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)
Difficulty: Medium
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
$F_{1} \times L_{1}$ = $F_{2} \times L_{2}$
$mg \times L_{1}$ = $F_{2} \times L_{2}$
$(10 \times10 )\times 0.2$ = $F_{2}\times0.5$
20 = $F_{2}\times 0.5$
$F_{2} =\frac{20}{0.5}$ = $\frac{200}{5}$= 40 N
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