Question

Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:
(A) mA*g
(B) mB*g
(C) μs*mA*g
(D) μs*mB*g
(E) μs(mA +mB)g

Answer 1

Answer:

(E) μs(mA +mB)g

Explanation:

We can apply for mB:

∑ Fx = mB*a   (→)

⇒  Ffriction = mB*a   ⇒  a = Ffriction / mB = μs*N / mB

⇒ a = μs*(mB*g) / mB  ⇒   a = μs*g    (acceleration of the system)

Now, for mA we have

∑ Fx = mA*a   (→)

F - Ffriction = mA*a      ⇒  F = mA*a + Ffriction  

⇒     F = mA*(μs*g) + μs*(mB*g)   ⇒   F = μs*g*(mA + mB)

We must know that the friction acts only between the two blocks