Question

When a force of 6.0 N is exerted at 30${}^{\circ }$ 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 6 N acting at an angle of 30${}^{\circ }$ is just able to loosen the wrench at a distance 8 cm from it.

Therefore, total torque acting at A about the point O.

$=6sin{30}^{\circ }\times \left(\frac{8}{100}\right)$

Therefore total torque required at b about the point O.

$=F\times \frac{16}{100}$

$\Rightarrow F\times \frac{16}{100}=6sin{30}^{\circ }\times \left(\frac{8}{100}\right)$

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