Tension between two rigid bodies

[Rikudo]

Homework Statement

A balance consists of a straight light rod A, a right-angled light rod B rigidly welded with rod A and a fixed fulcrum F. Four loads are suspended from the balance with the help of light threads. The rods have equidistant marks on them. Find the mass of the load C.

Relevant Equations

Torque and Newton's 2nd law

Ok. So, I already worked on this problem, and get $m_c$ = 2m/3, which is correct according to the book.
However, I also want to know the value of the tension (T) between rod A and B.

Note: Before we start working on my modified question, I want to point out that the force exerted by the fulcrum is F = 5m + 2m/3 (I get this by using Newton law)

If we only look at the forces which is working on rod A, then with using the Newton 2nd law, we will get:
$$F = mg + mg + T$$
$$T = 3m + \frac{2m} {3}$$

Strangely, I gained a different result when I tried using torque equation for rod A, with point F as the origin. (Here, L is the length between two marks)
$$0 = -3mgL + 4mgL-2TL$$
$$T = mg/2$$

Not only that, changing the location of the origin also changed the value of the tension. What is happening here?

 

[Orodruin]

These are rods, not ropes. The force between the rods is not the only interaction. In order for B to be in equilibrium, it must quite clearly also be affected by a net torque from A.

 

[Rikudo]

Sorry, but I don't understand what you mean. Could you please elaborate?

 

[Orodruin]

You have treated the interaction between the rods as if they were characterised by a single force between them. This is not accurate. In addition to the net force between the rods, there is also a net torque. You can see this easily by considering the torque equilibrium of rod B around the connection point. B will not be in equilibrium unless there is a torque acting on it from A.

 

[Rikudo]

Ah. For B to be equilibrium, rod A also must exert a force to B in the left direction. Is this what I overlooked?

 

[Orodruin]

Ah. For B to be equilibrium, rod A also must exert a force to B in the left direction. Is this what I overlooked?

No. Torque. Not an additional net force.

 

[Rikudo]

But,Torque is created by exerting force

 

[Orodruin]

But,Torque is created by exerting force

Yes, but if you idealise the connection with a point, then a force cannot create a torque around that point so you must also idealise the forces acting within the connection with a single force and a torque. In reality, the connection has a non-zero size. The torque is created by the force not being uniform across the connection. It will be more up on the side B extends to and is likely even pointing down on the other side. Summing up the forces acting across the connection gives you the idealised single force. Summing up the torques gives you the idealised torque.

Given any rigid body, the effect of any forces acting on it may be summarised as a total force and a total torque. The torque will generally depend on the point unless the force is zero.

You can act upon a body with forces and torques separately. You can create a torque without adding a resultant force by adding two forces equal in magnitude and opposite in direction that are offset by a distance. In your case, this all happens in the weld, which in reality has some finite size.

 

[Lnewqban]

But,Torque is created by exerting force

Considering the direction of gravity action, force T does not have any lever respect to the point at which the right-angled light rod B is rigidly welded with rod A.
Therefore, exerting force T can’t create any torque at that welded joint.

Nevertheless, the masses hanging from rod B enjoy two levers to load that welded joint with the summation of two clockwise torques: 3mL and (2/3)m5L.

Therefore, that welded joint is simultaneously loaded with a vertical downwards force and a clockwise net torque.