A ring of radius $30 m$ and mass $10 k g$ is pivoted at point 'A' to a perpendicular wall. Find the magnitude of force so that ring will be in equillibrium? (Take acceleration due to gravity $g$ = $10 m s^{-} 2$ )

10 N

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15 N

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20 N

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25 N

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Solution

The correct option is D
25 N

Given,

Mass of the ring $m$ = $10 k g$

Radius of ring $r$ = $30 m$

Acceleration due to gravity $g$ = $10 m s^{-} 2$

Draw all the forces acting on the ring.

Now, For equillibrium, net torque about A = $0$
Clockwise torque due to 'mg' = Anticlockwise torque due to 'F'.
$\Rightarrow m g \times 10 = F \times (10 + 30)$
$1000 = F \times 40$
$\Rightarrow F = \frac{1000}{40} = 25 N$

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A ring of radius 30 m and mass 10 kg is pivoted at point 'A' to a perpendicular wall. Find the magnitude of force so that ring will be in equillibrium? (Take acceleration due to gravity g = 10ms−2)

A ring of radius $30 m$ and mass $10 k g$ is pivoted at point 'A' to a perpendicular wall. Find the magnitude of force so that ring will be in equillibrium? (Take acceleration due to gravity $g$ = $10 m s^{-} 2$ )