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: 31

Answer

$$ M_P=\{-24 \mathbf{i}+24 \mathbf{j}+8 \mathbf{k}\} \mathrm{kN} \cdot \mathrm{m} $$

Work Step by Step

The coordinates of points $A$ and $P$ are $A(3,3,-1) \mathrm{m}$ and $P(-2,-3,2) \mathrm{m}$ respectively. $$ \begin{aligned} \mathbf{r}_{P A} & =[3-(-2)] \mathbf{i}+[3-(-3)] \mathbf{j}+(-1-2) \mathbf{k} \\ & =\{5 \mathbf{i}+6 \mathbf{j}-3 \mathbf{k}\} \mathbf{m} \end{aligned} $$ Moment of $\boldsymbol{F}$ About Point $\boldsymbol{P}$. $$ \begin{aligned} M_P & =\mathbf{r}_{A P} \times \mathbf{F} \\ & =\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 6 & -3 \\ 2 & 4 & -6 \end{array}\right| \\ & =\{-24 \mathbf{i}+24 \mathbf{j}+8 \mathbf{k}\} \mathrm{kN} \cdot \mathrm{m} \end{aligned} $$
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