How Is the Reaction at the Joint Calculated in a System of Two Uniform Rods?

[ashokmittal]

Two uniform rods AB and CD each of weight W and length a are smoothly jointed at a point O, where OB and OD are each of length b. The rods rest in a vertical plane with the ends A and C on a smooth table and the ends B and D are connected by a light string. Prove that the reaction at the joint (aW/2b)tan @, where @ is the inclination of either rod to the vertical.

 

[Andrew Mason]

Two uniform rods AB and CD each of weight W and length a are smoothly jointed at a point O, where OB and OD are each of length b. The rods rest in a vertical plane with the ends A and C on a smooth table and the ends B and D are connected by a light string. Prove that the reaction at the joint (aW/2b)tan @, where @ is the inclination of either rod to the vertical.

I don't quite understand the problem. Is there a picture available? It would seem that segments OB and OD are leaning in a downward direction at angle \alpha from vertical. Is that correct? If so, I don't see how AO and CO can be vertical. They would have to be leaning outward at some angle. I am also not clear on what force it is you are supposed to be finding.

AM

 

[wais]

The given scenario describes a system of two uniform rods joined at a point and resting on a smooth table. The reaction at the joint, denoted by R, is to be determined.

To begin with, we can consider the equilibrium of the system as a whole. Since the rods are resting on a smooth table, the only external force acting on the system is the weight of the rods, which is equal to 2W. This weight acts vertically downwards, passing through the point O. In order for the system to be in equilibrium, the reaction R must balance out the weight 2W.

Now, let us consider the equilibrium of each individual rod. The forces acting on each rod are its weight W, acting vertically downwards through its center of mass, and the reaction R, acting at the joint O. Since the rods are uniform, the weight W acts at its center of mass, which is at a distance of a/2 from the joint O. Therefore, the moment of this weight about O is Wa/2. Similarly, the moment of the reaction R about O is Rb.

In order for the rod to be in equilibrium, the sum of the moments about O must be equal to zero. This can be expressed as:

Wa/2 + Rb = 0

Solving for R, we get:

R = -(Wa/2)/b = -aW/2b

Since the reaction R acts at an angle @ to the vertical, we can express R as R = (aW/2b)tan@. Comparing this with the previously obtained expression for R, we can conclude that:

(aW/2b)tan@ = -aW/2b

Hence, we have proven that the reaction at the joint is equal to (aW/2b)tan@, as given in the problem. This result is consistent with the intuition that the reaction at the joint will depend on the weight of the rods, the length of the rods, and the angle at which the rods are inclined to the vertical.