The body in Fig. 11-38 is pivoted at O. Three forces act on it in the directions shown on the figure. FA = 10 N at point A, 8.0 m from O. FB = 16 N at point B, 4.0 m from O. FC = 20 N at point C, 3.0 m from O.
(a) What is the magnitude of the net torque about O?
b) What is the direction of the net torque about O?
(a) To calculate the net torque about point O, we use the formula:
Net torque = (Force A)(distance A) + (Force B)(distance B) + (Force C)(distance C)
Net torque = (10 N)(8.0 m) + (16 N)(4.0 m) + (20 N)(3.0 m)
Net torque = 80 N·m + 64 N·m + 60 N·m
Net torque = 204 N·m
Therefore, the magnitude of the net torque about point O is 204 N·m.
(b) To determine the direction of the net torque, we need to consider the directions of the individual torques produced by each force. The torque produced by each force is given by the formula:
τ = rFsinθ
where r is the distance of the force from the pivot point, F is the magnitude of the force, and θ is the angle between the force and the lever arm.
For force FA at point A:
τA = (8.0 m)(10 N)(sin90°) = 80 N·m
For force FB at point B:
τB = (4.0 m)(16 N)(sin180°) = -64 N·m
For force FC at point C:
τC = (3.0 m)(20 N)(sin270°) = -60 N·m
The negative sign indicates that the torque is in the opposite direction to the positive direction determined from the right-hand rule. Therefore, the net torque is counterclockwise.
Hence, the direction of the net torque about point O is counterclockwise.