Question

A block of mass $m = 10 \, kg$ is lying on an inclined plane $PQ$ at $30^\circ$. The mass is restrained from sliding down by a force $F$ applied up the plane. The coefficient of friction is $\mu = 0.3$. Take $g = 10 \, m/s^2$. The smallest force $F$ (in N) required to prevent the block from sliding down is ............ (rounded off to one decimal place).

Hint:

On inclined planes, the minimum restraining force equals the downslope component of weight minus the maximum available frictional resistance.

Answer 1

Correct Answer: 23

Step 1: Forces along the incline.
Component of weight down the incline: \[ W_{down} = mg \sin 30^\circ = 10 . 10 . 0.5 = 50 \, N \] Normal force: \[ N = mg \cos 30^\circ = 100 . 0.866 = 86.6 \, N \] Friction (up plane, maximum): \[ f_{max} = \mu N = 0.3 \times 86.6 \approx 26.0 \, N \]
Step 2: Equilibrium condition.
To prevent sliding: \[ F + f_{max} = W_{down} \] \[ F = W_{down} - f_{max} = 50 - 26.0 = 24.0 \, N \] Final Answer: \[ \boxed{24.0 \, N} \]