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.
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?
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