A block of mass `0.5` kg rests on a wedge of mass 2 kg as in Fig. 7.87 . The wedge is acted on by a horizontal force F and slides on a frictionless surface . If the coefficient of static friction between the wedge and the block is `mu_(s) = 0.8` , and the angle of inclination is `30^(@)` , find the maximum and minimum values of F for which the block does not slip . Take (g = `10 m//s^(2)`)

Answer

64 N , 31.8 N

Explanation

Writing force equations in horizontal and vertical directions respectively,

`F + N "sin" theta - f"cos" theta = ma " " ……. (i)`
N cos` theta + f "sin" theta = mg " " …… (ii) `
`a= (F)/(m + M) " " ....... (iii) `
and `" " f = mu_(s) N " " ........ (iv)`
Solving , `" " F = (m + M) g [(mu "cos"theta - "sin" theta )/(mu "sin" theta + "cos" theta)]`
= `3. 82 N ` (min)
If we increase F , the direction of friction will get reversed then again writting force equation in horizontal and vertical directions and solving , we get `F = (m + M) g [(mu "cos"theta + "sin" theta )/( "cos" theta - mu"sin" theta)] = 64.6 N` (max)