Question

I'm completely stumped as to how to do this.

Answer 1

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force $\vec{\textbf{R}}$ and F1 are completely along the x-axis, we can conclude that

$F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}$

We can now solve for the magnitude of $F_3:$

$F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}$

$\:\:\:\:=529.7\:\text{lb}$

The angle $\theta$ is then

$\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}$

or

$\theta = 49.2°$