1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculating the Width of a Triangle at Position x
Loading...