# Question #b93f4

Mar 28, 2017

(1) 100 N

#### Explanation:

Let the reactionary force exerted by each of the two fixed wall AB and AC on two ends of the rod (perpendicular to the plane of the walls) be $R$ N.

The line of action of each of these two forces will be inclined with the vertical line passing through the CG of the rod at an angle ${60}^{\circ}$

So the vertical component of each reactionary force will be $R \cos {60}^{\circ}$ and the total force of reaction (upward) will be $= 2 R \cos {60}^{\circ} = 2 R \times \frac{1}{2} = R$ N.

This total force of reaction will be just equal and opposite to the weight of the rod of mass 10kg.
So $R = 10 k g \times g = 10 k g \times 10 m \text{/} {s}^{2} = 100 N$

Hence the option (1) is the answer.