# Homework Help: Some issues with adding 2 vectors

1. Feb 1, 2012

### yesiammanu

1. The problem statement, all variables and given/known data
A force F1 of magnitude 6.00 units acts on an object at the origin in a direction of 30.0° above the positive x axis (Fig. P3.10). A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y axis. Find graphically the magnitude and direction of the resultant force F1 F2.

2. Relevant equations
I used the law of cosines R = √(A^2 + B^2 - 2ABcos∅)
I then used the law of sines sin β = A/R * sin∅

3. The attempt at a solution
I found that theta was 150 degrees by reversing the order of the vectors to flip the triangle, and subtracting from 30 degrees, as seen in the picture.
http://i.imgur.com/fA9zY.png
I found the answer to be 10.6 units through the equation but it does not seem like this is right at all; am I using the correct methodology?

I then found that the degrees were 73.6 by finding that sin β = .283, so β=16.4 but I wanted to put degrees above the x axis, so I subtracted 90 from 16.4 to get 73.6. This also seems wrong

I drew out the picture myself trying the "graphical method" and I came up with 8.1 with a 59 degree angle, which seemed much more reasonable. What am I doing wrong?

Last edited: Feb 1, 2012
2. Feb 1, 2012

### ehild

Check your second drawing: the 6-unit vector makes the 30° angle with the positive x axis.

ehild

3. Feb 1, 2012

### yesiammanu

Do you mean the triangle would go into the 2nd quadrant? I see that I might have actually decreased 30° instead of increasing by 30°, but if I redraw the triangle it seems like I get the same answer for magnitude

Else if you're saying that it actually should be 30° from a parallel line to the X axis, I'd get the angle 45 + 30 = 75°, which when plugged in gets me 6.74 units and 33.1°, which still seems to be off from my measurement

Last edited: Feb 1, 2012
4. Feb 1, 2012

### yesiammanu

I redid my drawing because of some mistakes I made (using the wrong units) and I got 10.1 units which seems pretty close to my 10.6 answer, but I got 58° for the angle measurement which is very off from my 73.6° angle calculation

5. Feb 1, 2012

### Opus_723

Your first drawing is correct, just use that one.

I'm not sure if I'm interpreting this right, but whenever my teacher told us to find the magnitude of a vector graphically, he meant to measure it by hand instead of do the trig.

But if you want to do the trig, you don't need the relations you listed, it's much simpler than that. Decompose F1 into its x and y components. F2 has only a y component. Add all of your like components together to get the x and y components of the resultant vector.

(x1, y1) + (x2, y2) = (x1+x2, y1+y2)

From this you can find the magnitude and direction.

Last edited: Feb 1, 2012
6. Feb 1, 2012

### yesiammanu

I'm supposed to both graphically measure it and do the trig unfortunately

Using that method with the equation:

Magnitude of →R is √((Asubx + Bsubx)^2 + (Asuby + Bsuby)^2))
And Vsubx is Vcos∅ and Vsuby is Vsin∅

I get

√((6cos30 + 0)^2 + (6sin30 + 5)^2) = √(27 + 64) = 9.53 which seems ok

7. Feb 1, 2012

### Opus_723

That looks good to me. You should also be able to find the angle the vector makes with the positive x-axis using the components of that resultant vector, and then you're all done.

8. Feb 1, 2012

### yesiammanu

Yep, I get 60.7° which is close. Thank you both for your help