# Question #ed224

Jan 6, 2017

Option -A

#### Explanation:

Considering the equilibrium of forces acting on the beam we can say that the moment of the applied upward force $F = 5 N$ acting at P about the pivot O will be equal to the moment of the weight $W$ about O, but they will be oppositely directed.

So

$O Q \times W = O P \times F$

$\implies W = \frac{O P \times F}{O Q} = \frac{O P \times F}{O P + P Q} = \frac{2 \times 5}{\left(2 + 3\right)} = 2 N$