Essential University Physics: Volume 1 (4th Edition)

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

Chapter 5 - Exercises and Problems - Page 93: 75



Work Step by Step

We know that the tension will be the largest for the line between the pot in the corner and the wall, so we solve for this tension. We also know that only the y-components of tension support the pots. Since they do not move, the y-components of tension must add to gravity. Thus: $3.85g = T_3sin54 - T_2sin13.9$ $9.28g=T_1sin68+T_2sin13.9$ Adding these gives: $13.13g = .81T_3 +.93T_1$ The pots are also not moving in the x-direction, so we find: $T_1cos54 = T_3cos68$ $T_1 = .64T_3$ Thus, using substitution, we find: $T_3 = 92 \ N$ Thus, the rope will work.
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