# Find depression in wire with hanging mass

• bodensee9
In summary, This conversation is about finding the depression of the midpoint of a wire when a mass is hung from it. The wire is extended by (1/1000) of its unstretched length when a mass is hung from its bottom end. The wire is then connected between two points on the same horizontal level, and the mass is hung from the midpoint. The equation 2Tsin(theta) = mg is used to find the depression, and the approximation sin(theta) = y/(the stretched length of l/2) is suggested. Another method using Pythagoras is also mentioned. The final answer is L/20, but it is unclear how the (1/1000) factor is incorporated in the solution.

#### bodensee9

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.

Hello bodensee9!

(have a theta: θ )
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.