# Consider the equilibrium situation on the picture attached. The wire

1. Dec 15, 2011

### aaaa202

Consider the equilibrium situation on the picture attached.
The wire supports a rod attached to an axle in the wall and the mass is suspended at about half the length from the point where the where the wire is attached (the exact position is unimportant for my point).
We neglect friction from the wall. My question is: Is this equilibrium possible? Because in order for the center of mass of the rod not to fall the wires vertical force component must equal mg. But then that would mean, that it exerted a bigger torque than the mass forcing the rod to rotate anticlockwise. What would happen in this situation and is it possible to model without friction?

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2. Dec 15, 2011

### AlephZero

Re: equilibrium

Think about the 3 forces acting on the rod. For equilibrium, the moment of the forces about every point in the plane must be zero.

Now think about the point where the lines of action of two of the forces intersect. If the moment of all three forces about that point is 0, the line of action of the third force must also pass through that point.

What does that tell you about equilibrium, with or without friction at the wall?

3. Dec 15, 2011

### technician

Re: equilibrium

there must be friction at the wall or.... more likely the rod is fixed to he wall in some way.
The line of action of the force from the wall at A must pass through the intersection of the line of action of the weight of the beam (Mg) and the tension in the wire CB
If there are only 3 forces acing and the system is in equilibrium then the lines of action of the three forces must pass through one point.
This is a way of saying there is no overall turning effect

4. Dec 15, 2011

### aaaa202

Re: equilibrium

Aleph: hmm I'm sorry but I don't really understand those terms - I'm not that familiar with english, especially not physicsoriented english-speaking. Can you please explain it in another way?
The only solution I could come up with was that the vertical component of wires force was such that the torques balanced and then the friction force did the rest of the job of keeping up the center of mass, since that would exert no torque. But then, I wanted to know if it was possible without friction..
Tech: the rod is attached to an axle so it is definately not fixed. So the conclusion would be that it is not possible without accounting for friction right?

In general what do both of you mean by "the line of action"? Just the direction of the force?

5. Dec 15, 2011

### AlephZero

Re: equilibrium

Draw the vectors showing the direction of the three forces acting on the bar

You know the force from the weight acts vertically.
You know the force from the wire acts along the wire.

Those two vectors intersect at a point (half way along the wire).

Take moments about that point. The moment of all 3 forces must be 0. So the direction of the force at the wall must also pass through the same point.

But the force at the wall can only act in that direction if there is friction at the wall......

6. Dec 15, 2011

### aaaa202

Re: equilibrium

hmm I tried that but doesn't really give me anything. But hey! When you say force at the wall are you then suggesting like an upwards normal force? Because when I said friction I was more or less also referring to normal forces of any kind. Indeed that would be able to explain since the string then only would have to lift such that the torques balanced and then the normal force would do the rest of the work. Is that what you are trying to say?

7. Dec 15, 2011

### AlephZero

Re: equilibrium

If the bar is horizontal, the the force at the wall must have a horizontal and a vertical component, because the resultant of those who components must pass through the same point at the other two forces (the middle point of the string). If there is no vertical force at the wall, because there is no friction, the bar will not be in equilibrium.

The bar can be in equilibrium without friction if it is sloping downwards, so the end touching the wall is at the same height as the middle point of the string.