To initiate an upward motion of a body along an inclined plane the minimum force required is twice the force required to keep the body at rest on the same incline . If the coefficient of friction is `mu` prove that the inclination of the plane is `theta = tan^(-1) (3mu)`.

Explanation

Let the mass of the body be m, and the minimum force required to keep the body at rest on the inclined plane be `F_(1)`. In this case the force of limiting friction f acts in the direction of the applied force.
`:. F_(1)`+f= mg sin `theta`
or, `F_(1) + mu "mg cos" theta = "mg sin " theta " ""or", F_(1) = "mg" (sin theta -mu cos theta ) `
If `F_(2)` is the minimum force needed to set the body in an upward motion then limiting friction acts downwards.
`:. F_(2) = "mg sin"theta +f= "mg sin"theta+mu "mg cos" theta`
= mg (sin `theta +mu cos theta) " " cdots (1)`
As `F_(2) = 2F_(1)` (given)
mg (sin `theta +mu cos theta ) = 2 mg (sin theta -mu costheta)`
or, sin`theta = 3 mu cos theta`
or, tan`theta = 3 mu " ""or", theta = tan^(-1) (3mu)` (Proved).