How Do You Calculate Tension and Wave Frequencies in a Supported Wire System?

[zalba]

Homework Statement

A wire is used to keep two uniform posts from falling, as in the figure. The wire is
5.00 m long, and has a mass of 0.732 kg. The posts are identical and are uniform in cross
section, with a mass of 25.0 kg each.
a. Find the tension in the wire (assume the wire is horizontal).
b. Find the fundamental frequency of waves on the wire, and the first two harmonics.

Also, theta is 57 degrees for both rods to the horizontal, but on opposite sides.

Homework Equations

F=ma
λ=2L
f=\frac{v}{λ}
v=\sqrt{\frac{T}{μ}}

The Attempt at a Solution

Part b seems fairly straightforward once I have solved a, but I am having some difficulty with it. My assumption would be that the tension in the wire has to equal the force of gravity to prevent it from falling over, such that T=2mgsinθ, but I have a feeling that is wrong. Also, since I am assuming the mass is uniform throughout the rope, that shouldn't have any effect on the tension right?

Thanks

 

[haruspex]

In the absence of a figure, please describe it.

 

[zalba]

Ok, so what there are are two identical rods, one leaning left with 53 degrees to the horizontal, and the other leaning to the right 53 degrees to the horizontal, both with pivot points on the ground. A wire is attached to the top of both of them, keeping them both up on the angle.

I think I figured out how to do this. I have a torque on the pivot points. If I say that L is the total length of the wire, and l as the length of one of the rod, then the torque on one pivot is

τ=\frac{TL}{2}sinθ-\frac{l}{2}mgcosθ-mgLcosθ

Is this right? I said the positive torque was pulling the rod upright, while the negative was pulling it down, and T is tension.

 

[ShouldStudy]

I said that, looking at one of the rods, that the torque = 0, since it's in static equilibrium. So therefore the wire applies a torque equal to the torque that gravity provides.

torque = 0

0 = 1/2(Tension)(length of rod)sin(θ) - 1/2(length of rod)(mg)cos(θ)


1/2(Tension)(length of rod)sin(θ) = 1/2(length of rod)(mg)cos(θ)

therefore

Tension = (mg)cot(θ)

Are you ready for the exam tomorrow? I'm guessing you're in 131 too ... godspeed.

P.S. Have you been able to solve the spinning wheel problem? I suspect it will be on the exam.

 

[haruspex]

0 = 1/2(Tension)(length of rod)sin(θ) - 1/2(length of rod)(mg)cos(θ)

Why the 1/2 in "1/2(Tension)(length of rod)sin(θ)" ?

 

[ShouldStudy]

It's half the tension in the wire. Should it just be T?

 

[haruspex]

It's half the tension in the wire. Should it just be T?

Yes. A wire in tension T is pulling what it's attached to at both ends with force T.