Solution

First find the acceleration of block 2

S Fx = ma

Fd1 - Fd2 = M2 ∙ a2

(M1 ∙ g ∙ μs1) -  [(M1 + M2) ∙ g ∙ μs2] = M2 ∙ a2

a2 = g [(M1)μs1 - (M1 + M2) μs2] / M2

Note: The problem only makes sense if

            (M1)μs1 > (M1 + M2) μs2

 

    = g [(M1)0.8 - (M1 + M1 / 2) 0.1] / (M1 / 2)

    = g [0.8 - (3 / 2) 0.1] / (1/ 2)

    = 2 g [0.8 - 0.0667

    = 1.47 g

Nest find the acceleration of block 1

S Fx = ma

F - Fd1 = M1 ∙ a1

F - (M1 ∙ g ∙ μs1) = M1 ∙ a1

a1 =  [F - (M1 ∙ g ∙ μs1)] / M1

Traditionally we tend to think of sliding friction as a force that resists motion. However, this problem clearly demonstrates that sliding friction can also cause motion.

Note also that the force F plays no part in accelerating block 2.

Metacognition Question: Is the acceleration of the top block higher than the one below? Obviously, the problem only makes sense if the top block slides further than one below it.