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Statics, equilibrium sphere in a groove on an incline plane

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The smooth homogeneous sphere rests in the 132° groove and bears against the end plate, which is normal to the direction of the groove. Determine the angle θ, measured from the horizontal, for which the reaction on each side of the groove equals the force supported by the end plate.
    20150708_173646_zpso7o7ozq3.jpg
    2. Relevant equations
    ΣF=0

    3. The attempt at a solution
    20150708_204719_zpso5sqvyu8.jpg

    edit: in the second equation F3sin(24) is supposed to be positive.
     
    Last edited: Jul 8, 2015
  2. jcsd
  3. Jul 8, 2015 #2

    haruspex

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    Your hand-drawn diagram is wrong, and your equations are wrong for both diagrams.

    This is very much a 3D problem. The printed diagram shows two different views of the same thing.
    The black and white view is looking down the slope, showing what the groove looks like in cross section.
    In the coloured view, the groove is seen side on, running up the slope from the end plate. The deepest line of the groove is shown by the lowest of the three sloping dashed lines.

    There are four forces on the ball, the three reaction forces and gravity.
     
  4. Jul 8, 2015 #3
    Ok, i dont see how 45 degrees wouldnt work then. because the sum of the two forces of the wedge = the force on the vertical plate.
    I will try writing out the sum of forces equations again.
     
  5. Jul 8, 2015 #4

    haruspex

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    It doesn't say the sum of forces from the groove equals the force from the end plate. It says the force on each side of the groove. I take this to mean three forces of equal magnitude.
     
  6. Jul 8, 2015 #5
    yea but i tried 22.5 deg
     
  7. Jul 8, 2015 #6

    haruspex

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    I don't see how my interpretation of the set-up leads to that angle.
     
  8. Jul 8, 2015 #7
    Fc = Fa+Fb at 45 degrees
    Fc=Fa=Fb at 45/2 degrees
    this is just my intuition though.
     
  9. Jul 8, 2015 #8

    haruspex

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    You can't add forces like that. Forces are vectors. These three forces act in three different directions.
     
  10. Jul 8, 2015 #9
    I edited my sketch above. I still have too many variables.
     
  11. Jul 8, 2015 #10

    haruspex

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    The very first equation there is wrong. You seem to be inconsistent over which way x and y point. The F1 term has x horizontal, while the W term (and the zero F2, F3 terms) has x parallel to the slope.
    Similarly in the second equation, y is vertical for the F1 and W terms but normal to the slope for the F2 and F3 terms. You also have a sign error there.
    You don't need any more equations. The actual values of the forces don't matter, and cannot be determined. All that matters is the ratios between them.
     
  12. Jul 8, 2015 #11
    im going to try changing my axis so that they are at the same angle as θ.

    edit: cancel that
    edit': ill change the side view to the same angle as theta and the front view will remain the same.
     
    Last edited: Jul 8, 2015
  13. Jul 8, 2015 #12
    Edited my sketch again.
    Would it be beneficial to solving the problem if i wrote out each force in unit vector notation?
     
  14. Jul 8, 2015 #13

    haruspex

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    The equations look good now. Don't forget, you also know F1=F2=F3. They're quite easy to solve - no need to go into vector notation.
     
  15. Jul 8, 2015 #14
    i dont see how this is easy to solve.
     
  16. Jul 8, 2015 #15

    haruspex

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    Look at your x and y equations. You know the three reaction forces are to have equal magnitudes, so just replace them all by F.
    That leaves you with W as the unknown you need to eliminate, so use the equations to do that. What one equation do you now have?
     
  17. Jul 8, 2015 #16
    w= (F+Fsin(24)+Fsin(24))/(sinθ + cosθ)
     
  18. Jul 8, 2015 #17

    haruspex

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    No, you still have W in there.
    You have two equations involving F, W and theta. Combine them into one equation in a way which eliminates W.
     
  19. Jul 8, 2015 #18
    I dont see how, because w is multiplied by two different trig functions?
     
  20. Jul 8, 2015 #19
    only thing i can come up with to eliminate w is to call is mg , but it makes m a variable
     
  21. Jul 8, 2015 #20

    haruspex

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    It's just standard handling of simultaneous equations. Get one equation into the form W = (some function of the other variables), then use that to substitute for W in the other equation.
     
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