Force exerted on a pivot point of a rigid bar

[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?
https://lh4.googleusercontent.com/MjprKKQrEIV7CuIYgPr4p6mVmgwciTI_6Hg5XV1gNgiL_0XKXsAFoo9nEXSdYP9JbenrfWd1WZ0MAC0=w1920-h918-rw

Homework Equations

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).

 

[PhanthomJay]

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?

 

[CWatters]

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?

 

[CWatters]

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 T_y 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.

 

[mixture]

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 T_y 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.

 

[haruspex]

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.

 

[mixture]

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

 

[haruspex]

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.

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.