A body is pivoted at some point. How much force is required to produce $40 N m$ torque, when it is applied at a distance of $200 c m$ from the pivot?

$2 N$

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$20 N$

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$4 N$

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$40 N$

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Solution

The correct option is B
$20 N$

Given, the moment of force = 40 N m
Distance from the pivot, r = 200 cm = 2 m
Let the force applied be F
Then ,the moment of force = force $\times$ distance from the pivot
or, $40 N m = F \times 2 m$
or, $F = \frac{40 N m}{2 m} = 20 N$

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A body is pivoted at some point. How much force is required to produce 40 N m torque, when it is applied at a distance of 200 cm from the pivot?

The correct option is B 20 N Given, the moment of force = 40 N m Distance from the pivot, r = 200 cm = 2 m Let the force applied be F Then ,the moment of force = force × distance from the pivot or, 40 N m=F×2 m or, F=40 N m2 m=20 N

A body is pivoted at some point. How much force is required to produce $40 N m$ torque, when it is applied at a distance of $200 c m$ from the pivot?

The correct option is B
$20 N$

Given, the moment of force = 40 N m
Distance from the pivot, r = 200 cm = 2 m
Let the force applied be F
Then ,the moment of force = force $\times$ distance from the pivot
or, $40 N m = F \times 2 m$
or, $F = \frac{40 N m}{2 m} = 20 N$