Question

Calculate the total torque acting on the body shown in figure (10-E2) about the point O.

Figure

Answer 1

Torque about a point = Total force × Perpendicular distance
Let the anticlockwise torque and clockwise acting torque be positive and negative, respectively.
Torque at O due to 5 N force is zero as it is passing through O.
Torque at O due to 15 N force,
$$\begin{gathered} {\tau }_1=15\times 6\times {10}^{-2}\times \sin 37° \\ =15\times 6\times {10}^{-2}\times \frac{3}{5} \\ =0.54\mathrm{N}-\mathrm{m}\left(\mathrm{anticlockwise}\right) \end{gathered}$$

Torque at O due to 10 N force,
${\tau }_2=10\times 4\times {10}^{-2}=0.4\mathrm{N}-\mathrm{m}\left(\mathrm{clockwise}\right)$

Torque at O due to 20 N force,
$$\begin{gathered} {\tau }_3=20\times 4\times {10}^{-2}\times \sin 30° \\ =20\times 4\times {10}^{-2}\times \frac{1}{2} \\ =0.4\mathrm{N}-\mathrm{m}\left(\mathrm{Anticlockwise}\right) \end{gathered}$$

Resultant torque acting at O,
$$\begin{gathered} \tau =\left(0.54-0.4+0.4\right) \\ =0.54\mathrm{N}-\mathrm{m}\left(\mathrm{anticlockwise}\right) \end{gathered}$$