Answer 1

In the first case,

\[\tau_1  = 6\sin30^\circ \times \left( \frac{8}{100} \right)\]

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

\[\tau_1  =  \tau_2\]

\[\Rightarrow F \times \frac{16}{100} = 6\sin30^\circ \times \frac{8}{100}\]

\[\Rightarrow F = \frac{\left( 8 \times 3 \right)}{16} = 1 . 5  N\]