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Moment of Force

A ring with r...

Question

A ring with radius $4 m$ hinged at point P. Two forces $F_{1}$ and $F_{2}$ each of magnitude $10 N$ act tangentially at the surface of the ring. Find the ratio of the torques produced by two forces $F_{1}$ to $F_{2}$ is

1:2

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2:1

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1:3

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3:1

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Solution

The correct option is B
2:1

Given,

Radius of ring $r$ = $4 m$

Force $F_{1}$ = $10 N$

Force $F_{2}$ = $10 N$

Torque due to force= Force $\times$ perpendicular distance between line of action of force and axis of rotation

Let perpendicular distance of point of rotation P from the Force $F_{1}$ is $d_{1} = 8 m$ while that of force $F_{2}$ is $d_{2} = 4 m$
Torque $T_{1}$ produced by $F_{1}$ = $d_{1} \times F_{1}$ = $8 \times 10$ = $80 N m$

Torque $T_{2}$ produced by $F_{2}$ = $d_{2} \times F_{2}$ = $4 \times 10$ = $40 N m$

Ratio of torques $\frac{T_{1}}{T_{2}} = \frac{2}{1}$

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A ring with radius 4 m hinged at point P. Two forces F1 and F2 each of magnitude 10 N act tangentially at the surface of the ring. Find the ratio of the torques produced by two forces F1 to F2 is

A ring with radius $4 m$ hinged at point P. Two forces $F_{1}$ and $F_{2}$ each of magnitude $10 N$ act tangentially at the surface of the ring. Find the ratio of the torques produced by two forces $F_{1}$ to $F_{2}$ is