Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 4 - Force System Resultants - Section 4.4 - Principle of Moments - Problems - Page 143: 46

Answer

$$ \begin{aligned} & y=1 \mathrm{~m} \\ & z=3 \mathrm{~m} \\ & d=1.15 \mathrm{~m} \end{aligned} $$

Work Step by Step

$$ \begin{aligned} -14 \mathbf{i}+8 \mathbf{j}+2 \mathbf{k} & =\left|\begin{array}{llc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & y & z \\ 6 & 8 & 10 \end{array}\right| \\ -14 & =10 y-8 z \\ 8 & =-10+6 z \\ 2 & =8-6 y \\ y & =1 \mathrm{~m} \\ z & =3 \mathrm{~m} \\ M_O & =\sqrt{(-14)^2+(8)^2+(2)^2}=16.25 \mathrm{~N} \cdot \mathrm{m} \\ F & =\sqrt{(6)^2+(8)^2+(10)^2}=14.14 \mathrm{~N} \\ d & =\frac{16.25}{14.14}=1.15 \mathrm{~m} \end{aligned} $$
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