Question

If several forces act on a particle, the total torque on the particle may be obtained by first finding the resultant force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect at a common point?

Answer 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{f_1},\overrightarrow{f_2},\overrightarrow{f_3},...\overrightarrow{f_n}\mathrm{be}\mathrm{the}\mathrm{forces}\mathrm{acting}\mathrm{on}\mathrm{a}\mathrm{point}P. \\ \mathrm{Let}O\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{along}\mathrm{which}\mathrm{torques}\left(\mathrm{moments}\right)\mathrm{will}\mathrm{be}\mathrm{taken}. \\ \mathrm{Let}: \\ \overrightarrow{f_1}+\overrightarrow{f_2}+\overrightarrow{f_3}+...+\overrightarrow{f_n}=\overrightarrow{R}...\left(1\right) \\ \mathrm{Moments}\mathrm{of}\mathrm{force}\left(\mathrm{torque}\right)\overrightarrow{f_i}\mathrm{about}O\mathrm{will}\mathrm{be}: \\ \overrightarrow{{\tau }_1}=\overrightarrow{\mathrm{OP}}\times \overrightarrow{f_1} \\ \mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{torques}\mathrm{about}O\mathrm{will}\mathrm{be}: \\ \overrightarrow{M}=\overrightarrow{\mathrm{OP}}\times \overrightarrow{f_1}+\overrightarrow{\mathrm{OP}}\times \overrightarrow{f_2}+...+\overrightarrow{\mathrm{OP}}\times \overrightarrow{f_n} \\ \Rightarrow \overrightarrow{M}=\overrightarrow{\mathrm{OP}}\times \left(\overrightarrow{f_1}+\overrightarrow{f_2}+\overrightarrow{f_3}+...+\overrightarrow{f_n}\right) \\ \Rightarrow \overrightarrow{M}=\overrightarrow{\mathrm{OP}}\times \overrightarrow{R}\left[\mathrm{From}\left(1\right)\right] \\ \mathrm{Thus},\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{torque}\mathrm{of}\mathrm{the}\mathrm{resultant}\mathrm{force}\overrightarrow{\mathrm{R}}\mathrm{of}\mathrm{the} \\ \mathrm{forces}\overrightarrow{f_1},\overrightarrow{f_2},\overrightarrow{f_3},...\overrightarrow{,f_n}\mathrm{gives}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{moments}\mathrm{of}\mathrm{the}\mathrm{torques}. \end{gathered}$$