Calculate tension in a rope (pulley-rope-mass system)

AI Thread Summary
The discussion centers on calculating the tension in a rope within a pulley-mass system involving a 2.00 kg textbook on a frictionless surface and a 3.00 kg hanging book. Initially, the tension is assumed to be equal throughout the rope at 29.4 N due to equilibrium. However, once the system is released, the tension changes, leading to different values for each section of the rope, which can be 7.5 N and 18.2 N. It is noted that the pulley can cause variations in tension, and the same-tension rule only applies to sections directly connecting objects. To accurately determine the tension, a force diagram for each mass should be drawn, relating net forces to acceleration.
MarliesM
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Mhmm okay thanks, so if the rope had a mass, the two people would exert a force to make sure the rope didn't fall..

But now what about this (still can't solve a tension problem); a 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150m to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20m in 0.800s. (a) What is the tension in each part of the cord?

My attempt; before the system is released from rest, the tension equals 3.00 x 9.81 = 29.4 N. This is because the rope is in equilibrium and no forces act except at its ends, so the tension is the same at both ends and throughout the rope. (Right?)

But I don't understand what to do after the system starts moving. There is only one force acting, so the tension should be 3.00 x 2.00 x 9.81 = 58.9 N, but the answer is 7.5 N, 18.2 N.
 
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Hi,

Since this is a specific physics problem, I have moved your post to the Homework & Coursework Questions area from the other thread where we were discussing rope tension.

... a 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150m to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20m in 0.800s. (a) What is the tension in each part of the cord?

My attempt; before the system is released from rest, the tension equals 3.00 x 9.81 = 29.4 N. This is because the rope is in equilibrium and no forces act except at its ends, so the tension is the same at both ends and throughout the rope. (Right?)
Well, the pulley could cause the two sections of rope to have different tensions.
But I don't understand what to do after the system starts moving. There is only one force acting, so the tension should be 3.00 x 2.00 x 9.81 = 58.9 N, but the answer is 7.5 N, 18.2 N.
A pulley can cause the two sections of rope to have different tensions, and that appears to be the case here. The same-tension-through-the-rope rule only applies to a section of rope that connects objects at each end; i.e. the 2.0kg-to-pulley section, and separately to the pulley-to-3.0kg section of rope.

To solve the problem, draw a force diagram for each mass, then relate the net force to the acceleration of that mass.
 
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