# Chapter 3 - Equilibrium of a Particle - Section 3.4 - Three-Dimensional Force Systems - Fundamental Problems - Page 111: 8

$F_{AD}=1.125KN$ $F_{AC}=844N$ $F_{AB}=506N$

#### Work Step by Step

We can find the required tension in the cables as follows: The sum of the forces in the z-direction is $\Sigma F_z=0$ $\implies F_{AD}(\frac{4}{5})-900=0$ $\implies F_{AD}=1125N=1.125KN$ The sum of the forces in the y-direction is $\Sigma F_y=0$ $\implies F_{AC}(\frac{4}{5})-F_{AD}(\frac{3}{5})=0$ $\implies F_{AC}(\frac{4}{5})-(1125)(\frac{3}{5})=0$ This simplifies to: $F_{AC}=844N$ The sum of the forces in the x-direction is $\Sigma F_x=0$ $\implies F_{AB}-F_{AC}(\frac{3}{5})=0$ $\implies F_{AB}-(844)(\frac{3}{5})=0$ This simplifies to: $F_{AB}=506N$

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