How is the torque about any two reference points (in the system) when a body is in translational equilibrium of same value?

Question

Here's what the textbook says

The equilibrium condition for the torques is true for any choice of the axis about which the torques are calculated. To prove this statement, we consider a rigid body on which many forces act. Relative to the origin O, force $\overrightarrow{F_1}$ is applied at the point located at $\overrightarrow{r_1}$, force $\overrightarrow{F_2}$ at $\overrightarrow{r_2}$ and so on. The net torque about an axis through O is therefore

$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{O} &=\overrightarrow{\boldsymbol{\tau}}_{1}+\overrightarrow{\boldsymbol{\tau}}_{2}+\cdots+\overrightarrow{\boldsymbol{\tau}}_{N} \\ &=\overrightarrow{\mathbf{r}}_{1} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{2} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{N} \times \overrightarrow{\mathbf{F}}_{N} \end{aligned}$

Suppose a point P is located at displacement with respect to O. The point of application of $\overrightarrow{F_1}$ with respect to P, is$(\overrightarrow{r_1} - \overrightarrow{r_P})$. The torque about P is

$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{P}=&\left(\overrightarrow{\mathbf{r}}_{1}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{1}+\left(\overrightarrow{\mathbf{r}}_{2}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{2} \\ &+\cdots+\left(\overrightarrow{\mathbf{r}}_{N}-\overrightarrow{\mathbf{r}}_{P}\right) \times \overrightarrow{\mathbf{F}}_{N} \\=&\left[\overrightarrow{\mathbf{r}}_{1} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{2} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{N} \times \overrightarrow{\mathbf{F}}_{N}\right] \\ &-\left[\overrightarrow{\mathbf{r}}_{P} \times \overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{r}}_{P} \times \overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{r}}_{P} \times \overline{\mathbf{F}}_{N}\right] \end{aligned}$

The first group of terms in the brackets gives $\tau_O$. We can rewrite the second group by removing the constant factor of $\overrightarrow{r_P}$

$\begin{aligned} \overrightarrow{\boldsymbol{\tau}}_{P} &=\overrightarrow{\boldsymbol{\tau}}_{O}-\left[\overrightarrow{\mathbf{r}}_{P} \times\left(\overrightarrow{\mathbf{F}}_{1}+\overrightarrow{\mathbf{F}}_{2}+\cdots+\overrightarrow{\mathbf{F}}_{N}\right)\right] \\ &=\overrightarrow{\boldsymbol{\tau}}_{O}-\left[\overrightarrow{\mathbf{r}}_{P} \times\left(\sum \overrightarrow{\mathbf{F}}_{\mathrm{ext}}\right)\right] \\ &=\overrightarrow{\boldsymbol{\tau}}_{O} \end{aligned}$

where we make the last step because $\sum \overrightarrow{F_{ext}}=0$ for a body in translational equilibrium. Thus the torque about any two points has the same value when the body is in translational equilibrium.

What is the physical meaning of this? How do we apply this in questions? An example too will help, because I'm having trouble visualizing this.

(source: Physics by Halliday, Resnick, Krane; 5th edition; Pg 188, Rotational Dynamics)

Answer 1

The statement simply means that the net torque of all forces acting on an object in translational equilibrium (i.e. resultant of all forces on the object is zero or $\sum\vec F_{ext}=0$ ) is the same regardless of the axis chosen.

This statement can be visualised in a better way if we consider two equal and opposite forces (say $F_1$) acting on a uniform rod as shown below:-

Let us consider the center of mass of the rod (at the midpoint) as the origin $O$ and the left corner of the rod as the point $P$. As the forces acting on the rod are equal and opposite as shown in the figures, the net force acting on the rod is zero and the rod is in translational equilibrium. Now, the net torque about the origin $O$ shall be:-$$\vert\vec \tau_O\vert = F_1l_1 + F_1l_2 = F_1(l_1+l_2)$$ since the torque of the two forces produce rotation in the same direction. The direction of $\vec \tau_O$ is into the plane of the paper (or the screen!). Now,

from the figure above, the torque about point $P$ can be written as:- $$\vert\vec \tau_P\vert = F_1(l/2+l_2) - F_1(l/2-l_1)$$ (since the torque of the two forces produce rotation in the opposite directions) $$\Rightarrow\vert\vec \tau_P\vert = F_1 (l/2+l_2-l/2+l_1) $$ $$\Rightarrow\vert\vec \tau_P\vert = F_1 (l_2+l_1) = \vert\vec \tau_O\vert $$ $$\Rightarrow\vert\vec \tau_P\vert = \vert\vec \tau_O\vert $$ Also, since the torque produced by the force to the right is greater than that produced by the force to the left about $P$, the direction of $\vec \tau_P $ is also into the plane of the paper or screen.

Therefore, $\vec \tau_O = \vec \tau_P$ as given in the statement. We can also visualise this statement when we consider an object at rest. In such a case, the object is in translational as well as rotational equilibrium. The torque about any axis remains zero which again verifies this statement that torque is constant irrespective of the choice of the axis in translational equilibrium.

Hope it helps.

Answer 2

The equipollent torque between two points A and B is defined as

$$ \boldsymbol{\tau}_B = \boldsymbol{\tau}_A + (\boldsymbol{r}_A-\boldsymbol{r}_B)\times \boldsymbol{F} \tag{1}$$

That is torque summed at B as a function of the torque summed at A in the presence of net force. If the net force is zero $\boldsymbol{F}=0$, then $$\boldsymbol{\tau}_B = \boldsymbol{\tau}_A \tag{2}$$ or the same torque components are measured (summed) regardless of location. This is a pure torque situation.

In statics, if we require the net force to be zero $\boldsymbol{F}=0$, then it is sufficient for torque measured at any point to be zero also. This means it will be zero is summed about any other point also.

But this is not the case in dynamics. Here the torque components are usually summed about the center of mass, but in general have known components at a different location, like along a hinge joint. In this case

NOTE: You have to be careful with words when it comes to torque. You apply a force and it causes a torque field that measures different values on different locations. You don't apply torque at any location, but torque is a result of the loading situation. In the case of a force couple (equal opposite forces), then the resulting torque field has the same value all over. This again is the pure torque situation.

There is an analogy here where a rotation causes each particle on the body to measure a different velocity. Even the equation for moving from one point to another to measure velocity is similar to (1)

$$ \boldsymbol{v}_B = \boldsymbol{v}_A + (\boldsymbol{r}_A-\boldsymbol{r}_B)\times \boldsymbol{\omega} \tag{3}$$

And the case of pure torque (zero force) is analogous to the body purely translating (zero rotation) and all the points moving with the same velocity.

One final note here is the interactions between motions and loading.

• In the case of zero net force and pure torque, a body is going to rotate about the center of mass.
• In the case of a force through the center of mass, a body is going to purely translate.

These two situations are two sides of the same story when it comes to rigid body dynamics.