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Homework Help: Calculating the Width of a Triangle at Position x

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Given an isosceles triangle
    - Length = L
    - Uniform Density = ρ
    - Width Varies from 0 at x = 0 to a at x = L

    I attached a picture of it.


    2. Relevant equations

    Have to show the width at position x is given by

    (a/L)x


    3. The attempt at a solution

    Now it is only two marks, but I don't have a clue where to even begin.
    Could anybody help?
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2012 #2

    tiny-tim

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    hi chris! :smile:

    hint: similar triangles? :wink:
     
  4. Feb 20, 2012 #3
    Cheers for the answer Tim, but I still don't get it :(
     
  5. Feb 20, 2012 #4

    tiny-tim

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    have you done "ordinary" geometry, with similar triangles, congruence, etc?
     
  6. Feb 20, 2012 #5
    In the past, yeah.

    edit: wait! I get it.
    God it is so simple.
    I couldn't get the idea of integrals out of my head, I guess because they are used later in the question!

    Cheers mate,
    I'll probably be back asking for help in the next question haha.
     
    Last edited: Feb 20, 2012
  7. Feb 20, 2012 #6
    1. The problem statement, all variables and given/known data
    Following on from this it asks you to find the mass ΔM contained in a strip of width w and length Δx.
    In terms of x, δx, p, t, a and L

    2. Relevant equations

    Mass = Density x Volume

    3. The attempt at a solution

    My initial reaction was to go along the lines of

    (ΔM/M) = (Area of strip/ Total Area)

    And then substitute in for mass as ρV, then substitute in for V, but then I ended up with

    Δm = Δx, which is clearly wrong.

    Is that along the right lines at all?

    Edit - This time I got Δm=wptΔx, which seems more right that before, but I'm still not sure.
     
    Last edited: Feb 20, 2012
  8. Feb 20, 2012 #7

    tiny-tim

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    hi chris! :wink:
    yes, the volume is w*t*∆x, so the mass is that times ρ :smile:
     
  9. Feb 20, 2012 #8
    Glad I'm finally getting somewhere then.

    So if I wanted total mass I'd have to then integrate that, which would be

    M = ∫ ρ w t Δx, with the limits of L and 0.

    Which would be

    M = ρt ∫w Δx, with the same limits.

    Right?

    Thank you so much for the help by the way.
     
  10. Feb 20, 2012 #9

    tiny-tim

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    that's right :smile:

    (and of course you write w as a function of x, from the first part)
     
  11. Feb 20, 2012 #10
    Which would give

    (pt a/L) ∫ x Δx ?
     
  12. Feb 20, 2012 #11

    tiny-tim

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    yup! :biggrin:

    (oh, except that with an ∫ , we write dx not ∆x :wink:)
     
  13. Feb 20, 2012 #12
    yeah I presumed so and ended up with

    1/2 * ptaL ?
     
  14. Feb 20, 2012 #13

    tiny-tim

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    that's right :smile:

    (and that agrees with area of a triangle = 1/2 aL)
     
  15. Feb 20, 2012 #14
    Awesome, thanks so much.

    Mind if I check one more answer with you?

    I've just wanted to know if the centre of mass I found is correct.

    Centre of mass is given by
    C.O.M = ∫(x*dM)/M
    = ∫(x wpt Δx)/M
    = (pt/m) ∫ ( x w dx )

    Substitute in for w, leading to
    = (pta/mL) ∫ x^2 dx with limits L and 0

    Leading to final answer

    C.O.M = (1/3M) * L^2 pta

    Would I have to substitute in for the total mass as well, or leave it as it is?
     
  16. Feb 20, 2012 #15

    tiny-tim

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    yes of course …

    the result should just be a number times L, shouldn't it? :wink:
     
  17. Feb 20, 2012 #16
    ah of course, which means

    (1/3).* ( pta L^2 / wpt Δx)
    (1/3).* ( a L^2 / w Δx )

    Substitute in for w = (a/L)x

    (1/3).* ( L^3 / x Δx)

    How do I get rid of the x's?
     
  18. Feb 20, 2012 #17

    tiny-tim

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    where did all those x and ∆x come from?? :confused:

    what happened to your …
    ?
     
  19. Feb 20, 2012 #18
    I substituted in
    M = wρtΔx
    So

    C.O.M = (1/3)*( ρta L^2/ wρt Δx )

    The ρt cancels, leaving

    C.O.M = (1/3)*( a L^2/ w Δx )

    I then substituted in
    w = (a/l) x

    So
    C.O.M = (1/3)*( a L^2/ (ax/l) Δx)

    The L on the bottom goes to the top as it is 1/L and the a cancels
    Leaving

    C.O.M = (1/3)*( L^2/ (x Δx)
     
  20. Feb 20, 2012 #19

    tiny-tim

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    i don't get it! :cry:

    you had …
    and, for the mass, M= …
    now combine them
     
  21. Feb 20, 2012 #20

    Argh no, I'm such an idiot, I was putting in the mass of a strip!

    Now its really simple!

    Now I get 2L/3
     
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