Equilibrium Problem

  • Thread starter Hollysmoke
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  • #1
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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

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Answers and Replies

  • #2
Office_Shredder
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I'm a bit confused here.... exactly what are you trying to solve for?
 
  • #3
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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
185
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I'm supposed to find the equilibriant force on the diagram I posted (see attachments)
 
  • #5
the force that give you equilibrium is 155.55N and its direction is 49.35 degress to the horizontal
 
  • #6
185
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uhmm...thanks for telling me the answer, but it's useless if I don't know how to get the answer =(
 
  • #7
-101.33 + (theforce)cosA = 0
-188 + (theforce)sinA = 0

tanA = -188 / -101.33


therefore A = 49.35 degrees to the horizontal
 
Last edited:
  • #8
if you need further explaination let me know
 
  • #9
185
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Wait how did you get -188 for the vertical?
 
  • #10
185
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nvm I think you meant 118. if you divided 188, you get 61 degrees.
 
  • #11
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
Hollysmoke said:
nvm I think you meant 118. if you divided 188, you get 61 degrees.

yes i meant 118
sorry about that
 

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