Question

A sphere of weight $W=100N$ is kept stationary on a rough inclined plane by a horizontal string AB as shown in figure. Find force of friction on the sphere.

• A. 26.8 N
• B. 30 N
• C. 75 N
• D. 50 N

Answer 1

The correct option is A 26.8 N

Balancing forces in the inclined-upward direction,

$Tcos{30}^0+f=Wsin{30}^0\to \left(1\right)$

Balancing torque about the center,

$T\times R=f\times R$, where $R$ is the radius of the sphere.

$⟹f=T$

From $\left(1\right)$, we get

$f(1+cos{30}^0)=Wsin{30}^0$

$⟹f=\frac{100\times 0.5}{1+0.866}=27.8N$

Option A is correct.