Sum

When a force of 6⋅0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?

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#### Solution

In the first case,

\[\tau_1 = 6\sin30^\circ \times \left( \frac{8}{100} \right)\]

In first case,

\[\tau_2 = F \times \left( \frac{16}{100} \right)\]

To loosen the nut, torque in both the cases should be the same.

Thus, we have

\[\tau_1 = \tau_2\]

\[\Rightarrow F \times \frac{16}{100} = 6\sin30^\circ \times \frac{8}{100}\]

\[\Rightarrow F = \frac{\left( 8 \times 3 \right)}{16} = 1 . 5 N\]

Concept: Torque and Angular Momentum

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