# If the torque caused by a force is cancelled out, is the force cancelled out as well?

I calculated the value of $\theta$ to be $53^{\circ}$ for (ii) by using the principle of moments ($6\sin\theta r 2 = 2.4 r$ ) as the disc is in equilibrium. Nothing wrong so far.

For (iii), I calculated the horizontal component of $\:\rm 6 N$ force ($6\cos 53^{\circ}= 3.6\:\rm N$) and since it is the only force that hasn't been cancelled, I deduced it must be the force (in opposite direction) of the pin on the disc.

But the answer is $6 \:\rm N.$

Isn't the vertical component of the force responsible for the anti-clockwise torque which is being cancelled by the clockwise torque? If so, why do I have to include it for my answer to (iii)?

Is it because despite being the cause of the anti-clockwise torque which has been cancelled, the force itself isn't cancelled and requires an opposing force ? But again, if this force is cancelled, how can it exert any torque? I'm confused.

I've just been introduced to this topic so I'd like to apologize if I'm missing out something very simple. Thanks in advance :).

Just look at the forces on the system. Suppose, there was no pin. Then, as evident, the only net force is due to $6 N$ force and thus the disc would accelerate in that direction and would also rotate due to a net torque.
Now, the disc is prohibited to move from it's place due to the pin( more technically it's center of mass must be stationary as there is no net force on the system). Therefore, the pin must provide an equal and opposite force of magnitude $6N$.