equilibriumrotationtorque
In the definition of rotational equilibrium for a rigid body, the net torque about a point has to be zero.
Now, about which point will the torque be calculated? Will it be the centre of mass or centre of gravity? Or, can it be calculated about any point?
I'm really confused. Somebody please help with proper reasons.
You can translate a force F along its vector (line of action) without changing the torque. For example, you can slide $F_1$ all the way to the vertical member - the one that force $F_2$ acts along.
Now if you look at the diagram I drew:
you can see two different ways to compute the torque.
Method 1: Extend $F_1$ along its direction, draw the perpendicular line to the origin (in turquoise), multiply. The answer is $F_1\cdot d$.
Method 2: Extend $F_1$ until it intersects the line of action of $F_2$. Now decompose it along perpendicular directions - one orthogonal to the radius vector to the origin, and one along it. I labeled the perpendicular component $F_3$ (red). Now the vector $F_2$ that you need to compute is given by looking for the reverse projection of $F_3$ along the line of action of $F_2$. I hope it is clear from the diagram. This is obviously a roundabout way to construct it. It would be quicker to compute torque according to the above, then just set $F_2 \cdot (x_1 + 1) = F_1 \cdot d$.
Note - it may be that you intended to show $d = x$, but that is not clear from the diagram. And if that is indeed not the case, then you need to know the angle of $F_1$ in order to solve this problem - you imply that you do know that. Even so - as presented your problem cannot be solved unless we can compute $d$ for which we need the height of your turquoise shape, or the angle of the dotted line $x$.
Equilibrium can mean different things depending on the context. For example think of a container of gas with a permeable barrier dividing the container in half. At equilibrium you would expect an equal concentration of gas molecules on either side of the partition even though there is still going to be a transfer of gas molecules between each side of the container. The net flow is $0$ though, so we say the system is in equilibrium.
In the context of torques, I would all we need for equilibrium is that the net torque is $0$. Whether or not the system in question has a net rotation depends on your "initial conditions". Typically then you see in many statics scenarios that the system in question is also at rest, but you wouldn't approach the problem any differently if there was rotation (unless the motion changes torques of the system, but then it turns into a dynamics problem instead of a statics problem).
In this snapshot of a book it says that For rotational equilibrium ,torque about the body's Center of Mass must be zero.But why only center of mass? About any other point also it is in equilibrium.It is confusing.
I think this is the distinction between "rotational equilibrium" and more of a complete "mechanical equilibrium". For example, if there is a single force acting on the center of mass of our system, then of course there is no torque about the center of mass of the system and there will be no rotation about the center of mass. However a single force cannot keep a system in mechanical equilibrium, as there will be a net force. But if we pick any other point not along the direction of the force we will find a net torque about that point.
For complete mechanical equilibrium we can choose any point about which to find the torque and it must be $0$. But for just rotational equilibrium we just need to consider torques about the center of mass.
Best Answer
The torque has to be zero about all points -- if it was nonzero for some point, the object would start to rotate around that point.
On the other hand, you often see that the torque is calculated around some specific point, and if it vanishes, the body is declaredto be in equilibrium. How does that fit together?
Here's a hint (in a simple case):
If you have shown that, you can conclude that it is sufficient for equibrium to show that the overall force and the overall torque around one point vanish. (Hence, you can pick a convenient point for the torque calculation, i.e. one where some torque is zero or some torques cancel out immediately.)