their common acceleration would be F/(m1+m2) = F/(2m)

(c) The only force that can make the upper block move is friction between the blocks. The upper block is also accelerating at a rate of F/(m1+m2), therefore, the frictional force acting on it must be (m2)(F) / (m1+m2) in the forward direction, which is mF/2m = F/2 in the forward direction

(b)From Newton's third law of motion, the friction acting on m1 due to m2 should be a reaction to this. THerefore friction acting on m1 due to m2 must be (m2)(F) / (m1+m2) in the backward direction (towards left), which is F/2 in the backward direction

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