Question

A body of mass $10kg$ lies on a rough inclined plane of inclination $\theta ={37}^{\circ }$ with the horizontal. When the force of $30N$ is applied on the block parallel to and upward the plane, the contact force by the plane on the block is nearby along

• A. OA
• B. OB
• C. OC
• D. OD

Answer 1

The correct option is A OA

The given arrangement can be reframed as shown :


So, the force down the incline is :
$mg\sin \theta =10g\sin {37}^{\circ }=100\times \left(\frac{3}{5}\right)=60\text{N}$
This $60\text{N}>30\text{N}$ which is applied on the block externally. So, the block will tend to slide down the plane. Hence, friction force $f$ will act up the incline as shown in the figure below.

Thus, contact force by the plane on the block is given by :
$R=\sqrt{N^2+f^2}$
And, the direction will be near by along $OA$ as per given in the question.