Question

When a force of 6$\cdot 0$ N is exerted at ${30}^{\circ }$ to a wrench at a distance of 8$cm$ from the nut, it is just able to loosethe nut. What force $F$ would be sufficient to loosen it It acts perpendicularly to the wrench at 16$cm$ from the nut?

Answer 1

In given figure,

$\begin{array}{c}\left(6\sin {30}^0\right)\left(8\right)=F\left(16\right)\\ \Rightarrow F=\frac{24}{16}=\frac{3}{2}=1.5N\\ \therefore F=1.5N\end{array}$