Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.2 Exercises - Page 823: 32

Answer

$\approx 28.66 N$; $\approx12.37 ^\circ$

Work Step by Step

Top vector $= (20 cos 45) i+(20 sin 45)j =10 \sqrt 2 i+10 \sqrt 2 j$ Bottom vector $= (16 cos 30) i+(-16 sin 30)j =8 \sqrt 3 i-8j$ Horizontal component : $10 \sqrt 2 i+8 \sqrt 3 i=(10 \sqrt 2 +8 \sqrt 3 )i$ Vertical component: $10 \sqrt 2 j-8j=(10 \sqrt 2 -8)j$ $\approx 28.66 N$; $\approx12.37 ^\circ$ Resultant force: $(10 \sqrt 2 +8 \sqrt 3 )i+(10 \sqrt 2 -8)j=\sqrt {(10 \sqrt 2 +8 \sqrt 3 )^2+(10 \sqrt 2 -8)^2}\approx 28.66 N$ $tan \theta =\dfrac{vertical}{horizontal}=\frac {(10 \sqrt 2 -8 )} {(10 \sqrt 2+\sqrt 3)}$ $\theta =arctan[\dfrac{vertical}{horizontal}=\frac {(10 \sqrt 2 -8 )} {(10 \sqrt 2+\sqrt 3)}]$ $ \theta \approx 12.37 ^\circ$
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