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Resistance of a Trapezium*

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data
    The shape is 3D trapezium with a base A and a length L. The front end is shorter than the back. The front is height h and the back height Y.

    What is the resistance of a current moving from h to Y?


    2. Relevant equations
    R=ρL/A


    3. The attempt at a solution

    First to determine the area:
    [itex]\frac{z-y}{x}[/itex]
    [itex]z=\frac{x}{L} (y-h)+h[/itex]
    That times A (the base) gives area

    The integral:
    [itex]\int \frac{\rho dx}{A\frac{x}{L} (y-h)+h}[/itex]

    evaluated from 0 to L I get:
    [itex]\frac{\rho L}{A(y-h)} ln(\frac{y}{h})[/itex]

    This seems straight forward but I have a problem. The next step of the problem asks that I set the limit of y to zero so that the previous solution will give the resistance of a rectangular solid. This doesn't seem to work as the first half will be negative and the natural log becomes undefined.

    I'm not sure that the integral is correct as my calculator, wolfram alpha and myself get differing answers. Though I think I made a mistake that puts me on wolframs side now. Also, the equation for the area may be incorrect though I am pretty sure that I got it correct.

    Any assistance would be appreciated.
     
  2. jcsd
  3. Nov 18, 2012 #2

    gneill

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    Staff: Mentor

    From your description of the geometry, if you let y go to zero won't you end up with a wedge rather than a rectangular solid? Perhaps the height of the 'back end' is meant to be y+h?
     
  4. Nov 18, 2012 #3
    I tried to draw it but it ended up looking terrible.


    So it is basically a right triangle sitting on top of a rectangle when you look at it from the side. Also it is 'A' deep from this perspective and both are 'L' long. The point of the triangle is (the short side) is height 'h' and the back side is height 'y.' So when 'y' goes to zero the triangle goes away. It is not quite y+h. Rather you could say it is h+(some other variable)=y.

    Trying to find a picture of one on the internet is surprisingly difficult.

    Also, for some reason my area equation got cut off:

    [itex]\frac{z-y}{x}=\frac{y-h}{L}[/itex]
     
  5. Nov 18, 2012 #4

    gneill

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    I still don't see how if, the back end has height y, when y goes to zero the back end can be anything but zero height. That makes it the thin end of a wedge.

    attachment.php?attachmentid=53088&stc=1&d=1353256880.gif
     

    Attached Files:

  6. Nov 18, 2012 #5
    Shoot, you're right. The back end is y+h. I don't know why I made that mistake. So, you'd be right if it is as I said. You're drawing is correct although except for that. The current runs right to left as pictured there.

    Sorry about that mistake.
     
  7. Nov 18, 2012 #6
    My area has to be wrong. I was using that Y as my total back end length in my area equation.

    Can I simply modify it as so or is it totally wrong now: [itex]z=\frac{x}{L} ((y+h)-h)+(y+h)[/itex]?
     
  8. Nov 18, 2012 #7
    Okay, here is where I'm at:

    Area:


    [itex]\frac{z-h}{x}=\frac{y+h-h}{L}[/itex]

    [itex]\frac{z-h}{x}=\frac{y}{L}[/itex]

    [itex]z=\frac{x y}{L}+h[/itex]

    [itex]\int \frac{\rho dx}{A\frac{x y}{L}+h}[/itex]

    [itex]\frac{\rho}{A}\int \frac{dx}{\frac{x y}{L}+h}[/itex]

    from 0 to L I get

    [itex]\frac{P}{A}\frac{L}{y}ln(\frac{y+h}{h})[/itex]

    So, in my final answer if y goes to zero I have a problem. But if y=0 in my integral it works out quite nicely to

    [itex]\frac{P}{A}\frac{L}{h}[/itex]

    Which is what I believe a rectangular solid to be. I'm just worried now that perhaps my original answer was intended to have the limit set to zero which would make it answer incorrect.

    Does this look ship shape to you?
     
  9. Nov 18, 2012 #8

    gneill

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    If h is the height of the rectangular slab, and y is the "back end" thickness of the triangular wedge sitting on it, then you can describe the height w.r.t. x (x being the distance from the "back end") as:

    z = (h+y) - x*(y/L)

    The cross sectional area is then z*A
     
  10. Nov 18, 2012 #9

    gneill

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    The problem is tricky because the straightforward approach leads to an indefinite form when y is allowed to go to zero after integration. It would be better to apply the limit before integrating.

    Note that the expression for the area can be rearranged:
    $$Area = z A = \left((h + y) - \frac{x}{L}y\right)A = \left( \left(1 - \frac{x}{L}\right)y + h\right)A$$
    If y is allowed to go to zero then the term on the left disappears without a struggle.
     
    Last edited: Nov 18, 2012
  11. Nov 18, 2012 #10
    I get a slightly different answer from your area equation.

    [itex]\int \frac{\rho dx}{A(h+y)-x\frac{y}{L}}[/itex]

    [itex]-\frac{\rho}{A}\frac{L}{y}ln(\frac{h}{h+y})[/itex]

    With my area:
    [itex]\frac{\rho}{A}\frac{L}{y}ln(\frac{y+h}{h})[/itex]
     
  12. Nov 18, 2012 #11

    gneill

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    Yeah, sorry, I missed carrying the base width "A" though the area rearrangement; I edited my previous post accordingly.

    Note that if you apply the limit y → 0 in your result you end up with the indefinite form 0/0.

    Either apply the limit before integration (simple) or deal with the indefinite form afterward (L'Hopital).
     
  13. Nov 18, 2012 #12
    I went ahead and inserted the A into the integral before. We're very similar but for the log and the negative sign.

    Actually, I think we're exactly the same except for your negative sign which actually cancels from the log since h/(h+y) is less than 1.

    That's sort of an interesting result. Are we both correct then or is that just a wildly unlikely coincidence?
     
  14. Nov 19, 2012 #13

    gneill

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    Staff: Mentor

    It's no coincidence that the results are the same since it was only an algebraic rearrangement of the integrand.
     
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