Solve Equilibrium Problem: Find Equilibriant Force

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Homework Help Overview

The discussion revolves around finding the equilibriant force in a physics problem involving forces in equilibrium. Participants are analyzing horizontal and vertical force components based on a provided diagram.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to set up equations for horizontal and vertical forces, questioning the correctness of their calculations and the relationships between these forces. Some express confusion about the definitions of the forces involved and their respective directions.

Discussion Status

The discussion includes various attempts to clarify the problem and verify calculations. Some participants have provided numerical values for forces, while others are seeking deeper understanding of the relationships between the forces and their components. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants mention attachments that presumably contain diagrams relevant to the problem, which may be critical for understanding the setup. There are indications of confusion regarding the signs and components of the forces involved.

Hollysmoke
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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?
 

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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 explanation let me know
 
Wait how did you get -188 for the vertical?
 
  • #10
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 vertical

but this does not mean that Fy will always be the vertical if you in a situation where Fx has to be sin0 then fx will be the vertical 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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