The body in Fig. 11-38 is pivoted at O. angle of C is 160 degrees, angle of A is 135 degrees, angle of B is 90 degrees. Three forces act on it in the directions shown on the figure, force of A, force of B, and Force of C. 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 need to first calculate the torque due to each force and then add them up.
The torque due to force A (TA) = FA * rA * sin(135)
TA = 10 * 8 * sin(135)
TA ≈ 56.57 Nm
The torque due to force B (TB) = FB * rB * sin(90)
TB = 16 * 4 * sin(90)
TB = 64 Nm
The torque due to force C (TC) = FC * rC * sin(160)
TC = 20 * 3 * sin(160)
TC ≈ 55.8 Nm
Now, the magnitude of the net torque is the summation of these torques:
Net torque = TA + TB + TC
Net torque ≈ 56.57 + 64 + 55.8
Net torque ≈ 176.37 Nm
Therefore, the magnitude of the net torque about point O is approximately 176.37 Nm.
(b) To determine the direction of the net torque, we need to consider the direction of each individual torque. Torque can be considered positive if it causes counterclockwise rotation and negative if it causes clockwise rotation.
From the calculation above, we see that torque from force A and force C are counterclockwise while the torque from force B is also counterclockwise. Therefore, the net torque would also be counterclockwise.
So, the direction of the net torque about point O is counterclockwise.