Question

When a force of 6.0 N is exerted at ${30}^0$ 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?

Answer 1

A force of $6N$ acting at an angle of ${30}^o$ is just able to loosen the wrench at a distance $8cm$ from it.

Therefore total torque acting at $A$ about the point $O$

${\tau }_1=6\sin {30}^o\times 0.08$

Therefore total torque required at $B$ about the point $O$ will be same as the torque produced at point $A$.
${\tau }_2=F\times 16/100$

$\Rightarrow F\times 0.16=6\sin {30}^o\times 0.08$

$\Rightarrow F=\frac{8\times 3}{16}=1.5N$.