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Is the net torque about any point zero for a body in equilibrium? Or the net torque only about the centre of mass zero?

Kruthik
  • 120

2 Answers2

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Ok let us see. The torque for the force $\vec F$ is given by: $$\vec r \times \vec F$$ Thus the total torque on a body with respect to the position $\vec p$ is given by: $$\vec \tau(\vec p)=\sum_i (\vec r_i -\vec p)\times \vec F_i$$ where all positions vectors are measured from the origin. Let us suppose that this is zero for one value of $\vec p$ then relative to the point $\vec p+\vec p'$ we have that: $$\vec \tau(\vec p+\vec p')=\sum_i (\vec r_i -\vec p-\vec p')\times \vec F_i$$ $$=\tau(\vec p)- \vec p' \times \sum_i \vec F_i$$ $$=- \vec p' \times \sum_i \vec F_i$$ But if we are in equilibrium $$\sum_i \vec F_i=0$$ so: $$\tau(\vec p+\vec p')=0$$ i.e. in equilibrium if $\tau(\vec p)$ for any $\vec p$ (e.g. the cofm) then it is zero for all $\vec p$.

Quantum spaghettification
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2

CM Center of mass

For a rigid body in equilibrium , is the net torque about any axis of rotation zero? \begin{align*} &\text{Static:}\\\\ &\text{Coordinate system center of mass $CM$}\\\\ &\vec{f}_1+\vec{f_2}=0\,,\text{Forces}\\ &\vec{f}_1\times\vec{r_1}+\vec{f}_2\times\vec{r_2}=0\,,\text{Torques}\\\\ &\text{Coordinate system $P$ : Torques }\\\\ &\vec{f}_1\times\vec{r'_1}+\vec{f}_2\times\vec{r'_2}= \vec{f}_1\times\left(\vec{r_1}+\vec{r_3}\right)+ \vec{f}_2\times\left(\vec{r_2}+\vec{r_3}\right) =\underbrace{\vec{f}_1\times \vec{r}_1+\vec{f}_2\times\vec{r}_2}_{=0}+ \underbrace{\left(\vec{f}_1+\vec{f}_2\right)}_{=0}\times \vec{r}_3=0 \end{align*} \begin{align*} &\text{General case}\\\\ &\text{Coordinate system $CM$}\\\\ &\sum_{i}^{k} \vec{f}_i=0\\ &\sum_{i}^{k} \vec{f}_i\times \vec{r}_i=0\\ &\text{Coordinate system $P$ }\\\\ &\sum_{i}^{k} \vec{f}_i\times \vec{r'}_i= \sum_{i}^{k} \vec{f}_i\times \left(\vec{r}_i+\vec{r}_{CP}\right)= \sum_{i}^{k} \vec{f}_i\times \vec{r}_i-\vec{r}_{CP}\sum_{i}^{k}\vec{f}_i=0 \end{align*}

Eli
  • 11,878