Question

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.

Answer 1

Given:

Weight of the beam, W = 120 N

The center of gravity (C.G.) is at a distance from the pivot =  0.20 m 

F = 0.80 m

Formula:

W × distance to C.G. = F × distance to F

Solution:

∴ Anticlockwise moment = Clockwise moment

W × distance to C.G. = F × distance to F

120 × 0.20 = F × 0.80

24 = 0.80 F

`F = 24/0.80`

= 30 N

The value of F required to keep the beam in equilibrium is 30 N.