Calculate angle of friction between wedge and block system is at rest M coefficient of friction between wedge and block.

Explanation

Let a be the common acceleration .
`F= (M +ma)" "` …(A)
Let us solve this question in the f.o.r of wedge , which is non- inertial .
When `F= F_(max)`
The tendency of block is to move up the plane of wedge . Hence friction will act down the plane of wedge .
From (A) ,
`F_(max) = (M+m)a_(max) " "` ...(i)
As the block is stationary on the wedge,
` sum F_(y)=0`
`rArr N = m (gcos theta + a_(max)sin theta) " "` ...(ii)
sum` F_(x) = 0 `
`rArr ma_(max) cos theta = muN + mgsintheta" " ` ...(iii)
From (i) and (iii)
`ma_(max) cos theta = mu m (gcos theta + a_(max) sin theta) + mg sin theta`
or `a_(max) [ cos theta - mu m sin theta] = g [mu cos theta + sin theta]`
` or a_(max) = (sintheta + mu cos theta)/(cos theta + mu sin theta)g`
or ` a_(max) = ((tan theta + mu)/(1-mu tantheta))g`
From (i) ` F_(max) = (M+m) ((tantheta + mu)/(1-mu tantheta))g`
When F = `F_(min)`
The tendency of block is to move downcard along the plane . Hence frictional force will act in upward direction .
From (A)
` F_(min) = (M + m) a_(min) " " `...(iv)
` sumF_(y) = 0 `
` rArr N = mg cos theta+ ma _(min) sin theta" " `...(v)
` sum F_(x) = 0`
` rArr mgsintheta = muN+ma_(min) cos theta " " `...(vi)
From (v) and (vi) ,
` mu (mg cos theta + ma_(min) sin theta) + ma_(min)cos theta = mg sin theta `
` rArr _(min) (cos theta + mu sin theta) = g (sin theta - mu cos theta)`
` rArr a_(min) = (sin theta - mu cos theta)/(cos theta +mu sin theta)g`
or ` or a_(min) = ((tan theta - mu)/(1+ mu tan theta))g`
From (iv),
` F_(min) = (M+m) ((tan theta - mu)/(1+mu tan theta ))g`
When `F= F_(min)`
The tendency of block is to move downward along the plane . Hence frictional force will act in upward direction.