# Statics - Coplanar Force Systems

by EtherealMonkey
Tags: coplanar, force, statics, systems
 P: 41 The problem statement: My relevant equation: $\phi$ will be the angle between the X axis and $F_{CO}$ $$\theta = \phi + \arcsin\left(\frac{3}{5}\right)$$ My attempt at a solution: $$\Sigma F_{x} = 0:$$ $$F_{CO}\cos\phi - F_{BO}\frac{4}{5} = 0$$ $$F_{CO} = \frac{F_{BO}\frac{4}{5}}{\cos\phi}$$ $$\Sigma F_{y} = 0:$$ $$F_{AO} - F_{BO}\frac{3}{5} - F_{CO}\sin\phi = 0$$ Combining terms and substituting the equation found for $\Sigma F_{x} = 0$ into $\Sigma F_{x} = 0:$ $$F_{AO} - \frac{3}{5}F_{BO} - \frac{4}{5}F_{BO}\tan\phi = 0$$ $$9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0$$ $$\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)$$ $$\phi = 65.12^{\circ}$$ $$\theta = 102^{\circ}$$ The published value of $\theta$: $$\theta = 70.1^{\circ}$$ I don't know what I did wrong. TIA for any response.
 Quote by EtherealMonke $$9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0$$ $$\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)$$