Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 3 - Equilibrium of a Particle - Section 3.4 - Three-Dimensional Force Systems - Problems - Page 112: 43


$F_{AD}=763N$ $F_{AC}=392N$ $F_{AB}=523N$

Work Step by Step

We can find the required forces as follows: $OA=\sqrt{(2)^2+(1.5)^2}=2.5m$ $OB=\sqrt{(2.5)^2+(1.5)^2}=2.91m$ Now, the sum of the forces in the x-direction is given as $\Sigma F_x=0$ $\implies F_{AC}-F{AD}(\frac{2.5}{2.91})(\frac{1.5}{2.5})=0$ $\implies F_{AC}=\frac{1.5}{2.91}F_{AD}$.eq(1) Now, the sum of the forces in the y-direction is given as $\Sigma F_y=0$ $\implies F_{AB}-F_{AD}(\frac{2.5}{2.91})(\frac{2}{2.5})=0$ $\implies F_{AB}=\frac{2}{2.91}F_{AD}$.eq(2) Now, the sum of the forces in the z-direction is given as $\Sigma F_z=0$ $\implies F_{AD}(\frac{1.5}{2.91})-W=0$ $\implies F_{AD}(\frac{1.5}{2.91})-40(9.81)=0$ This simplifies to: $F_{AD}=763N$ We plug in this value of $F_{AD}$ in eq(2) to obtain: $F_{AB}=\frac{2}{2.91}(763)$ $\implies F_{AB}=523N$ We plug in $F_{AD}$ in eq(1) to obtain: $F_{AC}=\frac{1.5}{2.91}(763)$ This simplifies to: $F_{AC}=392N$
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