Statics - Coplanar Force Systems

[EtherealMonkey]

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.

 

[PhanthomJay]

$$9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0$$

$$\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)$$

that plus 24/5 should be a minus 24/5.

I don't know what I did wrong.

TIA for any response.

you did well, just missed the sign.

 

[EtherealMonkey]

Jay, thank you sir!