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: 55



Work Step by Step

We can determine the required maximum weight 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 $ $F_{AC}=\frac{2}{7}F_{AC}\hat i-\frac{6}{7}F_{AC}\hat j+\frac{3}{7}F_{AC}\hat k$ 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) and 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 $\Sigma F_z=0$ $\implies \frac{2}{7}F_{AB}+\frac{3}{7}F_{AC}+\frac{4}{5}F_{AD}-W=0$..eq(3) From eq(1), we have $F_{AC}=\frac{3}{2}F_{AB}$.eq(4) We plug in this value and $F_{AD}=250lb$ in eq(2) to obtain: $-\frac{6}{7}F_{AB}-\frac{6}{7}(\frac{3}{2}F_{AB})+\frac{3}{5}(250)=0$ This simplifies to: $F_{AB}=70lb$ From eq(4), we have $F_{AC}=\frac{3}{2}(70)=105lb$ Now we plug in the known values in eq(3) to obtain: $\frac{2}{7}(70)+\frac{3}{7}(105)+\frac{4}{5}(250)-W=0$ This simplifies to: $W=265lb$
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