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If I have a 2D box with uniform mass of width $w$ and height $h$. I apply a torque $\tau_O$ to the centre of the box, labeled $O$. The box has a fixed point $A$ at the bottom left around which it may rotate but not move at all. Ignoring gravity, and assuming that the bottom-left lies at $(0, 0)$ with the x-axis being the ground, if I apply an anti-clockwise torque to point O, what will be the torque $\tau_A$ generated at point $A$? I did some (sketchy) maths and found that $\tau_A=\tau_O$. Is my reasoning correct? If it is, did I overcomplicate it?

Let $\ell$ be the distance / length between points $O$ and $A$ (more specifically $\ell=\sqrt{\left(\frac{w}{2}\right)^2+\left(\frac{h}{2}\right)^2}=\frac{1}{2} \sqrt{w^2+h^2}$. The force that $\tau_O$ creates at point $A$ will therefore be $F=\frac{\tau_O}{\ell}$. This will not move point $A$, but will generate an equal but opposite force, rotating point $O$ around $A$, with $\tau_A=\left(\frac{\tau_O}{\ell} \right) \ell$, the $\ell$'s cancel and I'm left with $\tau_A=\tau_O$.

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Moment and torque are vectors. A vector is not an arrow, vector is a quantity with magnitude and direction. We use an arrow to represent a vector. But the torque vector has no information about the arrow's point of origin.

So it doesn't matter where the torque is applied as long as it is represented by a parallel and same magnitude vector.

So your answer is, $$\tau_A = \tau_o$$

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  • $\begingroup$ therefore; the torque at A = torque at O $\endgroup$
    – Tiger Guy
    Commented Mar 22, 2023 at 13:21

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