Find depression in wire with hanging mass

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SUMMARY

The discussion focuses on calculating the depression (y) of a wire's midpoint when a mass (M) is hung from it. The wire has an unstretched length (l) and stretches by (1/1000)l when the mass is applied. The key equations used include 2Tsin(θ) = mg and sin(θ) = y/(l/2). The solution involves using Pythagorean theorem and approximations for small angles, ultimately leading to the conclusion that the depression y equals L/20.

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bodensee9
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Hello: I have the following:

Homework Statement



A wire of unstretched length l is extended by a distance (1/1000)l when a mass M is hung from its bottom end. If this same wire is connected between points A and B that are at a distance l from each other on the same horizontal level, and the mass M is hung from the midpoint of the wire, find the depression y of the midpoint.


Homework Equations


2Tsin(theta) = mg
sin(theta) = y/(the stretched length of l/2).

The Attempt at a Solution



Not sure where to go from there. Would it be safe to say that each half of the wire is stretched by l/2000? Thanks.
 
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Hello bodensee9! :smile:

(have a theta: θ :wink:)
bodensee9 said:
Would it be safe to say that each half of the wire is stretched by l/2000? Thanks.

How did you get that?

Use Pythagoras, and an approximation.
 
I need help with this problem too. 2Tsintheta=mg, and tantheta=2y/l, for small angles tan theta=sin theta, so if we could use that approximation then sin theta=2y/l,...I can also use Pythagoras to find the hypotenuse and then find the real sin theta and equate them to find y, I don't think that's a proper answer though...I'm not sure where the L/1000 comes in...

This isn't homework, I know the answer already is L/20, just don't know how to get that..
 
mmmboh said:
...I'm not sure where the L/1000 comes in...

L/2 = adjacent, (1 + 1/1000)L/2 = hypotenuse.
 

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