Magnitudes, Resultant Force (missing angle!)

[jegues]

1. The problem statement, all variables and given/known data
See attachement.


2. Relevant equations



3. The attempt at a solution

I would be able to solve it if I could somehow find \phi to describe the angle of F3 relative to the positive x axis. Can anyone see how to solve that specific angle?

Then and can simply sum as follows


F_{x} = F1cos(\theta) + F2cos(\alpha) + F3cos(\phi)

and


F_{y} = F1sin(\theta) + F2sin(\alpha) + F3sin(\phi)

and

F = \sqrt{F_{x}^{2} + F_{y}^{2}}

Then let \beta be the resultant angle,

\beta = tan^{-1}(\frac{F_{y}}{F_{x}} )

[LCKurtz]

You don't need to know the angle \phi itself, just its sine and cosine. And the picture gives x and y sides of a similar triangle to the one with F3 as the hypotenuse.

[jegues]

the picture gives x and y sides of a similar triangle

I think you are thinking of the similar triangle for F2 not F3. Am I correct in my assumption?

[vela]

No, the diagram gives you similar information for F3. It's just not explicitly drawn in like it is for F2. F3 lies on the hypotenuse of a triangle in the picture. You should be able to identify the lengths of the legs of that triangle.

[jegues]

Well it can't be the triangle with the sides 0.2 and 0.3 because there is a corner that intersects it at the bottom right. I don't see how the 0.1 near the joint helps us either.

Is it safe to assume that the corner in the bottom right doesn't effect our triangle so it will have sides of 0.2 and 0.3?

Otherwise, I'm just not seeing it! Any more help?

[LCKurtz]

Well it can't be the triangle with the sides 0.2 and 0.3 because there is a corner that intersects it at the bottom right. I don't see how the 0.1 near the joint helps us either.

Is it safe to assume that the corner in the bottom right doesn't effect our triangle so it will have sides of 0.2 and 0.3?

Otherwise, I'm just not seeing it! Any more help?

Truth is, I didn't see that little corner so you are correct. But I would bet money that it is safe, and you are expected, to ignore it, and use the .2 and .3.

[jegues]

Is there any other way I could solve that angle without using similar triangles (using sides 0.2 and 0.3)? If so how?

If there's no other way, then I guess I'm stuck assuming it's the safe(and correct) way to do it.

[vela]

I don't see why the corner is a problem.

[jegues]

Because if that corner is there then the sides aren't going to be 0.2 and 0.3 respectively.

[LCKurtz]

I just looked at the picture again. The corner is not a problem at all. It is only the dimension arrows that give it that appearance. The dimensions are OK as given.

[zgozvrm]

Well it can't be the triangle with the sides 0.2 and 0.3 because there is a corner that intersects it at the bottom right.
The thin lines are all dimension lines, so that is not a "corner." Look at the 0.3m measurement: there are arrows pointing up and down. The arrow pointing up points to a dimension line, as does the arrow pointing down. What you're seeing as a "corner" is this lower dimension line intersecting with the right-hand dimension line for the 0.2m measurement. The thin line extending out of the bold 1200N force line is just an extension of the force line which is being used to give us the slope of the force.

WI don't see how the 0.1 near the joint helps us either.
There are a couple of measurements that don't relate to this problem. My guess is that this diagram comes from a book with several questions relating to it.

[vela]

Because if that corner is there then the sides aren't going to be 0.2 and 0.3 respectively.
Draw a vertical line from the corner to the upper horizontal line. The vertical line, the upper horizontal line, the horizontal edge of the corner, and the measurement arrows form a rectangle. The opposite sides of the rectangle are the same length, so the vertical line is 0.3 m. Similarly, the horizontal distance from where F3 acts to the corner is 0.2 m.

[jegues]

Thank you for the clarification, I seem to have forgot that we were only using the similar triangle to solve the angle for F3 in relation to the positive x-axis.

I've solved the problem now!