# Force exerted on a pivot point of a rigid bar

• Gaeun Kim
In summary, a uniform rigid bar is supported by a rope at a 30° angle with the horizontal. The force exerted on the pivot point of the bar, represented by vector O, must have an upwards component in order to balance the vertical forces of tension and weight and maintain torque balance. This can be determined by taking moments about the point where the rope is attached to the bar and using the fact that the lines of action of three forces must intersect.
Gaeun Kim

## Homework Statement

A uniform rigid bar of weight W is supported in a horizontal orientation as shown by a rope that makes a 30° angle with the horizontal. The force exerted on the bar at point O, where it is pivoted, is best represented by a vector whose direction is which of the following?

## The Attempt at a Solution

I could understand that a horizontal force pointing to the right is needed at the pivot point to counteract Tx. (The tension force exerted by the rope is T.) There should also be a force pointing upward, which counteracts W, the weight of the rigid bar. Since there is already Ty, I thought that there is no need for another upward force on pivot point O. The answer sheet says (B) is the answer, but right now I'm confused if the answer should be (C) or (B).

I don't see your attachment. If you take a free body of the rod you have a x and y comp of T, where does the y comp go for equilibrium and what is T_y in terms of W?

Gaeun Kim
We can't access your google account to see a picture you have stored there. Use the upload button to upload it to the forum.

Gaeun Kim

Thanks for letting me know that the image wasn't posted! I did not find the upload button at first.
What I am confused about is why a downward force would be needed at point O when weight W is already counteracting T_y. Or is W not enough to counteract T_y? How do I know its magnitude when I am not given any information about it?

Gaeun Kim said:
What I am confused about is why a downward force would be needed at point O when weight W is already counteracting T_y

Downwards??

It's a statics problem so not only do the forces sum to zero but the moments/torques must as well. The rod isn't undergoing angular acceleration so the net moment about any point must be zero.

Lets try summing the moments about the middle of the rod...

W acts at the middle so can be ignored. You have Ty acting anti clockwise so there must be another force acting on the rod somewhere acting clockwise. eg the force at O must have an upwards component.

CWatters said:
Downwards??

It's a statics problem so not only do the forces sum to zero but the moments/torques must as well. The rod isn't undergoing angular acceleration so the net moment about any point must be zero.

Lets try summing the moments about the middle of the rod...

W acts at the middle so can be ignored. You have Ty acting anti clockwise so there must be another force acting on the rod somewhere acting clockwise. eg the force at O must have an upwards component.

No, moving the pivot point changes the entire question.

mixture said:
No, moving the pivot point changes the entire question.
Nobody said anything about moving the pivot point of the bar. For the purpose of analysing the torque balance, you can choose any point you like as your origin for determining the torques. They will all yield the same result, but some more readily than others.

Btw, the thread is five years old.

haruspex said:
Nobody said anything about moving the pivot point of the bar. For the purpose of analysing the torque balance, you can choose any point you like as your origin for determining the torques. They will all yield the same result, but some more readily than others.

Btw, the thread is five years old.

I'm just confused because if you choose the center of mass as the origin, suddenly you can determine that point O has a torque. In reality it doesn't because it is the pivot point, and applying any force on the pivot won't generate any torque. By pivot point it means the entire mass rotates about it I assume.
I think I know why this is the answer. You can't have the vertical force of T and W balancing each other out and their torques be the same because of the difference in moment arm. So if you equate their torques to be equal, the force on point O can have a upwards component to balance out the remaining vertical forces. It really depends on if Ty is greater than W

mixture said:
if you choose the center of mass as the origin, suddenly you can determine that point O has a torque
Yes, it has a torque about the centre of mass, no problem with that. A free pivot just cannot exert a torque about itself. Any force has a torque about every point that does not lie on its line of action.
mixture said:
It really depends on if Ty is greater than W
No, it does not depend on that.
Consider taking moments about the point where the rope is attached to the bar. T has no moment about there. W has an anticlockwise moment, so the pivot on the left must be exerting a clockwise moment.

There is a useful trick when only three forces act on a static body. If you see where the lines of action of two of those intersect, the third must also pass through that point. So take a straight line up from the bar's mass centre to where it hits the rope. The force from O must go through that point.

## 1. What is a pivot point?

A pivot point is a fixed point around which a rigid object can rotate or turn.

## 2. What is force exerted on a pivot point?

Force exerted on a pivot point is the amount of force applied to a rigid object at the point where it is attached to a pivot. This force causes the object to rotate around the pivot point.

## 3. How is the force exerted on a pivot point calculated?

The force exerted on a pivot point is calculated by multiplying the distance from the pivot point to the point where the force is applied by the magnitude of the force. This is known as the torque or moment of force.

## 4. How does the position of the force affect the pivot point?

The position of the force in relation to the pivot point affects the pivot point in two ways. First, the amount of force applied will determine the amount of torque or moment of force. Second, the direction of the force in relation to the pivot point will determine the direction of rotation of the rigid object.

## 5. What factors affect the stability of a rigid bar on a pivot point?

The stability of a rigid bar on a pivot point is affected by the magnitude and direction of the force exerted, the distance between the pivot point and the point where the force is applied, and the weight and distribution of weight on the rigid bar.

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