Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems - Page 253: 16

Answer

$F_A$ must be 6300N and $F_B$ must be 6100N.

Work Step by Step

Sum the torques about point B: $(+ \circlearrowleft) \sum \tau=0$ $-F_A*10m+4300N*8m+280kg*9.8m/s^2*5m+3100N*4m+2200N*1m=0$ $F_A=(4300N*8m+280kg*9.8m/s^2*5m+3100N*4m+2200N*1m)/(10m)$ $F_A\approx6300N$ Now sum the forces vertically: $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $F_A+F_B-4300N-2800N-280kg*9.8m/s^2-3100N-2200N=0$ $F_B=4300N+2800N+280kg*9.8m/s^2+3100N+2200N-F_A$ $F_B=4300N+2800N+280kg*9.8m/s^2+3100N+2200N-6300N$ $F_B\approx6100N$
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