Essential University Physics: Volume 1 (4th Edition)

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

Chapter 12 - Exercises and Problems - Page 225: 62


$T= 1,456 N $

Work Step by Step

We know that the tension is what opposes the torque due to gravity. Thus, we first find the distance from the base of the ligament to each center of mass: $d_1 = \frac{.28}{cos27}=.31 \ m$ $d_2 =\frac{.76}{cos27}=.85 \ m$ Thus, we find: $.31(29)(9.81)sin40^{\circ}+.85(68)(9.81)sin63^{\circ}=.85Tsin27^{\circ}$ $T= 1,456 N $
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