Help With Static Friction Problem?

Question

A 8.50 kg block (m1) and a 7.55 kg block (m2) rest on a horizontal frictionless surface with m2 resting on top of m1. The surface between the top and bottom blocks is roughened so that there is no slipping between the two blocks. A 42.6 N force is applied to the bottom block.

I determined the accelertation of the two block system to be 2.65 m/s^2, but help on how to solve these questions would be appreciated:

What is the force of static friction between the top and bottom blocks? (Friction is supplying the net force on the top block.)

What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block? (The frictional force is a function of the weight of the top block)

Answer 1

The acceleration of the system of the two blocks is:

a = F/(m1 + m2), a = 42.6/(8.5 + 7.55), a = 42.6/16.05, a = 2.65 m/s^2.

The moving force of the top block is the force of static friction Fr, so:

Fr = m2*a = 7.55*2.65 = 20.04 N

If n is the required minimum coefficient of static friction then:

Fr = n*N, where N is the force acted from the lower block to the upper block. But N = w2, where w2 is the weight of the upper block and w2 = m2*g (g = 9.81 m/s^2). So:

Fr = n*m2*g, n = Fr/(m2*g), n = 20.04/(7.55*9.81), n = 0.27

Answer 2

F = (m1+m2)a
a = 2,65m/s^2
It is expected that, according to the 3º Laws of mechanics of Newton, the force applied to the systems has a reaction of the same value and opposite side, Since both blocks move as a system, the friction force compensates the inertia of the second block on staying on the same place and not moving with block 1. So the force of static friction is m2*g*μs = m2*a
g*μs = a
μs = (2,65m/s^2)/(9,8m/s^2)
μs= 0,27
And therefore the force of friction on block 2 equals the impulse force on the opposite direction F = 7,55kg*2,65m/s^2= 20,0 N
Remember, the friction coefficient is an adimensional value!!