Question

A body of 6 kg rests in limiting equilibrium on an inclined plane whose slope is 30°. If the plane is raised to slope of 60°, then force(in kg-wt) along the plane required to support it is

(a) 3

(b) 2√3

(c) √3

(d) 3√3

Answer 1

Correct option: (b) 2√3

Explanation:

Let P be the frictional force required to support the body and µ be the coefficient of friction.

P = µR

Weight of the body(W) will be vertically downward.

W = 6g

(i) When the slope of the inclined plane is 30°

In this case,

On resolving 6g force along the plane and perpendicular to the plane

Under the equilibrium condition,

R = 6g cos 30° ...(i)

µR = 6g sin 30° ...(ii)

Dividing (ii) by (i)

(ii) When the slope of the inclined plane is 60°

Now,

Let a force P be applied to maintain equilibrium

On resolving 6g force along the plane and perpendicular to the plane

S = 6g cos 60° ...(iv)

P + µS = 6g sin 60°

From (iii) and (iv), we get