# Tension in a circular wire with force radiating outwards

1. Jan 14, 2013

### hyperddude

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

Let's say there's a uniform circular wire with radius $r$. There is a uniform force pushing outward from the center in all directions of the wire. This total force is 10N. What is the tension in the wire?

2. Relevant equations

N/A

3. The attempt at a solution

Let's say the circle is on an xy coordinate axis with its center at the origin. I think the tension is equal to the total force in the y-direction on the top semicircle. The y-component of the force on the bottom semicircle will be equal in magnitude and in the opposite direction.

To find the the force, we integrate. $dF = \frac{10}{2\pi r} dL$. For the component in the y-direction, $dF = \frac{10}{2\pi r} \sin{θ} dL$ by some simple geometry. $dL = rθ$, so we now have $dF = \frac{10}{2\pi} \sin{θ} dθ$. Integrating from $θ=0$ to $θ=\pi$, we get that $F=\frac{10}{\pi}$, which is the tension.

I'm moderately sure my math is correct, but my real question is whether I found the tension! I'm not very sure if this correctly defines the tension in the circle.

Last edited: Jan 14, 2013
2. Jan 14, 2013

### haruspex

You mean dL = r dθ, right?
Your method is right, but you have overlooked one detail. Draw the free body diagram for one half of the loop. What are all the forces?

3. Jan 14, 2013

### hyperddude

Oh, the force I found, $\frac{10}{\pi}$ wouldn't be the tension because the force gets divided evenly into 2. Then the tension would be $\frac{5}{\pi}$?

4. Jan 15, 2013

Yup.