A Weight `W` can be just supported on a rough inclined plane by a force either acting along the plane or horizontally. The angle of friction is `phi`. Find the angle of inclined plane and applied force.

Explanation

Let the angle of inclination of plane is `theta` and applied force is `F`.
When the force is acting along the plane
`_|_^(ar)` to the plane : `N = W cos theta`
`f = mu N = mu W cos theta`
Along the plane : `F + f = W sin theta`
`F = W sin theta - mu W cos theta`
`= W sin theta - tan phi W cos theta [ sincw mu = tan phi]`
` = W( sin theta - (sin phi)/(cos phi) cos theta)`
` = (W ( sin theta cos phi - cos theta sin phi))/(cos phi)`
`F = (W sin(theta - phi))/(cos phi)` (i)
`_|_^(ar)` to the plane : `N = W cos theta + F sin theta`
`f' = mu N = mu (W cos theta + F sin theta)`
Along the plane : `F cos theta + f' = W sin theta`
`F cos theta + mu(W cos theta + F sin theta) = W sin theta`
`F( cos theta + mu sin theta) = W ( sin theta - mu cos theta)`
`F( cos theta + (sin phi)/(cos phi) sin theta) = W (sin theta - (sin phi)/(cos phi) cos theta)`
`(F (cos theta cos phi + sin theta sin phi))/( cos phi) = (W sin ( theta - phi))/(cos phi)`
`F cos (theta - phi) = W sin (theta - phi)`
`F = (W sin (theta - phi))/(cos (theta - phi))` (ii)
Equating (i) and (ii) , we get
`cos (theta - phi) = cos phi`
`theta - phi = phi rArr theta = 2 phi`
`F = (W sin (theta - phi))/(cos phi) = (W sin phi)/(cos phi) = W tan phi`
Angle of inclined plane : ` theta = 2 phi`
Applied force : `F = W tan phi`