# Magnitudes, Resultant Force (missing angle!)

1. Jan 20, 2010

### 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,

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

#### Attached Files:

• ###### Q2.9.JPG
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2. Jan 20, 2010

### 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.

3. Jan 20, 2010

### jegues

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

4. Jan 21, 2010

### vela

Staff Emeritus
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.

5. Jan 21, 2010

### 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?

6. Jan 21, 2010

### LCKurtz

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.

7. Jan 21, 2010

### 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.

8. Jan 21, 2010

### vela

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

9. Jan 21, 2010

### jegues

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

10. Jan 21, 2010

### 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.

11. Jan 21, 2010

### zgozvrm

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.

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.

12. Jan 21, 2010

### vela

Staff Emeritus
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.

13. Jan 21, 2010

### 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!