Weight Distribution: Calculating Leg Weight

In summary, the conversation discusses ways to determine the distribution of weight on the legs of a table or a robot. The suggested methods include using scales to measure the forces and moments, using equations for force and moment balances, and taking moments about a line passing through two legs. The conversation also mentions the usefulness of textbooks or online resources for solving similar problems involving indeterminate structures and stresses. Finally, the conversation clarifies that this problem does not involve deformations and suggests considering symmetry when solving for weight distribution.
  • #1
rdx
50
0
Suppose I have a table with 4 legs and on this table I have a concentrated weight at an arbitrary location. How do I work out the distribution of weights on the legs?
 
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  • #2
put a scale under each leg and then read the scales.
 
  • #3
olgranpappy said:
put a scale under each leg and then read the scales.

LOL. However, when I say "work out" I mean mathematically.
 
  • #4
Sum the forces and moments.
 
  • #5
cyrusabdollahi said:
Sum the forces and moments.

Yeah, that works for three legs. Try it for four.
 
  • #6
You have 4 unknown forces.

(1)- Force Balance
(2-4) - moment balance

I don't see why it would be statically indeterminate as of yet. You should show some work so I can see if it is indeterminate or not.
 
  • #7
cyrusabdollahi said:
You have 4 unknown forces.

(1)- Force Balance
(2-4) - moment balance

I don't see why it would be statically indeterminate as of yet. You should show some work so I can see if it is indeterminate or not.

Okay, 3 legs:

w1 x1 + w2 x2 + w3 x3 = 0 ; moment in x-axis
w1 y1 + w2 y2 + w3 y3 = 0 ; moment in y-axis
w1 + w2 + w3 = W ; force balance.

These 3 equ can be solved for three unknowns, ie the w's.

Now throw in a fourth leg, a w4, you have a problem because there is no fourth eqn.
 
  • #8
Do another moment equation about a line that goes through two legs.
 
  • #9
cyrusabdollahi said:
Do another moment equation about a line that goes through two legs.

I'm from Missouri. You'll have to show me.
 
  • #10
I don't get it. Take the moments about a line that passes through two of the legs. Its one equation with two unknowns in it.
 
  • #11
cyrusabdollahi said:
I don't get it. Take the moments about a line that passes through two of the legs. Its one equation with two unknowns in it.

I don't get it. Write the eqn for me so I can see it.
 
  • #12
You know how to take the sum of moments, right? Sum the moments about a line that has an axis through two of the legs, any two. I expect you to come up with the eqn.
 
  • #13
cyrusabdollahi said:
You know how to take the sum of moments, right? Sum the moments about a line that has an axis through two of the legs, any two. I expect you to come up with the eqn.

In other words you don't know. This clearly won't work generally. How about 5 legs, or 6? No, I am doubtful. I listed the known eqns, yours is bogus.
 
  • #14
Stop being lazy, do you, or do you not, know how to take a moment balance around a line? They should have taught you this in statics 101.

Yes, I don't know for sure if it is statically indeterminate. Thats why I said to try some things out first to see if it is or not. I am not going to do the work for you.
 
  • #15
yes. with four legs you do not have enough equations, you need one more involving stresses or strains. or something, I can look up how to do it later, but I think this problem should be solved in most intro textbooks. look up stress or stain or young modulus in the index...
 
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  • #16
rdx said:
LOL, did you mean piss or bliss? Y R U trying to insult me?

calm down. lance armstrong is wrong. you are correct. the problem in textbook, though.
 
  • #17
olgranpappy said:
calm down. lance armstrong is wrong. you are correct. the problem in textbook, though.

Assuming I have such a text, great. If not, can you point me to a site? Thanks.
 
  • #18
How about a library? I looked up "Young's modulus" in my old intro textbook (Halliday Resnick and Walker, "Fundementals of physics") and within a few pages I found a section on "Indeterminate Structures." There is a picture of a pink elephant sitting on a four legged table... it's exactly the problem you are interested in...

to find webpages maybe just google "Elasticity" or "Indeterminate Structures" or "Pink Elephants." Maybe not that last one.
 
  • #19
Most tables are symmetrical, so assuming this one is, you could just split it into two one-dimensional center of mass problems. That's probably functionally equivalent to what cyrus said, but maybe makes it easier to understand how it could work.

I don't see anything particularly difficult about this problem either - it is just slightly more involved than the most basic one-dimensional problem.
 
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  • #20
olgranpappy said:
yes. with four legs you do not have enough equations, you need one more involving stresses or strains. or something, I can look up how to do it later, but I think this problem should be solved in most intro textbooks. look up stress or stain or young modulus in the index...
This problem does not involve deformations.
 
  • #21
olgranpappy said:
How about a library? I looked up "Young's modulus" in my old intro textbook (Halliday Resnick and Walker, "Fundementals of physics") and within a few pages I found a section on "Indeterminate Structures." There is a picture of a pink elephant sitting on a four legged table... it's exactly the problem you are interested in...

to find webpages maybe just google "Elasticity" or "Indeterminate Structures" or "Pink Elephants." Maybe not that last one.

