# Equilibrium Problem

As you can see in the attachmen, I'm supposed to find the equilibriant force and the force from the left side.

So is it something like this:

Fx = 0,

Fh + 200cos20-100cos30=0
Fh = -101.33N

Fy = 0
Fv + 200sin20 + 100sin30 = 0
Fv = -118N

But I get stuck on the rest. Can someone help me out please?

#### Attachments

• resolveso9.jpg
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Office_Shredder
Staff Emeritus
Gold Member
I'm a bit confused here.... exactly what are you trying to solve for?

Actually, I think I figured it out but are the vertical and horizontal forces correct? Or do the Fh and Fv should cancel out

I'm supposed to find the equilibriant force on the diagram I posted (see attachments)

the force that give you equilibrium is 155.55N and its direction is 49.35 degress to the horizontal

uhmm...thanks for telling me the answer, but it's useless if I don't know how to get the answer =(

-101.33 + (theforce)cosA = 0
-188 + (theforce)sinA = 0

tanA = -188 / -101.33

therefore A = 49.35 degrees to the horizontal

Last edited:
if you need further explaination let me know

Wait how did you get -188 for the vertical?

nvm I think you meant 118. if you divided 188, you get 61 degrees.

well

-188 +FsinA = 0
is the same as writing
-188 +Fcos(90-A) = 0

since its sin it means it is the verticle

but this does not mean that Fy will always be the verticle if you in a situation where Fx has to be sin0 then fx will be the verticle and it would be 101.33/118
instead of the other way around

if your still not 100% percent clear i ll draw a picture for you, its quite hard to explain without pictures

Hollysmoke said:
nvm I think you meant 118. if you divided 188, you get 61 degrees.

yes i meant 118