Question

A 3 m rod of weight 30 N has three additional forces acting on it. One is 40 N upward on the left side, one is 60 N upward on the right side, and one is 70 N downward 1 m from the right side. Is the rod in equilibrium?

A. Yes

B. No

Answer 1

The rod has zero net force but a non-zero net torque, due to which it is not in equilibrium. This is because the torques generated by the forces do not cancel out, resulting in a rotation around the pivot point.

"the correct option is approximately option B"

Step-by-step explanation:

To determine if the 3 m rod is in equilibrium, we must check two conditions: net force and net torque. First, calculating the net force, we have two upward forces of 40 N and 60 N and two downward forces of 30 N (weight of the rod) and 70 N. The total upward force is 40 N + 60 N = 100 N, and the total downward force is 30 N + 70 N = 100 N. Thus, the net force is zero, satisfying the first condition.

Now, for the rod to be in equilibrium, the net torque must also be zero. Assume the left side as the pivot point, and calculate the torque due to each force. The 40 N force on the left side has no lever arm, so its torque is zero. The weight of the rod (30 N) acts at the center of gravity, which is 1.5 m from the pivot, creating a counter-clockwise torque of 30 N × 1.5 m = 45 N·m. The 70 N force acts 1 m from the right end, thus 2 m from the pivot, creating a clockwise torque of 70 N × 2 m = 140 N·m. Lastly, the 60 N force on the right end creates a counter-clockwise torque of 60 N × 3 m = 180 N·m. Summing these, the net torque is 180 N·m + 45 N·m - 140 N·m = 85 N·m counter-clockwise, implying the rod is not in equilibrium.

Therefore, answering the student's question: No, the rod is not in equilibrium because, while the net force is zero, the net torque is not, resulting in a rotation of the rod around the pivot.