Mr. Rogers' AP Physics C: Friction Problems

Sliding Blocks Problem: Block 1 sits atop block 2. A force F = 2.5 M1 g is applied to block 1. Find the acceleration of both blocks. COF conditions are as shown at left.

Solution

First find the acceleration of block 2

S Fx = ma

Fd1 - Fd2 = M2a2

(M1gμs1) -  [(M1 + M2) ∙ g ∙ μs2] = M2a2

a2 = g [(M1s1 - (M1 + M2) μs2] / M2

Note: The problem only makes sense if
(M1s1 > (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 = M1a1

F - (M1gμs1) = M1a1

a1 =  [F - (M1gμs1)] / M1

Note: The problem only makes sense if
F > (M1gμs1) and a1 > a2

a1 =  [2.5 M1 g - (M1gμs1)] / M1

=  2.5 g - (0.8μs1)

=  1.7 g

Conclusion and Significance

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.

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