# Serway 9th ed – sliding friction and problem

1. The problem statement, all variables and given/known data

I am absolutely stumped on a problem from a trig based college physics text by Serway et. al. 8th edition. This is exercise 4.14 on page 107. The example attached to it is a 10.0 kg block M, on top of which rests a 5.0 kg block m. The coefficient of friction between the two blocks is 0.350 and the coefficient on the ground is zero (the 10.0 kg block has no friction on the ground surface). The example calculates the maximum force (tension T) that can be applied to a cable attached to the 10 kg block that will not cause the 5 kg one on top to slip. The answer is 51.5 N, and I got that part.

The exercise that follows has the same set-up, but now the cable is attached to the 5 kg block on top (see attached image), and it asks for the maximum force that can be applied to the top block without causing it to slip. The answer in the text is 27.715 N. I reasoned that the force cannot exceed the force of friction (17.5 N) between the blocks and this would pull the two blocks across the frictionless ground surface.

2. Relevant equations

$F_s = μ_s mg$

3. The attempt at a solution

Some reverse engineering got me

$\frac{M+m}{M}F_s = 27.715 N$

This is consistent with the answer in the text, but is it in fact the answer? How do I get there? If it is not the answer what is it? I don’t even know where to start.

Regards
David C.

Attached Images ex-4.14-p-107-03.jpg (16.1 KB)

http://ift.tt/M2J49z