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

[SherBear]

Homework Statement

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?

Homework 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)

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 theorem

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?

 

[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 =-)**

 

[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?