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?
To calculate the net torque about point O, we need to calculate the torque due to each individual force and then sum them up.
The formula for torque τ is given by τ = rFsinθ, where r is the distance from the pivot point O to the point where the force is applied, F is the magnitude of the force, and θ is the angle between the force and the lever arm.
For force FA:
rA = 8.0 m
FA = 10 N
θA = 180 degrees (sin180 = 0)
Therefore, the torque due to FA is τA = rA * FA * sinθA = 8.0 * 10 * 0 = 0 Nm
For force FB:
rB = 4.0 m
FB = 16 N
θB = 90 degrees (sin90 = 1)
Therefore, the torque due to FB is τB = rB * FB * sinθB = 4.0 * 16 * 1 = 64 Nm
For force FC:
rC = 3.0 m
FC = 20 N
θC = 90 degrees (sin90 = 1)
Therefore, the torque due to FC is τC = rC * FC * sinθC = 3.0 * 20 * 1 = 60 Nm
The net torque about point O is the sum of these individual torques:
Net torque = τA + τB + τC = 0 + 64 + 60 = 124 Nm
Therefore, the magnitude of the net torque about O is 124 Nm.