# Net torque on an object when when all the forces pass through a common point

I am confused about a problem in my textbook (Serway's College Physics, quick quiz 12.2 on the 9th edition). The problem considers an extended object that is acted on by three separate forces, all of which pass through the same point $P$: It is apparent that the net force is not zero so this object is not in force equilibrium. However, the answer to this quick quiz claims that the object is in torque equilibrium. I don't quite agree with the answer.

It is true that if we choose the reference axis to pass through point $P$ the net torque is zero. Nevertheless, if we choose other reference axes, the net torque will no longer be zero. It is also claimed in the book that the net external torque on the object about "any" axis must be zero in order to achieve static equilibrium.

Let me do the math. There are three forces and let me call them $\mathbf F_1$, $\mathbf F_2$, and $\mathbf F_3$. First let us choose point $P$ to be the reference point. The net torque is $$\mathbf r_1 \times \mathbf F_1 + \mathbf r_2 \times \mathbf F_2 + \mathbf r_3 \times \mathbf F_3= 0+0+0=0,$$ where $\mathbf r_1$, $\mathbf r_2$, and $\mathbf r_3$ are the position vectors from $P$ to the application points of $\mathbf F_1$, $\mathbf F_2$, and $\mathbf F_3$, respectively.

If I instead choose another origin located R away from point P, the net torque is $$(\mathbf r_1+\mathbf R) \times \mathbf F_1 + (\mathbf r_2+\mathbf R) \times \mathbf F_2 + (\mathbf r_3+\mathbf R) \times \mathbf F_3 = \mathbf R \times (\mathbf F_1+\mathbf F_2+\mathbf F_3)$$ is apparently not zero.

• – BowlOfRed Nov 8 '16 at 11:05

If I instead choose another origin located R away from point P, the net torque is $$(\mathbf r_1+\mathbf R) \times \mathbf F_1 + (\mathbf r_2+\mathbf R) \times \mathbf F_2 + (\mathbf r_3+\mathbf R) \times \mathbf F_3 = \mathbf R \times (\mathbf F_1+\mathbf F_2+\mathbf F_3)$$ is apparently not zero.

While we often assume a net torque will give us rotation, really a net torque means the angular momentum about that point is changing, and that doesn't always imply a rotation.

If your new point $P$ is fixed on the object (but not at the center of mass), then the non-zero net force means that your choice of axis is accelerating and is not at rest in an inertial frame.

If your new point $P$ is at rest, then the object will accelerate from it. If the velocity vector for the center of mass does not pass through this new axis, then the angular momentum of the object about that axis is changing. The non-zero net torque you calculate shows the change in angular momentum, not necessarily a rotation.

If you want to determine that an object is in rotational equilibrium, you want to find a net torque of zero and one additional constraint, either

• The axis coincides with the center of mass or
• The object and the axis are not accelerating
• Thanks, but what if the vector sum of F1, F2, and F3 is nonzero? – Chien-Te Wu Nov 8 '16 at 10:30
• Sorry, misread the non-zero net force part. Modified the answer. Don't try to calculate the torque of an accelerating object except through the center of mass axis if you want to find if it is in rotational equilibrium. – BowlOfRed Nov 8 '16 at 11:01
• Thanks for your reply. The answer given by Serway to his problem says that this object is in rotational equilibrium. According to your argument, is his answer incorrect? – Chien-Te Wu Nov 9 '16 at 9:39
• I don't see there's enough information. Having a net torque about an arbitrary point wouldn't matter. Having a net torque about the center of mass does. If the net torque about the center of mass is non-zero, then it is not in rotational equilibrium. – BowlOfRed Nov 9 '16 at 16:49