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Force of tabletop

  1. Mar 27, 2016 #1
    A ball of mass 15.47kg rests on a table of height 0.73m. The tabletop is rectangular with a mass of 12.11kg, supported by 4 thin legs. The width of it is 134.0cm and length of 63.7cm. If the ball touches the tabletop at point (67.0cm, 16.4cm) relative to corner 1, what is the force the tabletop exerts on each leg?

    2. Relevant equations

    F=ma
    ????

    3. The attempt at a solution
    I'm kinda lost as to how to solve this. I tried finding the area of each segment then getting a percentage of it and multiplying the total mass and gravity but that didnt make sense considering the ball and the table should exert more force on corner one(bigger area, bigger total force). So that was definitely wrong. I feel I'm missing something important equation wise
     
  2. jcsd
  3. Mar 27, 2016 #2
    As its a case of static equilibrium - your equation F=ma will not work.
    rather draw the force diagram and take moment about convenient axis/point so that you can find the answer.
     
  4. Mar 27, 2016 #3
    Yeah, static equilibrium sounds about right but I'm not sure how I should go about it.
     
  5. Mar 27, 2016 #4

    SteamKing

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    First thing, draw a sketch of the table top looking down on it and indicate the position of the ball relative to one of the corners as described.

    The table top is supported by one leg at each corner. The sum of the forces in each leg must equal the weight of the table top and the ball. That's the first equation of equilibrium you must work with.

    Taking corner 1 as a reference, the sum of the moments of the weight of the table top, the ball, and the forces in the legs must equal zero for equilibrium to occur.
    Write a moment equation for these items about each of the table sides which intersect at corner 1.

    Using these equations should allow you to solve for the forces in each leg holding up the table top.
     
  6. Mar 27, 2016 #5
    So would it be F - (mt+mb)g = 0 ?
     
  7. Mar 27, 2016 #6
    I'm kinda lost as to how to incorporate the coordinates with the equation and such
     
  8. Mar 27, 2016 #7

    SteamKing

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    Well, let's start with the basics. Do you know what the moment of a force is? Do you know how to calculate it?

    If F is supposed to be the total force of all four legs, then this equation is correct.

    However, because the ball is not located at the center of the table top, the force in each leg will probably be different.
     
  9. Mar 27, 2016 #8
    Yeah M=Fd right?

    Yes, sum F.

    I did think about dividing up by 4 initially but they are definitely uneven.
     
  10. Mar 27, 2016 #9
    i think you have insufficient data.
    one thing you can try take the table as of length and breadth say a and b now you can fix position of your 'mass, on the table as given.
    the weight of the table will act at the crossing of diagonals
    so its equidistant from the corners ,therefore each leg will support Mg/4. now you must do some geometry exercise to find the distance of your additional mass from the corners ,then you can take moments of this mass from the the corners /as well as centre to get relationships to find the loads felt by the legs.
     
  11. Mar 28, 2016 #10

    haruspex

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    Isn't that four unknowns but only three equations? I don't think taking additional axes will provide any more information.
    I agree with @drvrm, insufficient data. In the real world, it comes down to exact leg lengths and compression modulus.
     
  12. Mar 28, 2016 #11

    SteamKing

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    I think that given the location of the ball on the table, one can make a reasonable inference about which pair of legs have equal forces in them, and thus provide enough additional constraints to solve this problem.

    Sometimes, and Intro Physics problem is just an Intro Physics problem, no matter how badly phrased.
     
  13. Mar 28, 2016 #12

    haruspex

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    Unfortunately that inference is unwarranted. E.g. consider just the table's own weight. By symmetry, one might suppose all four legs get the same load, but you could reduce the load on one pair of opposite legs by some amount w, and increase the load on the other two by w, and still have a valid solution.

    That said, if we assume the table top and floor are perfectly rigid and flat, and model the legs as four identical springs, then a fourth equation emerges, and this does lead to the solution you imply.
     
  14. Mar 28, 2016 #13

    SteamKing

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    And the table could have one or more wobbly legs. The possibilities are endless.
     
  15. Mar 28, 2016 #14

    haruspex

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    You seem to be missing the point.
    I have no problem with making assumptions about the table being extremely accurately made, but that does still not provide enough equations. You have to make some further assumptions, such as my rigid floor and tabletop yet springy legs, to get a fourth equation. You could instead suppose rigid legs and a springy top. That would also allow a solution, but, in general, a different one.
     
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