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 217: 36


22,856 Newtons

Work Step by Step

We know that the net torque is 0, so it follows: $(15)(2500)(9.81)sin40^{\circ}+5(830)(9.81)sin40^{\circ}=15F_tsin50^{\circ}$ $F_t = \fbox{22,856 Newtons }$
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