Question

The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude

Answer 1

$\boxed{\sf \tau=rFsin\theta}$

Put values

$\\ \rm\hookrightarrow 40=2Fsin40$

$\\ \rm\hookrightarrow Fsin40=20$

$\\ \rm\hookrightarrow 0.64F=20$

$\\ \rm\hookrightarrow F=31.25N$