Answer
$F_B=570N$
Work Step by Step
$F_T=\frac{(23kg)g}{2\sin(15)}=435N$
$F_{T+B}=\frac{(23kg)g+F_B}{2\sin(27)}
=871N$
$F_B=570N$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 9 - Static Equilibrium; Elasticity and Fracture - General Problems - Page 257: 70
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