Question

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

I know the components of f1 are f1x= 1600, f1y=0, f2x= 433 f2y=250 and f3x= 467

Answer 1

Step-by-step 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_3cos(\theta) = 250\:\text{lb}`

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

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

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

The angle
`\theta` is then

`tan(\theta) = (F_(3y))/(F_(3x)) = (250)/(467)`

or

`\theta = 49.2°`