I have a Halliday, Resnick text called Physics. I will look for Young's. I also have a Mechanics by Symon. I will see what I can find. Thanks.
 
  • #22
russ_watters said:
Most tables are symmetrical, so assuming this one is, you could just split it into two one-dimensional center of mass problems. That's probably functionally equivalent to what cyrus said, but maybe makes it easier to understand how it could work.

I don't see anything particularly difficult about this problem either - it is just slightly more involved than the most basic one-dimensional problem.

Perhaps I should spell out the whole problem. I am designing software to simulate a 6-legged robot and part of this software analyzes the interaction of the robot with Newton's laws. As the robot walks or stands, I am trying to work out the loading on each leg. There are typically three legs down at a time while walking but they are not necessarily symmetrically located. Does this suggest anything in way of solutions?
 
  • #23
I'm probably way over-simplifying this, but can't you just consider the 3 legs that are down to be a tripod, and thus divide the weight by 3? All 6-legged robots that I've seen so far use an insect stride (2l and 1r, then 1l and 2r). That does mean symmetrical distribution, doesn't it? :confused:

edit: Okay, I know that there's something wrong with that somewhere...
 
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  • #24
russ_watters said:
This problem does not involve deformations.

Yes it does--if all four legs touch the ground. Haven't you ever sat at a four-legged table before? They wobble. Only three legs will hit the ground at a time. If you apply enough pressure the thing won't wobble.

A normal four-legged table that doesn't wobble only doesn't wobble because there is enough weight from the table itself to slightly deform the legs (which must of course have slightly different lengths) s.t. all four legs touch the ground.

Granted that in the case of a walking spider robot we are probably not interested in deformation, rather we are interested in which three of the six legs will hit the ground in any given position of the spider. But in the case of a table--the case which was originally presented--and which was answered incorrectly (repeatedly and rather abusively answered incorrectly at that)--one *does* need to consider deformations.
[insult deleted - watch it...]
 
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  • #25
cyrus, there are no more than two linearly independent direction vectors in a given plane - in this case the plane of the floor. Picking a third direction along this plane does not result in a third independent axis. In particular the direction you suggested is parallel to [itex]\hat{x} + \hat{y} [/itex], which were the two directions previously chosen. That makes this a statically indeterminate structure - and it's important to recognize this without having to go through a trial and error process - and the additional equation required to solve the problem must follow necessarily from appropriately set up deformation relationships.
 
  • #26
I have deleted a dozen posts of Cyrus and Olgranpappy bickering. Knock it off guys. Next time there well be warnings for annoying behavior for all involved.

I do not think any thing relating to the actual problem has been lost.
 
  • #27
I apologize for being so annoying with my pointless prodding.
 
  • #28
Hmmm, interesting.

I find it fascinating that an apparently simple problem like a table is so hard/impossible to solve. You wonder at all the weird theories about string theory and all, maybe people need to get back to basics and find some new techniques for evaluating the "simple" stuff. Anyway, thanks everybody.
 
  • #29
rdx said:
I find it fascinating that an apparently simple problem like a table is so hard/impossible to solve.

The problem is not "hard" to solve at all. You just need to make a CORRECT model of the situation, which involves the flexibilility of complete table (flexible legs and a flexible top, in general).

If you assume the table is a rigid body, there is no unique solution. Consider vertical forces of +1 on the two legs on one diagonal, and forces of -1 on the other two legs. The resultant forces and moments about any point are zero.

If you make some simplifying assumptions (e.g. the legs all have the same flexibility and the top is rigid), it's easy enough to find a closed form solution using energy methods. See any textbook on solid mechanics for the details of how to do it.

Some special cases are easy to solve by symmetry (e.g. if the mass is at the center of the table).

Statically determinate structures, where you CAN calculate reaction forces without considering the deflections of the structure, are a special situation that hardly ever occurs in reality. They are very common in exercises in elementary mechanics textbooks though, because you can practise working with forces and moments without worrying about anything else.
 
  • #30
You're funny

AlephZero said:
The problem is not "hard" to solve at all. You just need to make a CORRECT model of the situation, which involves the flexibilility of complete table (flexible legs and a flexible top, in general).

If you assume the table is a rigid body, there is no unique solution. Consider vertical forces of +1 on the two legs on one diagonal, and forces of -1 on the other two legs. The resultant forces and moments about any point are zero.

If you make some simplifying assumptions (e.g. the legs all have the same flexibility and the top is rigid), it's easy enough to find a closed form solution using energy methods. See any textbook on solid mechanics for the details of how to do it.

Some special cases are easy to solve by symmetry (e.g. if the mass is at the center of the table).

Statically determinate structures, where you CAN calculate reaction forces without considering the deflections of the structure, are a special situation that hardly ever occurs in reality. They are very common in exercises in elementary mechanics textbooks though, because you can practise working with forces and moments without worrying about anything else.

I defined the problem. Now you're telling me I need the "correct" model. Then you change the problem with simplifying assumptions; then you don't answer you wave your arms and claim text solve it? Whew, you must be tired out. But where's the solution? No, it's still hard, you just don't understand.
 

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