Equilibrium Problem

1. Jul 17, 2006

Hollysmoke

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?

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2. Jul 17, 2006

Office_Shredder

Staff Emeritus
I'm a bit confused here.... exactly what are you trying to solve for?

3. Jul 17, 2006

Hollysmoke

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

4. Jul 17, 2006

Hollysmoke

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

5. Jul 17, 2006

PhysicsApprentice

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

6. Jul 17, 2006

Hollysmoke

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

7. Jul 17, 2006

PhysicsApprentice

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

tanA = -188 / -101.33

therefore A = 49.35 degrees to the horizontal

Last edited: Jul 17, 2006
8. Jul 17, 2006

PhysicsApprentice

if you need further explaination let me know

9. Jul 17, 2006

Hollysmoke

Wait how did you get -188 for the vertical?

10. Jul 17, 2006

Hollysmoke

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

11. Jul 17, 2006

PhysicsApprentice

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

12. Jul 17, 2006

PhysicsApprentice

yes i meant 118