Question

A body of mass 2 kg rests on a rough inclined plane making an angle 30${}^{\circ }$
with the horizontal. The coefficient of static friction between the block and
the plane is 0.7. The frictional force on the block is:

• A. $9.8N$
• B. $0.7\times 9.8\times \sqrt{3}N$
• C. $9.8\times \sqrt{3}N$
• D. $0.7\times 9.8N$

Answer 1

Answer: (A)

$9.8N$

Since the body is at rest, the frictional force balances the component of gravitational force down the incline.

$\therefore mg\sin {30}^{\circ }-f=0$

$\Rightarrow f=mg\sin {30}^{\circ }=\left(2\right)\left(9.8\right)\frac{1}{2}=9.8N$

$\to QED.$