Finding F1 and F2 for particle equilibrium with three forces

[dumbengineer]

[SOLVED] Equilibrium of a particle - hibbeler , help please

Hi all
I was surfing the net looking for a physics / mechanic forum and thank god i found this.
I have problem with solving Equilibrium problems, hopefully you guys and girls would help me out.

1) Question. ( this is problem 3-1 , in engineering mechanic statics , 11th edition hibbeler)

- Determine the magnitudes of F1 and F2 so that the particle is in equilibrium
Given : F1 = 500N , θ1 = 45deg and θ2 = 30 deg

2) this is how far i went.

ΣFx = F1 cos θ1 + F2 cos θ2 - F = 0

ΣFy = F1 sin θ1 -F2 sin θ2 = 0
-
ΣFx = F1 cos 45 + F2 cos 30 - 500 = 0

ΣFy = F1 sin 45 -F2 sin 30 = 0 -> f1sin 45 = f2sin 30

so , f1 cos 45 + f1sin 45 - 500 = 0
f1 cos 45 + f1 sin 45 = 500
f1 1.41 = 500
500 / 1.41 = 354.60

well that answer is wrong , i know that f1 = 259 and f2 = 366 those are the correct answers

any help guys
thanks in advance

 

[PhanthomJay]

You need to be a bit more specific in the problem statement... are there 3 forces F1, f1, and f2? Direction of F1?? theta reference axis?? etc.

 

[Astronuc]

This so

f1 cos 45 + f1sin 45 - 500 = 0

is not consistent with

ΣFx = F1 cos θ1 + F2 cos θ2 - F = 0

Taking ΣFx = F1 cos θ1 + F2 cos θ2 - F = 0, and θ1 = 45° and θ2 = 30°, then

F1 cos (45°) + F2 cos (30°) = 0.7071 F1 + 0.866 F2 = 500 N.

The other equation is F1 sin 45° -F2 sin 30° = 0 = 0.7071 F1 - 0.5 F2 = 0.

Subtract the first equation from the second and one obtains

1.366 F2 = 500 N

Solve for F2 and substitute that value into either equation and solve for F1.

 

[dumbengineer]

Thank you very much for your help.