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Question
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
\[\tau_1 = \tau_2\]
\[\Rightarrow F \times \frac{16}{100} = 6\sin30^\circ \times \frac{8}{100}\]
\[\Rightarrow F = \frac{\left( 8 \times 3 \right)}{16} = 1 . 5 N\]
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