Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems: 45b

Answer

Tension in the cable $ =24.3$ $kN$

Work Step by Step

Let $W$ be the weight (Given as $27.8kN$) So, $W = mg = m\times 9.8 m/s^{2}$ This means that mass $ =27800 N \div 9.8 = 2836.7$ $kg$ Acceleration given in this case $= - ( 1.22$ $m/s^{2} )$ So, total Tension in cable is: $W+ma = 27800N+ [2836.7 \times (-1.22) ]$ $W+ma = (27800-3860.7) N$ $W+ma= 24.3$ $kN$
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