Consider the situation shown in the figure . The coefficient of friction between the blocks is `mu`. Find the minimum and the maximum force `F` that can be applied so that two blocks move together.

Explanation

`F= (M + m)a`
`a = (F)/((M + m))`
Now consider the block of mass `m` , if this block has endency to move down , for limiting equilibrium , `f = mu N`.
`uarr : N cos theta + f sin theta = mg`
`N cos theta + mu N sin theta = mg` (i)
`rarr : N cos theta - f cos theta = ma `
`N sin theta - mu N cos theta = ma` (ii)
Dividing (ii) by (i) , we get
`(a)/(g) = (sin theta - mu cos theta)/(cos theta + mu sin theta) = (tan theta - mu)/( 1 + mu tan theta)`
`a_(min) = ((tan theta - mu)/(1 + mu tan theta))g`
When large force is applied , the block has tendency to move up , the friction will be in downward direction.
Solving as above
`a_(max) = ((tan theta + mu)/(1 - mu tan theta))g`
Thus acceleration lies between
`((tan theta - mu)/( 1 + mu tan theta))g` and `((tan theta + mu)/( 1 - mu tan theta))g`
`F_(min) = (M + m) ((tan theta - mu)/(1 + mu tan theta)) g`
`F_(max) = (M + m) ((tan theta + mu)/(1 - mu tan theta))g`