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Calculating weights on a slope using virtual work.

  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A cart on an inclined plane is balanced by the weight w. All parts have negligible friction. Find the weight W of the cart using the principle of virtual work. Picture of set-up is attached.


    2. Relevant equations
    [tex] ∫Fdh = 0 [/tex] for virtual displacement dh in statics
    [tex] F_p = W cos \theta [/tex] the part of the weight of the cart parallel to the slope
    [tex] ∑F = 0 [/tex]
    [tex] W = 4w/sin\theta [/tex]
    This is the given solution, by the book.
    3. The attempt at a solution
    I tried two ways to get the solution: virtual work and forces but I got stuck in the virtual work solution, and I did not get the right answer with forces. So here is the force analysis first:

    There are two pulleys in the diagram: one attached to the slope at the bottom, holding w, and the other attached at the top holding W. Lets call the one holding W P, and the other p. The tension in P must be such that it balances the parallel component of W, as given by the formula in Relevant equations. Lets call that tension T.
    The perpendicular component of W can be ignored because it is balanced by the normal force. Because P supports p, the downwards force on p must be equal to T.

    I assumed that both p and P are massless. Then the force on p is w, plus the pull of the slope on the other side. Since the strings in p are not moving, the pull of the slope must equal that of w. The downwards force on p is the sum of the two, namely 2w. This must equal T, and since T is the parallel component of W, by replacing I end up with:

    [tex] 2w = W sin \theta [/tex]
    Which is missing a factor of 2 to get the correct answer. Where did I miss that factor?

    To apply the principle of virtual work, I let w fall a distance equal to the height of the slope, h. Then the point of the slope attached to p will rise by h, and the point attached to P rises by a h cos(theta). If P rises by h cos(theta), the string holding p will lengthen by h cos(theta), so the cart moves along the hypotenuse by this distance. The slope rotates around the point without any pulley by theta.
    I drew two triangles to represent the slope. In the rotated one, I drew another triangle to figure out the height difference, and it was proportional to h. I then used the change in height of the pulley to calculate the change in height due to the pulley pulling the cart. Or rather I tried. I have absolutely no idea of how to do that because it does not form a triangle. Of course, I could calculate the height change again using proportionality, but then I end up with two proportionality constants and I have no idea of how to find them.


    Anyone can help me? Is my approach correct, and if so how do I calculate the change in height for W? If not what would be the correct approach, that is, what do I vary to apply the virtual work? Also why is my Force analysis missing a factor of 2? Thank you for answering the questions.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2017 #2

    haruspex

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    Draw a free body diagram for the pulley attached to the cart.
    (Not sure what you mean by "tension in P". P is a pulley, with one string attached and one around it. Which string has the "tension in P"?
     
  4. Jan 20, 2017 #3
    the tension in the rope around the highest pulley must be 2w , so the pull on the cart must be 4w
     
  5. Jan 20, 2017 #4
    Is it 2w because there are two strings holding back the cart? If so, then I got it, thank you.
    I still need to solve this with virtual work though.
    Any ideas of how I could use virtual work to solve this?
     
  6. Jan 20, 2017 #5

    haruspex

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    You have to work out the change in height of w for a small change in height of W (or vice versa).
     
  7. Jan 20, 2017 #6
    I tried dropping w and going through the calculations, (there are in the original post), but I'm getting stuck because I need the proportionality constant, which is dependant on the horizontal position of W and I can't find it. Any ideas on how I could find it? Thanks
     
  8. Jan 20, 2017 #7

    haruspex

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    If the cart moves a distance s up the slope, how much does it rise in the process?
     
  9. Jan 20, 2017 #8
    It should rise a distance s sin ø from trig.
    However, in doing so, the string attached to the cart must shorten by s, and conversely, the string holding the other pulley should lengthen by s. The pulley holding the weight w is lowered by s as well. Because of this the slope will rotate anticlockwise by a certain amount, and the weight W will fall a little because of this. I'm having difficulty computing by how much the weight will fall. I set up similar triangles to calculate the drop in height, but I can't calculate the proportionality constant. How could I proceed with this?
     
  10. Jan 20, 2017 #9

    haruspex

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    You mean the string that goes around the pulley attached to the cart?
    There are two portions to that string parallel to the slope, one above the pulley and one below. Which portion will shorten by s?
     
  11. Jan 20, 2017 #10
    Both I would think, because they would have to cover less distance to the cart, and since there must be tension in them, they must be taut, but I am not sure. Does it matter though? Because the one below the pulley won't do any work right? The length on top will have to decrease anyway because otherwise there is no tension
     
  12. Jan 20, 2017 #11

    haruspex

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    Right. So if the cart moves distance s, how much rope that was parallel to the slope no longer is?
    At the moment we're just trying to figure out how far things move. We'll deal with work later.
     
  13. Jan 20, 2017 #12
    2s I would think, since it would be s for each string, one above the pulley, and the one below. I'm not sure if I understood your question well though. I'm supposing you mean the amount of string in all that moves out of the way. But since only the upper half of the string is attached to the pulley holding w, w would drop by only s as well.
     
  14. Jan 20, 2017 #13

    haruspex

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    Right, the total length of string parallel to the slope has reduced by 2s. Where has it all gone?
     
  15. Jan 20, 2017 #14
    I think the string holding the pulley holding w increases by 2s. It's hard to tell what would happen to the string below the pulley, but I believe it should run across the pulley attached to the cart, then the one holding the string attached to the cart, so that the string holding the pulley holding w increases by s, and since I'm shortening the string above the pulley as well by s, the overall increase in length would be 2s.
     
  16. Jan 20, 2017 #15

    haruspex

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    The string as a whole does not change length, but yes, the vertical portion, from the top pulley to the lower pulley, would increase by 2s.
    Not sure about the rest of what you wrote in that post, but I think you have the idea.

    We have established that the lower pulley descends by 2s, yes? There is a string running over that pulley, with two vertical portions. One portion is fixed to the ground. If the pulley descends 2s, what happens to the length of that portion of string? Where does it go?
     
  17. Jan 21, 2017 #16
    Never mind what I wrote. Just my ramblings about how I calculated that 2s.
    As to the string attached to the ground, I think it decreases by 2s. If the pulley is lowered, the string attached to the ground goes slack since it cannot move. If it goes slack there can't be any tension, and if there is no tension, w falls. It will fall until the string is taut again.

    The string is taut again if w falls by 2s. So w falls by 4s in all, and that's the solution.

    Thanks a lot for taking the time to help me. I know it was long, several days in fact, and I am not easy to deal with. Thank you for bearing with me and guiding me through the solution.
     
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