Answer
1.00 $\times$ $10^{2}$ N, 53.1$^{\circ}$ south of east
Work Step by Step
$F_{x}$ = 60 N east
$F_{y}$ = 80 N south
F = $\sqrt 60^{2}$ + $\sqrt 80^{2}$
= 100 N, $tan^{-1}$($\frac{80}{60}$)
$\approx$ 100 N, 53.1$^{\circ}$ south of east
Physics (10th Edition)
by Young, David; Stadler, Shane
Published by
Wiley
ISBN 10:
1118486897
ISBN 13:
978-1-11848-689-4
Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 53
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