Two block , with masses `m_(1)` and `m_(2)` are staked as shown in fig and placed on a frictionless horizontal surface . There is a friction between the two block .An external force of magnitude `F` is applied to the top block at an angle `alpha` below the horizontal . The coefficient of friction between `m_(1)` and `m_(2)` are `mu_(s)`
b. If the two blocks move together, find their acceleration
b . Calculate the maximum value of force so that blocks will move together

Answer

a.`(F cos alpha)/(m_(1) +m_(2))` b. `(mu_(2)(m_(1) +m_(2))m_(2)g)/([m_(2) cos alpha - mu_(s)(m_(1) +m_(2)) sin alpha)]`

Explanation

a. The only horizontal force on the two - block combination is the horizontal copmonent of `barF, bar F cos alpha`. The block will acceleration
with `a = (F cos alpha)/((m_(1) + m_(2))`
b. The normal force between the block is `m_(2) g + F sin alpha` for the block to move together the prodect of this force and `mu_(s)` must be greater than the horizontal force that the lower block exerts on the upper than block .That horizontal force is one of an acting reaction part , the reaction to this force acceleration the lower block . Thus for the blocks to stay together
`m_(2) a le mu_(s) (m_(1)g + F sin alpha)`.
Using the result of part (a) we get
`m_(2) (F cos alpha)/(m_(1) +m_(2)) le mu_(s) (m_(2)g + F sin alpha)`
Solving the inequality for F given the desired result.