Two forces each of $5 N$ act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint

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Solution

Given: $F_{1} = F_{2} = 5 N, d_{1} = d_{2} = 0.5 m$
Torque $τ = F \times d$
$t a u_{1} = F_{1} \times d_{1} = 5 N \times 0.5 m$
$= 2.5 N m$ (Anti-clockwise)
$τ_{2} = F_{2} \times d_{2} = 5 N \times 0.5 m$
$= 2.5 N m$ (Anti-clockwise)
The resultant moment of force $= τ_{1} + τ_{2}$
$= 2.5 + 2.5 = 5 N m$ in Anti clockwise direction.

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