Statics: Mechanical Equilibrium

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SUMMARY

The discussion centers on determining the loads acting on a frame in mechanical equilibrium, specifically at points A and D, with a 200 lb weight hanging from it. The user initially miscalculated the forces, believing the x-component at point A to be zero, while the correct value is 400 lb. The analysis involves using Free Body Diagrams (FBDs) for both two-force and three-force elements, applying the principles of static equilibrium, including the sum of forces in both x and y directions and the sum of moments about point A.

PREREQUISITES
  • Understanding of mechanical equilibrium principles
  • Familiarity with Free Body Diagrams (FBDs)
  • Knowledge of two-force and three-force member analysis
  • Basic trigonometry, specifically 45-45-90 triangles
NEXT STEPS
  • Study the principles of static equilibrium in detail
  • Learn how to construct and analyze Free Body Diagrams (FBDs)
  • Explore the characteristics of two-force and three-force members in structural analysis
  • Review trigonometric applications in engineering mechanics
USEFUL FOR

Students and professionals in mechanical engineering, structural engineers, and anyone involved in analyzing forces in static systems.

teknodude
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I need some help in this problem to see if my game plan is right. My answer doesn't make sense, well to me it doesn't.

A weight of 200 lb hangs from a frame of negligible weight, as shown. If the frame is in equilibrium, determine the loads acting on the frame at A and D. Use the fact that the frame acts as a three-foce element to check your answer.

-they are all pin connections
-there's a 45 45 90 degree triangle there.

First off i drew a FBD for just DB, and found that it is a two force element, so the line of action passes through the force points of application.

Then when i draw a FBD for ABC the forces at B, it is just a force pointing at a 45 degree angle. Then there's a force Ay and Ax at A and the 200 lb force acting downward at C.

the sum of the forces in the x direction

-Ax -B cos45 = 0

The sum of the force in the y direction

Ay + B sin45 - 200 lb = 0

The sum of the moments in the z direction at point A

-(6ft)(200lb) + (3ft)(Bsin45) = 0
 

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ok i think I'm using the idea of a 2-force element wrong.


The diagram that i have there needs to be zoomed in a bit, doesn't seem to show up right cause some browsers resize it.
If you need a better diagram then i'll try to scan it this time instead of using paint.
 
Member DB is a two force member, that makes the analysis simple.
 
At A, for the force i get

0 for the x component
200 lb for the y component

anyone else get these values at point A?
 
no, Fax=400lb, Fay= 200N down
 
Answer

If anyone ever needs the solution, I'd be more than happy to help you come to it
 
I got it. I was making this problem more harder than it seemed :cry:
 

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