Spring constant of a horizontal wire

Syd096
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Homework Statement



When a mass "m" is hung from the centre of a horizontal wire, length "L". How far will the centre of the wire stretch downwards, in terms of m, L, E (youngs modulus) and A (cross sectional area of wire)

Homework Equations



mg = Tcosθ
When m is the mass hung from the wire
And T is tension in the wire

F=EAΔL/Lo

Where E is youngs modulus
F will be the tension in the wire?
ΔL is the amount the wire stretches
Lo is the original length of wire

The Attempt at a Solution



mg = Tcosθ
T=mg/cosθ

mg/cosθ=EAΔL/Lo

but is it posible to change this equation so it is only in terms M,Lo,E,A

Or find the effective spring constant of the wire
 
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This can be a wickedly hard problem. How tight is the horizontal wire before the mass is hung on it?
 
No value given. Possible add another variable (Original tension)
 
Syd, welcome to PF!

I have a few comments and questions.

1. It sounds like we assume zero tension initially, since none was given.

2. How are you defining θ? I'm guessing it's the angle of the wire w.r.t. horizontal, but I'm having trouble with this equation:
mg = Tcosθ
Did you draw a free-body diagram first? How many forces does it have?

3. After eliminating T, we are left with 2 unknowns, θ and ΔL. Can you find another equation (using geometry) to relate them?
 
Also note that the linear stress-strain relation is only valid within the elastic regime, which means that it is typically applied for very small values of ΔL/L (ΔL/L<<1).
 
It is a spring only withing the elastic range, so by definition, asking for a spring constant restricts the discussion to the linear elastic range.
 
Draw the picture, first with the wire horizontal and then with the wire stretched to the extended position with the weight in place. Then think about how much the wire has been stretched, and also draw an FBD in the drooped position.
 
Dr.D said:
It is a spring only withing the elastic range, so by definition, asking for a spring constant restricts the discussion to the linear elastic range.
There is no reason to speak in terms of a spring constant if the stress-strain relation is used (and required by the problem statement). Linearity and smallness are typical assumption that the OP needs to be aware of; the spring constant is not.
 
Actually, the linear range is a part of the stress-strain relation.
 
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Dr.D said:
Actually, the linear range is a part of the stress-strain relation.
If you mean "the stress-strain relation" that Syd should probably use in the solution, then I agree. If you mean "the" general "stress-strain relation", then I disagree.
 
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