Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 4 - Exercises and Problems - Page 73: 74

Answer

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

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.