# Resolution of Forces Help Please

1. Feb 20, 2013

1. The problem statement, all variables and given/known data
A particles is under the influence of two forces, 500N at 50° from north and 350N at 350° from north. Find the magnitude and direction of the resultant pull on the particle.

2. Relevant equations

?

3. The attempt at a solution
$x=500cos40-350cos80\\ x=322.25N\\ y=500sin40+350cos10\\ y=666.08N \\ \sqrt{322.25^2 + 666.08^2}=739.94N \\ arctan(322.25/666.08)=25.82° from north. \\$
I am a bit confused on when to use cos or sin. The above is how a classmate pursaded me to change it to but on my first go I had
$x=500cos40-350sin10 \\ y=500sin40+350cos10 \\$

Last edited: Feb 20, 2013
2. Feb 20, 2013

### rude man

Draw the vectors on an x-y plot (+y = North). It should be obvious from that.

3. Feb 20, 2013

Thanks for your reply. I have already done that and the figure and direction obtained seem correct but I dont know for sure and if my method is correct.

EDIT: Sorry just realised my mistake. They are both the same but doing it a different way, my calculator was set on radians instead of degrees and that is why I was getting different figure to earlier sorry.

Last edited: Feb 20, 2013
4. Feb 20, 2013

### Redbelly98

Staff Emeritus
A very common mistake. Remember to check the setting whenever you have an exam