Two bodies 1 (at the top) and 2 (the rightmost one) with same masses are connected to each other with lightweight non-expandable thread thrown over sheave. The block A has an acceleration a. I need to find the minimal value of acceleration a, which allows bodies 1 and 2 to stay still relative to the block A. Coefficient of friction between both bodies and block A equals µ = 0,20.
In order to find this a I need to write down equations of Newton's 2nd law for every body. I'm struggling with friction for body 1 (the one that's on top of the block A).
On one hand we have thread tension force which pulls body 1 to the right. The friction that prevents the body being pulled to the right is pointed to the left. Ok, So far so good. But on the other hand block A is moving to the right. Since Body 1 pushes down on block A with it's weight the friction between them acts on block A which is pointed to the left (against direction of movement of block A). But according to 3rd Newton's law, the exact same force is acting on body 1 at the same time but pointed to the opposite direction - to the right.
Thus we have two vectors of force of friction seemingly pointed in opposite directions acting on body 1. What would Newton's second law equation look like for 1 body then? Will there be two vectors related to force of Friction or will there be just one resulting? If so, what will it be equal to? In what direction will it act?