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I am trying to understand this question, but I keep coming to a dead end:

Two identical rectangular masses rest on top of one another, the top is tied to a wall on the left, while the one on the bottom has a roped attached to it on the right side. What is the max force that can be applied to the rope on the bottom before the block will begin to slide out from underneath the top one. (the coefficient of friction is equal between the block and the ground, and the bottom block with the top block).

Ok, so I am pretty sure this question is just asking me to find the net force on the bottom block, as applying a force larger then this will result in an acceleration of the lower block:

$$ F_{net}=ma$$ $$ 2F_f=ma$$ $$2F_n\mu=ma$$ $$2mg\times2\mu=ma$$

Ok, I am totally lost as to how solve this question. My intention is to find the net force acting on the lower block. I am pretty sure the the acceleration will be zero, resulting in the right side of the equation equaling zero. But I am not sure how to handle the coefficient of friction.

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2 Answers 2

up vote 2 down vote accepted

$f$ is the coefficient of friction, right?

Each block weighs $mg$ right?

If you are going to pull out the bottom block, the total friction is the friction on its top plus the friction on its bottom, right?

So what is the friction on its top? The weight of one block times $f$, right?

What is the friction on its bottom? The weight of two blocks times $f$, no?

You take it from there.

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It may help to explicity include the tension $T$ applied to the rope attached to the bottom mass, and also to note that for static friction, the "no-slip" condition is that $$F_f \leq \mu_s F_N$$ so you will need to include an inequality.

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