UnkleRhaukus
  • UnkleRhaukus
Tension in a rope
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
|dw:1380875928995:dw|
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
is my diagram right?
ganeshie8
  • ganeshie8
sag in rope due to mass of rope ?
UnkleRhaukus
  • UnkleRhaukus
yes
ganeshie8
  • 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
  • UnkleRhaukus
that's the tension at the end of the rope , but i'm looking for the tension in the middle
ganeshie8
  • ganeshie8
oh tension can be different thru out the rope in this system ?
ganeshie8
  • ganeshie8
since the rope has mass... things may not be ideal hmm
anonymous
  • anonymous
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Kainui
  • Kainui
Is this a rope of uniform density?
Kainui
  • Kainui
Is it in a vaccuum with a neutral charge?
UnkleRhaukus
  • UnkleRhaukus
yes the rope is uniform density , and has a total mass of m
anonymous
  • anonymous
Tension at the middle? Where is the middle?
UnkleRhaukus
  • UnkleRhaukus
yes i think we can make these assumptions
UnkleRhaukus
  • UnkleRhaukus
yeah in the middle of the symmetric rope
anonymous
  • anonymous
It would depend on the length between tethers and the length of the rope.
Kainui
  • Kainui
So if we're leaving this general, it's just based on the angle, which doesn't really matter to us though.
UnkleRhaukus
  • UnkleRhaukus
|dw:1380883258783:dw|
Kainui
  • 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
  • UnkleRhaukus
the tensions should be a function of (m,θ,g)
Kainui
  • 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
  • anonymous
That's assuming all the mass is in the center?
anonymous
  • anonymous
|dw:1380883547495:dw|
anonymous
  • anonymous
|dw:1380883559040:dw|
Kainui
  • Kainui
Since it has a uniform density, center of mass is there, so it's safe assumption.
anonymous
  • anonymous
|dw:1380883578211:dw|
Kainui
  • Kainui
|dw:1380883594219:dw|
Kainui
  • Kainui
Yay necklaces!
UnkleRhaukus
  • UnkleRhaukus
dental plan
anonymous
  • anonymous
If you can solve for \(n\) beads, then you can solve by \(n\to \infty\).
UnkleRhaukus
  • UnkleRhaukus
integrals?
anonymous
  • anonymous
Assuming that the 'center of mass' thing works, then it's a matter of finding the right angle.
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @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.
anonymous
  • anonymous
Do you think you can get that with \(\theta\)?
Kainui
  • 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
  • anonymous
But is that angle even the right angle?
Kainui
  • Kainui
If it was a right angle, then there'd be no sag...
anonymous
  • anonymous
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