In the given the wedge is acted upon by a constant horizontal force 'F'. The wedge is moving on a smooth horizontal surface A ball of mas 'm' is at rest relative to the wedge The ratio of forces exerted on 'm' by the wedge when 'F' is acting and 'F' is withdrawn assuming no friction between the edge and the ball is equal to .

Explanation

When force F is applied
`N_(1) = "mg cos" theta + F "sin" theta`
`(F "cos" theta = "mg sin" theta implies F = "mg tan" theta )`
If F = 0 , `" " N_(2) = "mg cos" theta , " " (N_(1))/(N_(2)) = 1 + (F"sin" theta)/("mg cos" theta)`

`(N_(1))/(N_(2)) = 1 + ( "mg tan" theta "sin" theta)/( "mg cos" theta) = 1 + tan^(2) theta = sec^(2) theta`
Two blocks of mass m and M are placed on a rough inclined plane as shown , when `(theta gt alpha)`.
(i) Minimum value of M for which m slides upwards is `M = m("sin" theta + mu_(s) "cos" theta)`

(ii) Maximum value of M for which m slides downwards :
`M = m ("sin" theta - mu_(s) "cos" theta)`
A body is released from rest from the top of an inclined plane of length 'L' and angle of inclination `'theta'`. The top of plane of length `(L)/(n) (n gt 1)` is smooth and the remaining part is rough . if the body comes to rest on

reaching the bottom of the plane then find the value of coefficient of friction of rough surface .
For smooth part :
Using `" " v^(2) - u^(2) = 2as , V^(2) = 2a_(1) (L)/(n)`
`a_(1) = " g sin" theta , a_(2) = g ("sin" theta - mu "cos" theta)`
For rough part `" " 0 - V^(2) = 2a_(2)((n-1)/(n))L`
`2a_(1)(L)/(n) = - 2a_(2) ((n-1)/(n))L`
`"g sin" theta = -g["sin" theta - mu "cos"theta] (n-1)`
`mu = "tan" theta((n)/(n-1))`
A body is pushed down with velocity 'u' from the top of an inclined plane of length 'L' and angle of inclination '`theta'` . The top of plane of length `(L)/(n)(n gt 1)` is rough and the remaining part is smooth . If the body reaches the bottom of the plane with a velocity equal to the initial velocity 'u' , then the value of coefficient of friction of rough plane is `mu_(k) = n ("tan" theta)`.