2 people pulling on ropes or cables with a third person trying to equal out

1. Oct 7, 2012

SherBear

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

We have a person pulling on a string or rope in the negative y negative x quadrant (call it left) this person is called Wesson, they have a force of 300N at 30°

Same situation but pulling the other way, positive y and positive x quadrant (call it right), this person is called Canola, they have a force of 200N at 30°

The third person pulling is the boss called Olive, they have a force of 250N at 60°

What I'm seeking is this: What is the magnitude and direction of the force the boss needs to apply to make the system at equilibrium?

2. Relevant equations

First make a FBD then..

I have the sum of Fx = -tw + tc + to = 0

Pithagoram's theory

arc tan........Tan-1 (Fy/Fx)

3. The attempt at a solution

from the sum of Fx I have = -300N sin 30° + 200N sin 30° + 250N sin 60°

-150 + 100 + 216.50=

Fx=166.5

To get the magnitude i'm using pithagorm's theorm

so I have sqrt 166.5^2 + 0 (because the sum in the y dir is zero)

magnitue= 52.17N

To get the direction I have

Tan -1 (0 / 166.5N) = 0 N = Direction

Am I on the money or way off base?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 7, 2012

SherBear

We have a person pulling on a string or rope in the negative y negative x quadrant---**Sorry about that, to be left, I meant positive y and negative x quadrant =-)**

3. Oct 8, 2012

SherBear

For this one I have
wesson= -300cos30 ihat + 300sin30 jhat=-259.80 ihat + 150 j hat
canola= 200 cos 30 ihat + 200 sin 30 jhat= 173.20 i hat + 100 j hat
then add.............................................= 433 i hat - 50 j hat

magnitude= sqrt 433 ihat ^2 + -50 jhat ^2=

435.87N=magnitude

Direction= Tan-1 50jhat / 433 i hat=
Direction=6.58 degrees above the positive x axis or north of the positive x axis

is this right?