A non uniform beam of weight 120 N pivoted at one end is shown in the diagram below. Calculate the value of F to keep the beam in equilibrium.

Given,

Weight (W) = 120 N

From the figure,

Distance of C.G. from the pivot point O = r = 0.20 m

Distance of F from the pivot point O = R = 0.80 m

Now,

Moment of force about pivot point O due to C.G. = r x W = 0.20 x 120 = 24 Nm (clocKwise direction)

and

Moment of force about pivot point O due to F = R x F = 0.80 x F (anticlocKwise direction)

For equilibrium,

Anticlockwise moment of force due to F about pivot point O = Clockwise moment of force due to C.G. about pivot point O

$$\begin{aligned}0.80 \times \text F = 24 \\[1 em] \Rightarrow \text F = \frac{24}{0.80}= \frac{240}{8} =30\ \text N\end{aligned}$$

So, value of F = 30 N.