Constructing forces (mechanics)

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The discussion revolves around determining the reactions at points A and C in a statics problem involving a metal frame. Participants are exploring how to accurately interpret the direction of forces and their angles, noting that different interpretations can yield varying results. They emphasize the importance of using equilibrium equations, specifically that the sum of forces must equal zero. A suggestion is made to analyze the equilibrium of member AB by taking moments about points A or B to clarify force directions. The conversation highlights the nuances of statics, particularly regarding pinned joints and moments of reaction forces.
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Homework Statement


I've attached a picture of my problem (with diagram): 4.70 .

"For the frame and loading show, determine the reactions at A and C."

From the diagram, you can see that the metal object is made up of 2 parts:
1) A to B
2) B to D


Homework Equations





The Attempt at a Solution


how do you arrive at the correct direction of the forces (ie: angle of the force; direction of its "application"). I have 2 interpretations and they will give different angles with which the force is applied (same magnitudes though). Which one is the correct interpretation? Because this is a statics problem (bodies at equilibrium), the sum of the forces = 0 (illustrated by the triangles).

How do you logically determine which interpretation (of force directions) is correct when you do these problems?

Thank you.
 

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The simple minded thing to do is just assume there is a horizontal and vertical compoent of force at A B and C. There are no moments at A B or C because the joints are all pinned.

Then write the equations of equilibrium for each two components. There are 3 equations for each component, that's 6 equations for the 6 unknown forces.

If you want top do it by a "neater" method, start by looking at the equlilbrium of AB. Take moments about A (or B) and it should be clear what is the direction the forces at A and B.

Hint: one of the options you drew is right.
 
It is worth also mentioning, the member AB is the special case of a two forces body, and the member BCD of a 3 forces (no parallel) body of statics.
 
Thanks guys. Even though something is pinned though, there can still be a moment of the reaction forces at that point. I have a test tomorrow so when I'm done with it (may take a day or 2 for me to get remotivated, but I'll post something that shows that). I'll also try your way to figure out the prob. Hopefully this won't be asked before I figure it out :)

I'm appending what I wrote: above... the moment of the forces where something in "pinned" = 0.
 

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