Two forces $F_{1} = 5 N$ and $F_{2} = 8 N$ acting at points A and B on a rod pivoted at a point O, such that OA = $2 m$ and OB = $4 m$ . Calculate the total moment of the two forces about O.

22 Nm anticlockwise

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42 Nm clockwise

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22 Nm clockwise

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42 Nm anticlockwise

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Solution

The correct option is C
22 Nm clockwise

Moment of force = Force $\times$ perpendicular distance between axis of rotation and line of action of force
Moment of force due to $F_{1} = F_{1} \times O A$
= $5 N \times 2 m = 10 N m$ (anticlockwise)
Moment of force due to $F_{2} = F_{2} \times O B$
= $8 N \times 4 m = 32 N m$ (clockwise)
$∴$ Net Torque = $32 N m - 10 N m$
= $22 N m$ (clockwise)

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Two forces F1=5 N and F2=8 N acting at points A and B on a rod pivoted at a point O, such that OA = 2 m and OB = 4 m. Calculate the total moment of the two forces about O.

Two forces $F_{1} = 5 N$ and $F_{2} = 8 N$ acting at points A and B on a rod pivoted at a point O, such that OA = $2 m$ and OB = $4 m$ . Calculate the total moment of the two forces about O.