Question:
When a force of $6.0 \mathrm{~N}$ is exerted at $30^{\circ}$ to a wrench at a distance of $8 \mathrm{~cm}$ 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 $16 \mathrm{~cm}$ from the nut?
Solution:
Torque in both the cases will be equal
$\tau_{1}=\tau_{2}$
$\left(6 \sin 30^{\circ}\right)(8)=(F)(16)$
$F=1.6 \mathrm{~N}$
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