A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long.

Question

A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar? (40 N)

Accepted Answer

Solution:

Mass of the block = m = 10 kg
Length of the bar = l = 1 m
Moment arm of $w_{1} = L_{1}$ = 20 cm = 0.2 m
Moment arm of $w_{2} = L_{2}$ = 50 cm = 0.5 m
Did force require to balance the bar $F_{2}$ =?

By applying principle of moments:

Clockwise moments = anticlockwise moments
$F_{1} imes L_{1}$ = $F_{2} imes L_{2}$
$mg imes L_{1}$ = $F_{2} imes L_{2}$
$(10 imes10 ) imes 0.2$ = $F_{2} imes0.5$
20 = $F_{2} imes 0.5$
$F_{2} =\frac{20}{0.5}$ = $\frac{200}{5}$= 40 N