Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 218: 56


$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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