physics problem dealing with balance

[rwguinn]

Yes, exactly. For example A, if the weights are imbalanced, the bar will be tilted vertically with the heavier weight lowest. If it's 40kg vs 40kg, the weights exert no net torque, so the bar can now remain horizontal---or indeed at any position you put it in. (Tilt it to 30 degrees, steady it, and let go---it'll stay there. Tilt it to horizontal, same behavior.)

For example B, any given choice of weights will give a different equilibrium angle. It's gradual and continuous. At X=0 it'll equilibrate at (say) 45 degrees clockwise (how far depends on the details). At X=10kg it'll tilt less, say 30 degrees, and at X=40kg, equilibrium requires the bar to be horizontal.

Um... no. unless the central support is a pin joint.
It is only in equilibrium if the beam is horizontal (perpendicular to the "g" vector. Draw a free-body diagram.
That's why it's called a balance...

 

[ben m]

Um... no. unless the central support is a pin joint.
It is only in equilibrium if the beam is horizontal (perpendicular to the "g" vector. Draw a free-body diagram.
That's why it's called a balance...

You're right that it depends on the central support. If the pivot is at the center-of-mass of the beam, then any angle can be maintained at equilibrium. If the pivot is above the CoM (which is how you'd usually build a balance) then the equilibrium is horizontal.

 

[rwguinn]

You're right that it depends on the central support. If the pivot is at the center-of-mass of the beam, then any angle can be maintained at equilibrium. If the pivot is above the CoM (which is how you'd usually build a balance) then the equilibrium is horizontal.

that should be the system CoM, should it not?

 

[Furcifer]

I have a unique problem that I cannot find an answer to.

[qimg]http://www.internationalskeptics.com/forums/picture.php?pictureid=5304&albumid=304&dl=1324057954&thumb=1[/qimg]

In the diagram, Ex. A shows a simple balance, with the weights on both sides attached to the cross beam with string. To simplify the problem, both strings are attached equidistant to the pivot point, lets say at 40 centimeters. When both weights are 40 kilograms, the whole thing is balanced, with the cross beam parallel to the ground.

In Ex. B, the weight on one side is replaced with another beam, 80 centimeters long, that is fixed to the crossbeam at a 90 degree angle. The distance from the pivot point to the center (width wise) of the fixed beam is still the same as the distance to where the string of the other weight is attached. If the fixed beam is also 40 kilograms, will the cross beam still balance out parallel to the ground? If not, what amount of weight is needed at the string side to balance the cross beam?

As you have it written you have to add the weight of the string to the side with the fixed beam. Usually they include a "neglect the mass of the string" caveat.

 

[Isentropic Frontkick]

Not quite. The sum of the moments about the center point is zero when it is in static equilibrium. The sum of the moments about all other points between the weight attachments is non-zero (and zero for the small segments beyond the weight attachment points).

Draw a free body diagram (don't forget the tension of the string that suports the crossmember and acts in the positive y-direction) and you'll see that the sum of moments about any given point is zero if the system is in static equilibrium.

Or you can just trust your colleagues...

In static equilibrium, the net torque about any point must be zero.

naughty, naughty...
Don't mix internal moment with external (applied) moment.

 

[ben m]

that should be the system CoM, should it not?

Well, sort of. If you're looking at a beam-with-no-weights, or generally a rigid object of any sort, then yes, I mean the system CoM. And that's what I was thinking of.

If you're in the system with weights on strings, the "center of mass" by itself isn't a useful quantity to begin with, because the object (bar+strings+weights) is flexible; the relationship between the COM position and the rotation of the bar is nontrivial. (For example, if the strings are very long the center of mass is obviously far below the fulcrum, but that fact doesn't tell you anything about the stability in the same way that it would for a rigid bar itself.)

 

[rwguinn]

Well, sort of. If you're looking at a beam-with-no-weights, or generally a rigid object of any sort, then yes, I mean the system CoM. And that's what I was thinking of.

If you're in the system with weights on strings, the "center of mass" by itself isn't a useful quantity to begin with, because the object (bar+strings+weights) is flexible; the relationship between the COM position and the rotation of the bar is nontrivial. (For example, if the strings are very long the center of mass is obviously far below the fulcrum, but that fact doesn't tell you anything about the stability in the same way that it would for a rigid bar itself.)

D'oh! yer rite...