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Tension in a circular wire with force radiating outwards

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Let's say there's a uniform circular wire with radius [itex]r[/itex]. 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. [itex]dF = \frac{10}{2\pi r} dL[/itex]. For the component in the y-direction, [itex]dF = \frac{10}{2\pi r} \sin{θ} dL[/itex] by some simple geometry. [itex]dL = rθ[/itex], so we now have [itex]dF = \frac{10}{2\pi} \sin{θ} dθ[/itex]. Integrating from [itex]θ=0[/itex] to [itex]θ=\pi[/itex], we get that [itex]F=\frac{10}{\pi}[/itex], 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. jcsd
  3. Jan 14, 2013 #2

    haruspex

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    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?
     
  4. Jan 14, 2013 #3
    Oh, the force I found, [itex]\frac{10}{\pi}[/itex] wouldn't be the tension because the force gets divided evenly into 2. Then the tension would be [itex]\frac{5}{\pi}[/itex]?
     
  5. Jan 15, 2013 #4

    haruspex

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