Tension in a rope

- UnkleRhaukus

Tension in a rope

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- UnkleRhaukus

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- UnkleRhaukus

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- UnkleRhaukus

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## More answers

- UnkleRhaukus

is my diagram right?

- ganeshie8

sag in rope due to mass of rope ?

- UnkleRhaukus

yes

- ganeshie8

then u wud take 'mg = 2T cos(theta)' is it
im not sure how the entire mass of rope gets manifested at center ?

- UnkleRhaukus

that's the tension at the end of the rope , but i'm looking for the tension in the middle

- ganeshie8

oh tension can be different thru out the rope in this system ?

- ganeshie8

since the rope has mass... things may not be ideal hmm

- anonymous

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- Kainui

Is this a rope of uniform density?

- Kainui

Is it in a vaccuum with a neutral charge?

- UnkleRhaukus

yes the rope is uniform density , and has a total mass of m

- anonymous

Tension at the middle? Where is the middle?

- UnkleRhaukus

yes i think we can make these assumptions

- UnkleRhaukus

yeah in the middle of the symmetric rope

- anonymous

It would depend on the length between tethers and the length of the rope.

- Kainui

So if we're leaving this general, it's just based on the angle, which doesn't really matter to us though.

- UnkleRhaukus

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- Kainui

Unless they're telling us the length of the rope and the length of the distance it's stretched across, then we're screwed.

- UnkleRhaukus

the tensions should be a function of (m,Î¸,g)

- Kainui

So the rope's not moving right? That means the forces up and forces down should be equal, right? So...
mg=2Tcos(theta)
Solve for T, problem solved.

- anonymous

That's assuming all the mass is in the center?

- anonymous

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- anonymous

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- Kainui

Since it has a uniform density, center of mass is there, so it's safe assumption.

- anonymous

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- Kainui

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- Kainui

Yay necklaces!

- UnkleRhaukus

dental plan

- anonymous

If you can solve for \(n\) beads, then you can solve by \(n\to \infty\).

- UnkleRhaukus

integrals?

- anonymous

Assuming that the 'center of mass' thing works, then it's a matter of finding the right angle.

- anonymous

\(\color{blue}{\text{Originally Posted by}}\) @UnkleRhaukus
the tensions should be a function of (m,Î¸,g)
\(\color{blue}{\text{End of Quote}}\)
We need to know the amount of slack in the rope.

- anonymous

Do you think you can get that with \(\theta\)?

- Kainui

Luckily, it's left as a variable. No fuss, no muss! Now suppose you have a necklace with finite volume and infinite surface area...

- anonymous

But is that angle even the right angle?

- Kainui

If it was a right angle, then there'd be no sag...

- anonymous

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- Kainui

mg/[2cos(theta)]=T as theta approaches 90 is... Uh oh.

- Kainui

I didn't realize, but this is gonna be infinitely heavy guys.

- Kainui

Sorry, infinitey tensioned

- UnkleRhaukus

how am i meant to find \(\alpha\) @wio ?

- Kainui

http://en.wikipedia.org/wiki/Catenary#Mathematical_description

- anonymous

I think @Kainui has the right idea.

- Kainui

@wio impossible I don't believe you. I'm going to bed.

- Vincent-Lyon.Fr

Horizontal component of tension is constant throughout because
\(T_x(x+dx) - T_x(x) = dW_x=0\) hence \(T_x(x)=T_o\) whereas
\(T_y(x+dx) - T_y(x) = dW_y = \mu g. dl\)
Solving the equations leads to equation of a catenary:
\(y(x) = \Large \frac{T_o}{\mu g} \normalsize \cosh \,(\Large \frac{\mu g}{T_o}x)\)
See http://en.wikipedia.org/wiki/Catenary

- anonymous

Tend=((m/2)*g)/cos(theta)
Tmid=m/2*g*tan(theta)

- anonymous

And if you must know how you got it.
j: |T_end | cos(theta) - (m/2)*g = 0
i: -|T_end|sin(theta)+ |T_ mid|=0
we know Tend , substitute that we get
[-(m/2)g sin(theta)]/cos(theta)+ |T_mid| =0
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