balance beam question

[newmacnow]

I have a balance beam question. See sketch. with the beam supported in the middle and a weight hung across it approximately 24" long. the left side drops down roughly 1/2". Is there a way to calculate the weight difference from one end to the other? Even if it is only a ratio. total weight is unknown.

 

[MiketheEngineer]

First - double posting frowned upon.

Sure, simple statics 101.

 

[newmacnow]

I know about the double post. I realized I posted it in the wrong forum, so I reposted it here.
How do you account for the fact that it isn't in equilibrium? How do you take into account that the bar tilts 1/2". Obviously if the bar tilts 1/2" with this uneven load, but say an 1" with another uneven load, how do you calculate the weight from side to side.

 

[SnTMan]

Why do you say it isn't in equilibrium? Does it continue to drop?

Regards,

Mike

 

[gruntguru]

Start with the knowledge that the centre of gravity lies vertically below the hanger for any equilibrium state, with or without added weights.

Engineering is the art of creating things you need, from things you can get.

 

[chicopee]

what is that green stuff under the beam?

 

[chicopee]

Discounting the green stuff, the center of gravity of the beam is slightly left of the hanging member pivot point and that center is realigning itself with the hanging member pivot point by slighlty rotating CCW thereby explaing the 1/2" drop on the left end.

 

[bradleyelwood]

I really dont understand the green stuff either. But let me try to answer what I think is your question. I assume your beam is 12 inches, (more or less) either side of the knife edge pivot point. You put the beam on the pivot and the left side drops 1/2 inch relative to the right side, so the beam "rotates" 0.020833 radians out of level. Now, lets say that the vertical C of G of the beam is one inch below the pivot. The C of G hangs below the pivot and the actual horizontal C of G is 0.0208 inches off of the apparent geometric center.

 

[newmacnow]

Thank you.

The green represents the weight being measured. The grey portion represents the balanced beam that hangs perfectly level when nothing is hanging from it. When the green object is hung from the beam, the system tilts 1/2" down on the left side. So the green object which appears to be balanced, is not actually balanced. There is more weight on the left side. How much more weight is on the left side? If the object weighs 10 lbs and it was balanced, then each side of the beam it hangs from would be seeing 5 lbs. Since the left side drops 1/2" the weight must be something like 7 lbs left and 3 lbs right? I would assume that if it was 8 lb/2 lb then it might drop say 1" and if it was 6 lb/4lb it might drop 1/4"?
What we are trying to figure out is that since all we know is that the left side drops 1/2" what does that mean the weight is on the left compared to the right?
Thanks

 

[rb1957]

bradley has calc'd the cg of the weight causing the 1/2 deflection. note, this is the cg of the arm weight and the weight added,; so the CG of the weight added alone is further away from the mid-point. from this you can determine the distribution of the weight added (of the green stuff); you don't have to guess.

 

[chicopee]

So before I analyze the situation, I would like to know if the green area is a uniformly increasing load from right to the left?

 

[rb1957]

you've got W1, the weight of the arm, CG1 = 0 (ie balanced)
you've got W2 = the weight of the green stuff, CG2
you've got (courtesy of Bradley) the total CG is -0.0208, so that W2*CG2 = -0.0208*(W1+W2) ... CG2 = -0.0208*(W1+W2)/W2
now if you want to express this as Wl = the weight of W2 on the left of the balance and Wr, the weight on the right, then
Wl+Wr = W2 and Wl*CGl+Wr*CGr = W2*CG2

and you can see you can't solve this without assuming something ... like CGr = -CGl = L/2

 

[BAretired]

If the c.g. of the balance is e below the suspension point, it moves e/48 to the right in the diagram. If the balance beam weighs W, it exerts a moment of W*e/48 about the suspension point.

Balancing that is 12*d where d is the difference between the left and right load.

So 12*d = W*e/48

and d = W*e/576

BA