## Essential University Physics: Volume 1 (3rd Edition)

We find the sum of the torques about point P: $\tau_p=(F_{1p}\times r_{1p})+(F_{2p}\times r_{2p})+(F_{3p}\times r_{3p})$ Using substitution, we find: $\tau_p=\vec{0}+\vec{0}\times R = \vec{0}$ We now consider point O: $\tau_O=(F_{1O}\times r_{1O})+(F_{2O}\times r_{2O})+(F_{3O}\times r_{3O})$ $\tau_O=\vec{0}$ Thus, regardless of the pivot point, the net torque is 0.