# Sag in wire due to a weight (given the Young's modulus)

1. Apr 23, 2016

### ezioauditore

1. The problem statement, all variables and given/known data
A mild steel wire of length 1 m and cross-sectional area 0.5*10^-2 cm^2 is stretched within elastic limit horizontally between two pillars.A mass of 100 g is suspended from mid-point.Depression at mid-point?

2. Relevant equations

Sag in metal rod is=WL^3/(4BD^3Y) where W=weight, l=length,b=breadth,d=depth,Y=Young's modulus.But here its a wire.
3. The attempt at a solution
I tried to calculate the k of the wire since its stretched within its elasticity limit and found out the increase in length of the wire.But could not relate the sag to increase in length.

2. Apr 24, 2016

### SteamKing

Staff Emeritus
Your sag formula for the metal rod appears to be developed for a rod with a rectangular or square cross section, which is not the cross section of a typical steel wire.
The wire is not going to be in bending; it supports the load by remaining in tension.

You should analyze this problem from first principles by drawing a free body diagram showing the weight suspended between two supporting points. Without getting into catenaries and stuff, you can assume each part of the wire suspending the weight is straight. You want the tensile stress in each part of the wire to be less than the elastic limit, whatever that number is.

3. Apr 24, 2016

### billy_joule

I think it's fair to assume the wire has zero stiffness, treat it as a string not a beam (or rod).
Draw a free body diagram of the weight and show how tension varies with angle (note how tension goes to infinity as theta goes to zero). Extension/sag depends on tension and tension depends on sag...

4. Apr 28, 2016

### CWatters

+1

I would make TWO drawings..

1) A free body diagram showing the forces acting on the weight. This should lead you to equations - hint it's a statics problem.
2) A diagram showing the wire straight and deflected by the weight. Some geometry will give you an equation for the change in length of the wire(s).

Eventually you will have enough equations to solve it all.