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Horizontal cable with one vertical point force

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Hang a weight (F) from the middle of a horizontal wire of length L, attached on both sides. The wire has an effective modulus (E) and effective area (A). Find an equation for the new length of the wire LF. (NOTE: Do not try to solve for a form LF =, this is too hard).

    HW23.jpg

    2. Relevant equations
    To see the equations I thought were relevant, see my attempt at the solution.


    3. The attempt at a solution
    Force Balance Equation: T = [itex]\frac{Fcosθ}{2}[/itex]

    Compatibility Equation: ∂ = ∂1 + ∂2. ∂1 = ∂2.

    Force-Displacement Equation: ∂1 = [itex]\frac{TL}{2EA}[/itex] = [itex]\frac{FLcosθ}{4EA}[/itex].

    This gives: ∂=2∂1 = [itex]\frac{FLcosθ}{2EA}[/itex]

    Which gives a final answer of: LF=L+∂ = L + [itex]\frac{FLcosθ}{2EA}[/itex].

    But I'm not confident in my work, because professor said it would be too difficult to solve for the solution in this form.
     
  2. jcsd
  3. Jan 29, 2012 #2
    From where do you measure θ? What is θ when F goes to zero?
     
  4. Jan 29, 2012 #3
    Theta is the angle made at the support ends. Theta is the same on the left and right side. When F=0, Theta=0, theoretically.
     
  5. Jan 30, 2012 #4
    Shouldn't you have then,

    2Tsinθ = F
     
  6. Jan 30, 2012 #5
    Okay I've got the same problem as part of a homework.
    I agree that you need 2Tsin(A)=F
     
  7. Jan 30, 2012 #6
    So then, with the new force balance equation, we get:

    1 = [itex]\frac{TL}{2EA}[/itex] = [itex]\frac{FL}{4EAsinθ}[/itex]

    Which gives us:

    ∂=∂1 + ∂2 = 2∂1 = [itex]\frac{FL}{2EAsinθ}[/itex]

    So the final answer then would be:

    LF=L+[itex]\frac{LF}{2EAsinθ}[/itex]
     
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