A tetrahedron has its corners $A=(1,1,1), B=(2,0,-1), C=(0,1,-1), D=(3,1,2)$. There is a point mass of weight $G=(0,0,-mg)$ placed on all of its corners. What is the total torque $M$ from the weights with respect to ...
- corner $A$.
- the tetrahedrons center of gravity.
Given a force $\vec f = (0,0,-mg)$ and an application point $A$ the torque from this force regarding a point $O$ is given by
$$ \vec M_O = \vec f \times (A-O) $$
for the tetrahedron we have
$$ \vec M_A = \vec f \times (A-A)+\vec f \times (B-A)+\vec f \times (C-A)+\vec f \times (D-A) = \vec f \times\left(B+C+D-3A\right) $$
Regarding the barycenter we have
$$ \vec M = \vec 0 $$
because being $P = \frac 14\left(A+B+C+D\right)$ (barycenter) we have
$$ \vec M_P = \vec f \times (A-P)+\vec f \times (B-P)+\vec f \times (C-P)+\vec f \times (D-P) = \vec f \times\left(A+B+C+D-4\frac 14\left(A+B+C+D\right)\right)=\vec 0 $$