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Tension in a rope

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Other answers:

is my diagram right?
sag in rope due to mass of rope ?
yes
then u wud take 'mg = 2T cos(theta)' is it im not sure how the entire mass of rope gets manifested at center ?
that's the tension at the end of the rope , but i'm looking for the tension in the middle
oh tension can be different thru out the rope in this system ?
since the rope has mass... things may not be ideal hmm
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Is this a rope of uniform density?
Is it in a vaccuum with a neutral charge?
yes the rope is uniform density , and has a total mass of m
Tension at the middle? Where is the middle?
yes i think we can make these assumptions
yeah in the middle of the symmetric rope
It would depend on the length between tethers and the length of the rope.
So if we're leaving this general, it's just based on the angle, which doesn't really matter to us though.
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Unless they're telling us the length of the rope and the length of the distance it's stretched across, then we're screwed.
the tensions should be a function of (m,θ,g)
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.
That's assuming all the mass is in the center?
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Since it has a uniform density, center of mass is there, so it's safe assumption.
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Yay necklaces!
dental plan
If you can solve for \(n\) beads, then you can solve by \(n\to \infty\).
integrals?
Assuming that the 'center of mass' thing works, then it's a matter of finding the right angle.
\(\color{blue}{\text{Originally Posted by}}\) @UnkleRhaukus the tensions should be a function of (m,θ,g) \(\color{blue}{\text{End of Quote}}\) We need to know the amount of slack in the rope.
Do you think you can get that with \(\theta\)?
Luckily, it's left as a variable. No fuss, no muss! Now suppose you have a necklace with finite volume and infinite surface area...
But is that angle even the right angle?
If it was a right angle, then there'd be no sag...
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