# Relationship between net force and net torque

Why are the following two statements false?

1. If net torque is zero then net force must be non-zero.
2. If net torque is non-zero then net force must also be non-zero.

1. If net torque is zero then net force must be non-zero.

The torque is defined as $$\vec{\tau} = \vec{r} \times \vec{F}$$, where $$\times$$ denotes vector (cross) product, and $$\vec{r}$$ is vector from the axis of rotation to the point where the force acts. When there is no external force acting on the body ($$\vec{F} = \vec{0}$$) or one or more forces act in the direction through or parallel to the axis of rotation, the torque will be zero.

In addition to this, for an object to be in equilibrium, it must have no tendency to accelerate or to start rotating. The former means that the net force is zero, and the latter means that the angular momentum and the net torque about any point are zero.

Hence, zero torque does not require non-zero net force. They both can be zero at the same time.

1. If net torque is non-zero then net force must also be non-zero.

Imagine a rod pivoted at the center. One force acts on the far left end downwards and the other force acts on the far right end upwards. If the two forces have the same magnitude then the net force is zero but the resulting net torque is not zero. Hence, there could be a zero net force that will produce non-zero torque. Source: H. D. Young, R. A. Freedman, "University Physics with Modern Physics in SI Units", 15th ed., 2019.

• Interesting. Net force equal to zero does not imply power equal to zero. I hadn't thought about the fact that opposite forces can cancel each other out while work is still being done, but it makes sense. Mar 5, 2022 at 0:18
• So the net force on a bicycle's cranks will be zero if you're pushing equally on both pedals, but you are still doing work. Mar 5, 2022 at 0:19
• @DuncanC In translational motion the work is $F \Delta s$, and in rotational motion the work is $\tau \Delta \theta$. In case of a bicycle, the translational power by the legs is zero, but the rotational power is $\tau \omega$ which is converted to translational motion. Mar 6, 2022 at 9:00

Just have a look at the formula:

$$\vec\tau=\vec r\times \vec F\qquad\Leftrightarrow\qquad \tau=rF_\perp.$$

• The torque $$\tau$$ can be zero, if $$r=0$$. So, the $$F$$ component doesn't have to be zero. That corresponds to you trying to turn af thin strew by just grapping it a twisting - the lever arm-length is so tiny that your applied torque is basically zero, even when you apply large force.

• But if any of the two multiplied parameters $$r$$ or the $$F$$ component are zero, then mathematically the torque becomes zero as well. That corresponds to you finally applying a proper wrench to turn your screw with a long lever arm-length $$r$$, but you aren't pulling... So no torque is exerted.