A nut just opens by applying a force of 150 N using a wrench of length 30 cm. If we wants to open it by a force of 50 N then the length of wrench should be:

45 cm

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75 cm

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90 cm

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100 cm

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Solution

The correct option is C 90 cm
Given:
Force, $F = 150 N$
Distance, $d = 30 c m = 0.3 m$

Moment of force required to open the nut is:
$τ = F d$
$τ = 150 N \times 0.3 m = 45 N m$

New force applied, $F_{2} = 50 N$
Let the length of the wrench be $d_{2}$
Torque applied:
$τ = F_{2} d_{2}$
$45 = 50 d$
$d = \frac{45}{50} m$
$d = 0.9 m = 90 c m$

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