A rod is pivoted about its center. A 5-N force is applied 4 m from the pivot and another 5-N force is applied 2 m from the pivot, as shown. The magnitude of the total torque about the pivot (in N.m) is:
A) 0
B) 5
C) 8.7
D) 15
E) 26

Question

A rod is pivoted about its center. A 5-N force is applied 4 m from the pivot and another 5-N force is applied 2 m from the pivot, as shown. The magnitude of the total torque about the pivot (in N.m) is: 
A) 0 
B) 5 
C) 8.7 
D) 15 
E) 26

Quizplus · Accepted Answer

Torque is given by the formula: Torque = Force x Distance x sin(angle between force and lever arm)
For the 5-N force applied at 4 m from the pivot, the torque is (5 N) x (4 m) x sin(90 degrees) = 20 Nm.
For the 5-N force applied at 2 m from the pivot, the torque is (5 N) x (2 m) x sin(90 degrees) = 10 Nm.
The total torque about the pivot is the sum of these torques, which is 20 Nm + 10 Nm = 30 Nm. 
Therefore, the magnitude of the total torque about the pivot is 15 N (assuming the pivot is at the center of the rod). 
The closest answer choice is D, which is 15.

A Rod Is Pivoted About Its Center