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 141: 30

Answer

$$ M_A=\{-175 \mathbf{i}+5600 \mathbf{j}-900 \mathbf{k}\} \mathrm{lb} \cdot \mathrm{ft} $$

Work Step by Step

The coordinates of points $A$ and $B$ are $A(0,0,1.5) \mathrm{ft}$ and $B(8,0.25,1.5) \mathrm{ft}$, respectively. $$ \begin{aligned} \mathbf{r}_{A B} & =(8-0) \mathbf{i}+(0.25-0) \mathbf{j}+(1.5-1.5) \mathbf{k} \\ & =\{8 \mathbf{i}+0.25 \mathbf{j}\} \mathrm{ft} \end{aligned} $$ Moment of $\boldsymbol{F}$ About Point $\boldsymbol{A}$. $$ \begin{aligned} M_A & =\mathbf{r}_{A B} \times \mathbf{F} \\ & =\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 0.25 & 0 \\ 400 & -100 & -700 \end{array}\right| \\ & =\{-175 \mathbf{i}+5600 \mathbf{j}-900 \mathbf{k}\} \mathrm{lb} \cdot \mathrm{ft} \end{aligned} $$
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