# Forces in Equilibrium

1. Nov 22, 2007

### Trail_Builder

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

A set of forces are in equilibrium. Find "F" and angle "x"

The following are forces actually on a particle in the center (N being newtons):

4N to the right along the horizontal
(3 x root3)N downwards down the vertical
6N diagonally up right, 60deg from the horizontal
2N diagonally up left, 60deg from the horizontal
"F"N diagonally down left, angle "x" is the actual between force "F" and the force pointing directly down.

2. Relevant equations

3. The attempt at a solution

Resolve going right:

4+(6cos60)=(2cos60)+(Fcos(90-x))
Fcos(90-x)=6

Resolve going down:

(3*(root3))+(Fsin(90-x))=(6cos30)+(2cos30)
Fsin(90-x)=3*(root3)+1-3*(root3)

Sub one into the other

6*Fsin(90-x)=Fcos(90-x)
6*sin(90-x)=cos(90-x)

this is where I get stuck lol.

not sure If Im doing it right but Im stuck ahhhh

hope you can help

2. Nov 22, 2007

### HallsofIvy

Staff Emeritus
"6*sin(90-x)=cos(90-x)"

so 6*cos(x)= sin(x) and then sin(x)/cos(x)= tan(x)= 6.

3. Nov 22, 2007

### Trail_Builder

o rite, i didnt realise sin(x)/cos(x) = tan(x) :D

thought I was missing a trig identity or something lol

thanks :D