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physics problem dealing with balance

Careyp74

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Careyp74
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Feb 29, 2008
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I have a unique problem that I cannot find an answer to.

picture.php


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?
 
I'm pretty sure they'd balance out. The upper beam has the same amount of force pulling it down on each side, at the same distance from the center point (assuming we neglect the weight of the string on the hanging weight side).

Even in the first one, if one string were twice as long as the other, assuming the total weight/mass stayed the same on each side, I believe it would remain level.
 
Hellbound said:
I'm pretty sure they'd balance out. The upper beam has the same amount of force pulling it down on each side, at the same distance from the center point (assuming we neglect the weight of the string on the hanging weight side).
Click to expand...

I agree. The vertical position of the mass isn't relevant, only its horizontal distance from the fulcrum, so the fact that the mass is spread out vertically doesn't change its effect on that side of the lever.
 
If you pushed down on the side with the string weight, you would feel resistance because of the vertical beam now starting to stick out past the point it is connected, right? Or not, I am not sure.
 
What you have there is exactly how those dial scales in the grocery store work. More weight in the basket on the string side pulls it down, the solid leg swings out enough to balance things. The amount of tilt to the cross beam shows on the meter face as units of weight.

Everything vertical, weights are even. Even weights, beam horizontal.
 
It appears the beam balances when counterclockwise torque equals clockwise torque (at the pivot point).

The torques from the horizontal beam already balance in any case. The clockwise torque from the vertical arm is 1.6 kilogram-meters (assuming 40 cm from pivot point to the vertical arm's center line). That is indeed balanced by a 40 kg weight hanging from the left of the beam.

What makes the problem seem more complicated is that the torque from the rigidly attached vertical arm varies in a different (and somewhat more complex) way with the angle of the balance arm, when it is out of horizontal, than the torque from the hanging weight. (In fact, there doesn't appear to be enough information in the problem to say exactly how it varies for the second case, since the length and mass of the balance arm are not given.) But the problem only requires achieving a balance with the arm horizontal, so that potential complexity doesn't come into play.

Respectfully,
Myriad
 
Careyp74 said:
If you pushed down on the side with the string weight, you would feel resistance because of the vertical beam now starting to stick out past the point it is connected, right? Or not, I am not sure.
Click to expand...

Yes, that's true. The two systems will not act identically when the beam is not vertical, for the reason you pointed out.
 
Thirded.

Assume that the weights are suspended from a point and are free to rotate about that point, so that the centre of mass of each weight reaches equilibrium directly below that point. The only quantity that then matters is the mass suspended from each point; the spatial distribution of that mass (which is what this question is getting at) has no effect. The beam on the right hand side will hang vertically downwards; when the central beam is horizontal, the right hand beam will be at 90 degrees to it. The angle may now be fixed without changing anything.

Dave
 
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Careyp74 said:
If you pushed down on the side with the string weight, you would feel resistance because of the vertical beam now starting to stick out past the point it is connected, right? Or not, I am not sure.
Click to expand...

Yes, you would, and that's what makes the difference between the two cases interesting. In the former case, with weights on both ends, the bar is in equilibrium at any angle, not just when it's horizontal. In the latter, it's only in equilibrium when it's horizontal.

Dave
 
OK, I think I see where my thought processes started to drift. Let me change it around a little to fully understand it.

If you start out with the weight present on the right side of each example, and on the left side started with 1 kilo, the apparatus would be rotated clockwise. Then, adding one more kilo at a time, when you reach 40 kilos, example A would be at equilibrium, but with no reason for the crossbar to move horizontal yet. Would B start moving before 40 kilos?
 
Careyp74 said:
OK, I think I see where my thought processes started to drift. Let me change it around a little to fully understand it.

If you start out with the weight present on the right side of each example, and on the left side started with 1 kilo, the apparatus would be rotated clockwise. Then, adding one more kilo at a time, when you reach 40 kilos, example A would be at equilibrium, but with no reason for the crossbar to move horizontal yet. Would B start moving before 40 kilos?
Click to expand...

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.
 
Careyp74 said:
OK, I think I see where my thought processes started to drift. Let me change it around a little to fully understand it.

If you start out with the weight present on the right side of each example, and on the left side started with 1 kilo, the apparatus would be rotated clockwise. Then, adding one more kilo at a time, when you reach 40 kilos, example A would be at equilibrium, but with no reason for the crossbar to move horizontal yet. Would B start moving before 40 kilos?
Click to expand...

Yes, I think.

Because with 1K on the left side, the bar will swing clockwise, meaning the 40K rigid bar is tilted to the inside. That would change the center of gravity on that side of the fulcrum, meaning it essentially has a shorter "lever" to pull it horizontal with.

Someone check me on that, please.
 
ok, thanks guys. They never took it this far in Physics class, for some reason.
 
Sum of forces in the y direction is the same in both cases, and sum of moments about any given point is also the same: all equal zero.

What is relevant about the beam when it replaces the weight is the resultant force it creates. Since it is vertical, the centroid is along its center axis, thus it behaves the same as the weight. As others have said - if the beam were at an angle, then it's centroid would move, thus the vector of its resultant would change, and likely there would be forces in the x direction too, thus the sum of moments changes and the apparatus shifts.
 
Isentropic Frontkick said:
Sum of forces in the y direction is the same in both cases, and sum of moments about any given point is also the same: all equal zero.
Click to expand...

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).
 
Myriad said:
<snip>

What makes the problem seem more complicated is that the torque from the rigidly attached vertical arm varies in a different (and somewhat more complex) way with the angle of the balance arm, when it is out of horizontal, than the torque from the hanging weight. (In fact, there doesn't appear to be enough information in the problem to say exactly how it varies for the second case, since the length and mass of the balance arm are not given.) But the problem only requires achieving a balance with the arm horizontal, so that potential complexity doesn't come into play.

Respectfully,
Myriad
Click to expand...

I would think that as long as the horizontal support has its center of mass at the pivot point, its length and mass should be irrelevant, since it will contribute no net torque about the pivot point. If the vertical arm is skewed out so that its center of mass is further from the pivot point than the ball on the left, the system will not be in equilibrium and will rotate clockwise.
 
Newtons Bit said:
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).
Click to expand...

In static equilibrium, the net torque about any point must be zero.
 
Newtons Bit said:
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).
Click to expand...
naughty, naughty...
Don't mix internal moment with external (applied) moment.
 
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