Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.4 The Cross Product - 12.4 Exercises - Page 862: 41

Answer

$\approx 417 N$

Work Step by Step

$| \tau|=|r \times f|= |r||f|sin \theta$ Given: $|r|=30 cm=0.30 m$;$r=30 j$, $r \cdot \lt 0,3,-4 \gt =|r| (\sqrt {0+9+16}) cos \theta= (30) (5) cos \theta$ Hence, $90=150 cos \theta$ $ cos \theta=\frac{90}{150}=\frac{3}{5}$ $sin \theta=\sqrt {1-cos^2 \theta}=\sqrt {1-\frac{9}{25}}=\frac{4}{5}$ $| \tau|=|r \times f|= |r||f|sin \theta$ $ 100 N m=0.30 \cdot |f| \cdot \frac{4}{5}$ $ |f| =\frac{1250}{3} \approx 417 N$
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