Two equal masses (1 and 2) are connected by the rope across a pulley and placed on the larger block A as shown on the picture. Find the minimal acceleration a of the block A along the horizontal so that masses 1 and 2 remain stationary with respect to the block A. Coefficient of kinetic friction between each of the masses and block A is 'm', rope is massless and non-stretchable.

Question

anonymous · Answer

is there a picture? (-:

anonymous · Answer

http://i49.tinypic.com/14l0gmp.png

anonymous · Answer

do you have solutions to this @stargirlzilin , but what I could come up with until now includes too many variables to be a solution I guess.

anonymous · Answer

$$a = \frac{mg(1-m)}{M}$$

anonymous · Answer

I got a=g, which is very strange.

anonymous · Answer

$$ a= \frac{mg}{M}$$ is also something I could think of for two frictions, is a=g the solution from your book?

anonymous · Answer

I don't have the solution to this problem....unfortunately...

anonymous · Answer

It's strange that they mention nothing about the Mass of the block A, because obviously this is not the same mass as the little masses have, since since force = ma, hence I denoted a big M there