Physics Tension Q

[cclawfjj]

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A thick uniform cable, suspended from the ceiling, hangs straight down. How does the tension at the top of the cable compare to the tension at the bottom of the cable?

 

[cclawfjj]

 

[LifetimeDoc]

How about in terms of this image? 😀

I'd definitely say the "tension" is much greater at the bottom of the cable.

Note: in theory, yes a massless cable will have the same tension at the top and bottom.

 

[future_dr_house]

In theory, tension is the same throughout the cable.

 

[xanthomondo]

In theory, tension is the same throughout the cable.

i think thats true if the weight of the cable is ignored....wouldnt the weight of the cable cause there to be more tension at the top? (as there is more weight from the cable below it)

thats why the slinky is pulled out more towards the top than the bottom

im probably wrong

 

[degoo_]

tension is equal

 

[degoo_]

I assume this is a physics I question, where the mass of the cable/other forces are ignored.

 

[future_dr_house]

i think thats true if the weight of the cable is ignored....wouldnt the weight of the cable cause there to be more tension at the top? (as there is more weight from the cable below it)

thats why the slinky is pulled out more towards the top than the bottom

im probably wrong

you are totally right. but if it's massless, then T is equal.

if the cable has mass...then this page helps: http://www.cord.edu/dept/physics/p128/lecture99_13.html

T1 = T2 + mstring*a.


where a = gravity in the hanging string case.

 

[cclawfjj]

so if we assume the mass is present, the top of the cable has more tension? If its a massless spring, uniform tension thruout?

 

[em783]

I think in terms of the MCAT, all cables, strings, etc. are considered massless (unless otherwise noted in the question), making T equal throughout. But in reality, it's changing throughout and I would assume there is greater T at the top of the cable as someone had mentioned.

 

[deleted103644]

You can tell what the right answer will be just by looking at the picture. The tension at any point is proportional to the space between links on the slinky (Hooke's law).
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Since this cable is not accelerating, at any point the tension up is equal to the tension down, which is the weight of the cable beneath it. (You can calculate it the other way as well, as the force of the ceiling minus the weight of the cable above the point.)
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So if the cable has mass M and length L, at a point x units from the ceiling the tension will be equal to..
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T(x) = Mg(L-x)/L = Mg(1-x/L)
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e.g. T(0) = Mg, T(L/2) = Mg/2, T(L) = 0.
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Looking at the picture you will see the links on the slinky follow a similar pattern in spacing.