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Homework Help: Weight distribution

  1. Jul 10, 2007 #1

    rdx

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    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?
     
  2. jcsd
  3. Jul 10, 2007 #2

    olgranpappy

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    put a scale under each leg and then read the scales.
     
  4. Jul 10, 2007 #3

    rdx

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    LOL. However, when I say "work out" I mean mathematically.
     
  5. Jul 10, 2007 #4
    Sum the forces and moments.
     
  6. Jul 10, 2007 #5

    rdx

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    Yeah, that works for three legs. Try it for four.
     
  7. Jul 10, 2007 #6
    You have 4 unknown forces.

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

    I dont 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.
     
  8. Jul 10, 2007 #7

    rdx

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    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.
     
  9. Jul 10, 2007 #8
    Do another moment equation about a line that goes through two legs.
     
  10. Jul 10, 2007 #9

    rdx

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    I'm from Missouri. You'll have to show me.
     
  11. Jul 10, 2007 #10
    I dont get it. Take the moments about a line that passes through two of the legs. Its one equation with two unknowns in it.
     
  12. Jul 10, 2007 #11

    rdx

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    I don't get it. Write the eqn for me so I can see it.
     
  13. Jul 10, 2007 #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.
     
  14. Jul 10, 2007 #13

    rdx

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    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.
     
  15. Jul 10, 2007 #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 dont 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. Im not going to do the work for you.
     
  16. Jul 10, 2007 #15

    olgranpappy

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    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 text books. look up stress or stain or young modulus in the index...
     
    Last edited: Jul 10, 2007
  17. Jul 10, 2007 #16

    olgranpappy

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    calm down. lance armstrong is wrong. you are correct. the problem in textbook, though.
     
  18. Jul 10, 2007 #17

    rdx

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    Assuming I have such a text, great. If not, can you point me to a site? Thanks.
     
  19. Jul 10, 2007 #18

    olgranpappy

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    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.
     
  20. Jul 10, 2007 #19

    russ_watters

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    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.
     
    Last edited: Jul 10, 2007
  21. Jul 10, 2007 #20

    russ_watters

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    This problem does not involve deformations.
     
  22. Jul 10, 2007 #21

    rdx

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    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.
     
  23. Jul 10, 2007 #22

    rdx

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    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?
     
  24. Jul 10, 2007 #23

    Danger

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    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...
     
    Last edited: Jul 10, 2007
  25. Jul 10, 2007 #24

    olgranpappy

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    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...]
     
    Last edited by a moderator: Jul 10, 2007
  26. Jul 11, 2007 #25

    Gokul43201

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    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.
     
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