# Tensile stress given theta and force

1. Nov 8, 2012

### plz

Tensile stress given theta and force (I'm kinda desperate)

1. The problem statement, all variables and given/known data

Aluminum wire is lightweight. You can hang a piece of it nearly horizontally with very little tension. After having done so, you then hang a HEAVY (25 kg) block from the wire. The wire sags to make an angle of 12 degrees with the horizontal. Determine the radius of the wire.

2. Relevant equations

Stress / Strain = Modulus

Stress = Force / area = 25*9.8 / (pi)r^2

Strain = Delta L/L = [(L/cos12) - L / L] ?

Young's Modulus for aluminum = 70 x 10^9

3. The attempt at a solution

(25)(9.8) / [(pi)(r^2)] = (0.022)(70*10^9)

r = 2.25 * 10^-4

This is wrong, Please give me solid advice and not simple hints please. I'm really tired of this problem and I'd like to solve it with as little guesswork as possible. I need to have a solution in a few answers. If you can help me I will love you till the ends of the earth.

Last edited: Nov 8, 2012
2. Nov 8, 2012

### Staff: Mentor

The tension in each of the two sections of the wire is not 25 kg. Assume that the weight is suspended from the mid point of the wire, and draw a free body diagram to get the tension in each of the two sections. The vertical components of the tensions must sum to the weight of the weight.

3. Nov 8, 2012

### Sheridans

I had the same problem. Could someone just tell me where the 0.022 came from? That was the only part i was missing.

Anyway, like Chestermiller said, just do a free body. The force in the stress equation is tension, not the weight of the block.

4. Nov 8, 2012

### PhanthomJay

It actually is all there in plz's relevant equations, and represents the unit strain in the wire. The wire must stretch from its original length to the new length as measured along the diagonals of the wire in its loaded stretched position.