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 114: 54


$F_{AC}=119lb$ $F_{AB}=79.2lb$ $F_{AD}=283lb$

Work Step by Step

We can write the tension developed in each cable as follows: $F_{AB}=-\frac{3}{7}F_{AB}\hat i-\frac{6}{7}F_{AB}\hat j+\frac{2}{7}F_{AB}\hat k$ and $F_{AC}=\frac{2}{7}F_{AC}\hat i-\frac{6}{7}F_{AC}\hat j+\frac{3}{7}F_{AC}\hat k$ and $F_{AD}=\frac{3}{5}F_{AD}\hat j+\frac{4}{5}F_{AD}\hat k$ The sum of the forces in the x-direction is given as $\Sigma F_x=0$ $\implies -\frac{3}{7}F_{AB}+\frac{2}{7}F_{AC}=0$.eq(1) The sum of the forces in the y-direction is given as $\Sigma F_y=0$ $\implies -\frac{6}{7}F_{AB}-\frac{6}{7}F_{AC}+\frac{3}{5}F_{AD}=0$.eq(2) The sum of the forces in the z-direction is given as $\Sigma F_z=0$ $\implies \frac{2}{7}F_{AB}+\frac{3}{7}F_{AC}+\frac{4}{5}F_{AD}-300=0$eq(3) Solving eq(1), eq(2) and eq(3), we obtain: $F_{AC}=119lb$ $F_{AB}=79.2lb$ and $F_{AD}=283lb$
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