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Missing step in engineering problem

  1. Mar 30, 2006 #1
    Hi there,

    I'm reading over a section in my statics text that describes an analysis of a cable with a distributed load, but it's been a long time since I've done this stuff, and I'm stuck on just one step. :grumpy:

    Here's how it goes. I'm fine with everything up to this equation:
    -T∙cos(θ) + (T + ΔT)∙cos(θ + Δθ) = 0

    That just comes from balancing the forces in one direction on a Δx long segment of the cable. So far so good. But then the text says "dividing this equation by Δx and taking the limit as Δx→0, Δθ→0 and ΔT→0, we obtain:" and then gives this:
    d(T∙cos(θ))/dx = 0

    Now, I'm trying to see how they got that, and failing.:redface: Likewise, for the forces in the y direction, they go from:
    -T∙sin(θ) - ω(x)∙Δx + (T + ΔT)∙sin(θ + Δθ) = 0

    To:
    d(T∙sin(θ))/dx - ω(x) = 0

    Now, the ω(x) component I have no problem with, but again, how do they go from "[-A∙f(B) + (A + ΔA)∙f(B + ΔB)] / Δx" to "d(A∙f(B))/dx" simply by taking the limits as the Δ's go to 0?
     
  2. jcsd
  3. Mar 30, 2006 #2
    I just had a thought... could it be that:
    (T + ΔT)∙cos(θ + Δθ) - T∙cos(θ) ≡ Δ[T∙cos(θ)]

    Or is there still something I'm missing? :rolleyes:
     
  4. Mar 30, 2006 #3
    I try something

    -T∙cos(θ) + (T + ΔT)∙cos(θ + Δθ) = 0 (1)

    but

    (T + ΔT)∙cos(θ + Δθ) = (T + ΔT)[cos(θ)∙cos(Δθ)-sin(θ)∙sin(Δθ)] (2)

    equivalently:

    (T + ΔT)∙cos(θ + Δθ) =
    T∙cos(θ)∙cos(Δθ)-T∙sin(θ)∙sin(Δθ) +
    ΔT∙cos(θ)∙cos(Δθ)-ΔT∙sin(θ)∙sin(Δθ) (3)

    when Δθ is small, the sine is practically Δθ and the cosine is one. So we have from (3):

    (T + ΔT)∙cos(θ + Δθ) =
    T∙cos(θ)-T∙Δθ∙sin(θ) +
    ΔT∙cos(θ)-ΔT∙Δθ∙sin(θ) (4)

    ΔT∙Δθ∙sin(θ) can be neglected (product of very small quantities). So we have from equation (4):

    (T + ΔT)∙cos(θ + Δθ) =
    T∙cos(θ)-T∙Δθ∙sin(θ) +
    ΔT∙cos(θ) (5)

    Therefore, from equation (5) and (1) we deduce:

    -T∙cos(θ) + (T + ΔT)∙cos(θ + Δθ) =
    -T∙cos(θ) + T∙cos(θ)-T∙Δθ∙sin(θ) +
    ΔT∙cos(θ) = 0

    simplifying:

    -T∙cos(θ) + (T + ΔT)∙cos(θ + Δθ) =
    -T∙Δθ∙sin(θ) + ΔT∙cos(θ) = 0

    When you "divide" by Δx you get:

    -T∙Δθ/Δx∙sin(θ) + ΔT/Δx∙cos(θ) = 0

    Observing that

    Δθ/Δx-->dθ/dx

    ΔT/Δx-->dT/dx

    we have:

    -T∙dθ/dx∙sin(θ) + dT/dx∙cos(θ) = 0 (6)

    Finally, observing that

    d(T∙cos(θ))/dx = dT/dx∙cos(θ) -T∙dθ/dx∙sin(θ),
    we get your answer.
     
  5. Mar 30, 2006 #4
    Of course! :cry: I got as far as Eq3, but completely forgot about the small angle trick.

    And it took me almost an hour of thinking before I finally realized what happened here too. :blushing:

    I've been out of school for almost 5 years now, and getting back into the mindset is a real pita. It's all slowly coming back though! Thank you for your help!
     
  6. Mar 30, 2006 #5
    I am glad I could help :)
     
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