Constructing forces (mechanics)

  • #1
17
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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.
 

Attachments

  • problem_4.70.JPG
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  • interps_4.70.JPG
    interps_4.70.JPG
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Answers and Replies

  • #2
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.
 
  • #3
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.
 
  • #4
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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