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 ona 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 doesnot slip . Take (g = 10 m//s^(2))

A block of mass 0.5 kg rests on a wedge of mass 2 kg as in Fig. 7.87 .

1.	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 ona 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 doesnot slip . Take (g = 10 m//s^(2))
Answer» SOLUTION :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 horizontaland vertical directions and solving, we get `F = (m + M) g [(mu "cos"theta + "sin" theta )/( "cos" theta - mu"sin" theta)] = 64.6 N` (MAX)