Tension in rigid bodies

Why does the rod AB not have a tension directed towards A?

7D question 13 of Edexcel Applied 2

I don't have access to specific exam questions or materials. However, in general, a rod AB might not have tension directed towards point A if the rod is under compression or bending forces rather than tension forces. It's also possible that the question is referring to a different scenario that requires more context to understand. If you could provide more information or context, I might be able to help you better.

The full question is as follows:
A uniform rod AB of length 2a m and mass m kg is smoothly hinged at A. It is maintained in equilibrium by a horizontal force of magnitude P acting at B. The rod is inclined at 30 degrees to the horizontal with B below A".

I found P using the principle of moments, but am struggling with part b of the question: "Find the magnitude and direction of force exerted on the rod by the wall".
Actually, I have given more thought to the question and understand it better, but now have a different query:
I've read that I should resolve vertically and horizontally to find the components of the reaction at the hinge A. Why does the reaction not just act normal to the wall, like in the case of a ladder? How come there is also a vertical component?

Thank you.

For your first question, the reason why the tension in the rod AB is not directed towards A is because there is no force pulling on the end A. The only force acting on the rod is the horizontal force P at point B, which causes the rod to be in tension. Since there is no force pulling on point A, there is no tension directed towards it.

For your second question, the reason why the reaction force at the hinge A has both horizontal and vertical components is because the rod is inclined at an angle to the horizontal. When a rod is inclined, the force exerted on it by the wall will not be perpendicular to the rod's length. Instead, the force will be resolved into two components: a perpendicular component that acts normal to the wall, and a parallel component that acts parallel to the wall.

To maintain the equilibrium of the rod, we need to balance the horizontal component of the force exerted by the wall with an equal and opposite force at the hinge A, and we need to balance the vertical component of the force exerted by the wall with an equal and opposite force at the hinge A. This is why the reaction force at the hinge A has both horizontal and vertical components.

Thank you, my understanding is better now! So the rod is in fact under tension but at point B, not A. But why then can we ignore this tension at B when resolving the whole system vertically and horizontally?

Thank you, my understanding is better now! So the rod is in fact under tension but at point B, not A. But why then can we ignore this tension at B when resolving the whole system vertically and horizontally?

The rod is in tension (or compression) which internally transmits a force along the rod, but its probably better to focus on the forces acting on the rod. Simply, the force at B is a horizontal force which tries to pull the rod away from the wall and which must be countered by a force at A (the wall/hinge) pulling it into the wall as the rod is in equilibrium. Similarly, the weight of the rod is a force pulling the rod down. It must be countered by a vertical force at A (the wall/hinge) pulling it upwards.

A sketch usually helps, but here tension is not an external force acting on the rod so should not be included on the diagram/balance equations. It internally transmits forces along the rod so you can relate/balance what happens at A and B.

Yes, that's correct - the rod is under tension at point B due to the horizontal force acting on it.

When resolving the whole system vertically and horizontally, we can ignore the tension at point B because the rod is assumed to be in equilibrium. This means that the sum of the forces acting on the rod in both the horizontal and vertical directions must be zero.

In the horizontal direction, there are two forces acting on the rod - the horizontal force P at point B and the horizontal component of the reaction force at point A. These two forces must be equal and opposite, otherwise, the rod would not be in equilibrium.

In the vertical direction, there are three forces acting on the rod - the weight of the rod, the vertical component of the reaction force at point A, and the tension force at point B. The weight of the rod and the vertical component of the reaction force at point A must be equal and opposite in order to balance each other out. This means that the tension force at point B must also be equal in magnitude to the weight of the rod plus the vertical component of the reaction force at point A, otherwise, the rod would not be in equilibrium.

Therefore, we can ignore the tension force at point B when resolving the whole system vertically and horizontally because it is already taken into account by the other forces acting on the rod in both the horizontal and vertical directions.