The mass of the wedge shown in fig is `M = 4 kg` and that of block is `m = 1 kg` The horizontal surface beneath the wedge is smooth while the wedge and block is equal to `mu = 0.1` Taking `g = 9.8 ms^(-2)` and assuming pulley to be massless and friction . calculate maximum possible value of force `F` upto while the block will remain stationary relative to the wedge

Answer

10

Explanation

Since thread is pulled to the force `F` therefore tension in it is also equal to F. This tension acts on the block vertically upward Hence , it has the tendency to slip up along the vertical surface of the wedge hence the friction on the block will act vertical downward .
Since the horizontal force acting on the system on the wedge and block is F therefore the system will acceleration horizontally rigthward and block remain stationary ralative to wedge therefore the acceleration of block and the wedge is identical this acceleration `a = (F)/(m + M)`
Maximum value of force `F` will correpond to limiting friction hence `FBD` will be as shown in figure

For horizontal force on the block `N = ma`
`N = (mF)/(m + M)` or `N = (F)/(5)`
For vertical equilibrium of the block
`F = mu N + mg or F = 10 N`