Horizontal cable with one vertical point force

AI Thread Summary
The discussion focuses on deriving an equation for the new length of a horizontal wire (LF) when a weight (F) is hung from its center. Participants explore force balance and displacement equations, highlighting the relationship between tension (T), angle (θ), and the wire's elongation. One participant expresses uncertainty about their solution, particularly regarding the angle measurement and its implications when the force is zero. Another contributor suggests an alternative approach using sine functions in the force balance equation, leading to a revised expression for LF. The conversation emphasizes the complexity of the problem and the need for clarity on the angle's role in the calculations.
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Homework Statement


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


Homework Equations


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


The Attempt at a Solution


Force Balance Equation: T = \frac{Fcosθ}{2}

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

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

This gives: ∂=2∂1 = \frac{FLcosθ}{2EA}

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

But I'm not confident in my work, because professor said it would be too difficult to solve for the solution in this form.
 
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From where do you measure θ? What is θ when F goes to zero?
 
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.
 
Shouldn't you have then,

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

1 = \frac{TL}{2EA} = \frac{FL}{4EAsinθ}

Which gives us:

∂=∂1 + ∂2 = 2∂1 = \frac{FL}{2EAsinθ}

So the final answer then would be:

LF=L+\frac{LF}{2EAsinθ}
 
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