Equilibrium of Forces: Finding F and Angle X | Homework Help

[Trail_Builder]

Homework Statement

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.

Homework Equations

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 I am doing it right but I am stuck ahhhh




hope you can help

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

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 I am doing it right but I am stuck ahhhh




hope you can help

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

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

 

[Trail_Builder]

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

thought I was missing a trig identity or something lol

thanks :D