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?

Figure

Answer 1

In the first case,
${\tau }_1=6\sin 30°\times \left(\frac{8}{100}\right)$
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:
${\tau }_1={\tau }_2$
$\Rightarrow F\times \frac{16}{100}=6\sin 30°\times \frac{8}{100}$
$\Rightarrow F=\frac{\left(8\times 3\right)}{16}=1.5\mathrm{N}$