# Torque created at bottom left corner due to torque in centre of 2D box

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 $$\tau_A=\tau_O$$.

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