Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 4 - Exercises and Problems - Page 73: 74


$T = Mg + (1-\frac{y}{L})mg$

Work Step by Step

No matter where you are on the rope, there will be a gravitational force due to the mass, which is: $Mg$. However, how much rope tension there needs to be depends on how far from the top of the rope you are. Thus, it follows: $T = Mg + (1-\frac{y}{L})mg$ Note, using this equation, when at the top of the rope (y=0), tension equals $Mg+mg$, which makes sense because it has to support both masses. However, at the bottom of the rope (y=L), tension equals Mg. This makes sense, for at the bottom of the rope, the force of tension no longer has to support the rope.
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