Question

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

Options

• `2/sqrt(3)` kg

• `2sqrt(3)` kg

• `4/sqrt(3)` kg

• `4sqrt(3)` kg

Answer 1

`2sqrt(3)` kg

Explanation:

Let p be the force required to support the body and `mu` be the coefficient of friction.

Case - I: When plane make inclination of 30°

In this case R = 6 g cos 30°

`mu` R = 6 g sin 30° ......[Limiting equilibrium]

∴ `mu` = tan 30° = `1/sqrt(3)`

Case - II: When plane raised to the slope of 60°

In this case.

S = 6 g cos 60° and P + `mu` s = 6 g sin 60°

∴ P + `1/sqrt(3)` (6g cos 60°) = 6 g sin 60°

P = `6  g(sqrt(3)/2 - 1/(2sqrt(3))) = 2sqrt(3) g = 2sqrt(3)` kg wt.