Question

A body of mass $10\text{kg}$ lies on a rough inclined plane of inclination $\theta ={\sin }^{-1}\frac{3}{5}$ with the horizontal. When a force of $30\text{N}$ is applied on the block parallel to and upward the plane, the total reaction by the plane on the block is nearly along (take $g=10m{s}^{-2}$)

• A. $OA$
• B. $OB$
• C. $OC$
• D. $OD$

Answer 1

The correct option is A $OA$

Consider the following diagram,

$mg\sin \theta =10\times 10\times \frac{3}{5}=60\text{N}>30\text{N}$.

Thus, frictional force $\left(f\right)$ acts up the incline.

The normal reaction by the plane on the block is $N=mgcos\theta$ which acts along OB.

The total reaction by the plane on the block is the vector sum of the frictional force and the normal reaction by the plane on the block, i.e.;

$R=\sqrt{f^2+N^2}$

This acts along OA.