# What's wrong with this?

1. Sep 16, 2010

I have to find the resultant of:

[PLAIN]http://img17.imageshack.us/img17/8263/68709011.jpg [Broken]

So I decided to break it up into components:

[PLAIN]http://img37.imageshack.us/img37/7032/71250823.jpg [Broken]

(The black arrow in the first picture is the blue arrow in the second, that's the way I did it!)

I found the blue line in the picture using the cosine rule:

$c \ = \ \sqrt{30^2 \ + \ 40^2 \ - \ (2 \cdot 30 \ \cdot \ 40 \ \cdot \ \cos(120))$

c = 60.827 and it's just horizontal along the x-axis.

Then I added c = 60.827 to the third vector:

[PLAIN]http://img96.imageshack.us/img96/1917/92586648.jpg [Broken]

I found the white arrow, the total resultant, using the same technique:

$d \ = \ \sqrt{25^2 \ + \ 60.827^2 \ - \ (2 \cdot 25 \ \cdot \ 60.827 \ \cdot \ \cos(105))$

d = 71.5

My book says the answer is 67.6

I got the angle of the resultant using the sine rule:

$\theta \ = \ \sin^{-1} ( \frac{25 \cdot \sin(105)}{60.827} ) \ = \ 23.39$

My book says the angle is: 11.3°

What did I do wrong?

Last edited by a moderator: May 4, 2017
2. Sep 16, 2010

### Mindscrape

Where did you get 105? Clearly the most straightforward way to do this is to break it up into components, i.e. x=30cos45+40cos15+25sin15
y=30sin45-40sin15-25cos15
r^2=x^2+y^2
theta=arctan(y/x)

3. Sep 16, 2010

The 105° comes from adding the 90° angle under the red vector in third picture with the
15° the blue vector is making, giving 105°.

The problem was asked to be solved in component form but I tried it this way & got
the wrong answer, I can't see why & am totally stumped & would just like to know
why.

4. Sep 16, 2010

### Mindscrape

I don't know, round off errors most likely. Components are the way to go, less calculations.

5. Sep 16, 2010