A square metal plate 0.180 m on each side is pivoted about an axis through point O? at its center and perpendicular to the plate (?Fig. E10.3?). Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 18.0 N, F2 = 26.0 N, and F3 = 14.0 N. The plate and all forces are in the plane of the page.

Solution 3E Before solving this question, let us recall that torque is taken as positive if the force acts counterclockwise and negative if the force acts clockwise. Given, F 1 = 18.0 N Its perpendicular distance from the point O is = 0.180 m/2 = 0.09 m But this force acts clockwise, so the torque will be negative. Therefore, torque due to this force 1 = 18.0 N × 0.09 m = 1.62 N.m Given, F 2 = 26.0 N This force is also at a perpendicular distance of 0.09 m from the point O. This force acts counterclockwise and so its torque will be positive. Torque due to the force F , 2 2 = 26.0 N × 0.09 m = 2.34 N.m Given, F 3 = 14.0 N Let us now find the perpendicular distance of the force F from O. 3 Let the distance be x. 2 2 2 2 From the figure, x = 0.09 + 0.09 = 0.0162 m x = 0.127 m Now, F ac3s counterclockwise, therefore its torque will be taken as positive. Therefore, torque due to the force F is = 14.0 N × 0.127 m = 1.78 N.m 3 3 Therefore, the net torque about the axis passing through O is = + + 1 2 3 = 1.62 N.m + 2.34 N.m + 1.78 N.m = 2.50 N.m Therefore, 2.50 N.m is the net torque.