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Tension in rope between falling objects

  • Thread starter team31
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  • #1
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Two point-lie objects, each with mass m, are connected by a massless rope of length l. The objects are suspened vertically near the surface of Earth, so that one object is hanging below the other. Then the objects are released. How can I find the tension in the rope?
I know there 3 force, tension, Graviational Froce on each block. but how can I set them up?
my first thought was (M*m*G)/(R^2)+(M*m*G)/((R+l)^2)-T-(m*m*G)/(l^2)=(2m)G/(R^2). I know it's wrong but I don't know where's my mistake.
 

Answers and Replies

  • #2
214
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First of all, I think it is sufficient to assume R^2 = (R+l)^2.

Secondly, there are two equations of motion (one for each particle), not just one, and in each equation you consider only the forces acting the given particle. On the lower particle you have 2 forces: tension up and gravity down. Similarly for the upper particle.

The most important observation to make is that if there is tension in the rope at all, the particles will accelerate with the same acceleration. This can be seen by noticing that the z-positions of the bodies differ by a constant (the length of the rope).
 
  • #3
10
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so is that the graviational force between the tow mass is counted as part of the tension
 
  • #4
214
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I see.... I would tend to neglect the mutual gravitation of the two particles. It seems to me to be much smaller than the other quantities involved.
 

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