Analyzing Structural Forces: Understanding Reactions and Moments in Mechanics

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SUMMARY

The discussion focuses on problem 6.123 from "Vector Mechanics for Engineers" by Beer et al., which involves analyzing structural reactions and moments. The user calculates the reactions at node A, identifying Ax and Ay as the horizontal and vertical components, respectively. However, the user encounters a discrepancy when comparing their calculated Ay of 250 N upwards with the provided solution, which states that the reaction has an angle of 61.3°. The issue arises from neglecting the contributions of forces at nodes B and D, which are essential for a complete analysis.

PREREQUISITES
  • Understanding of equilibrium equations in mechanics, specifically ∑Fx=0 and ∑Fy=0.
  • Familiarity with vector decomposition of forces.
  • Knowledge of moment calculations in structural analysis.
  • Experience with analyzing reactions in static systems.
NEXT STEPS
  • Review the principles of static equilibrium in mechanics.
  • Study the method of joints in truss analysis.
  • Learn about calculating moments about a point in structural systems.
  • Examine the role of additional nodes in force analysis, particularly in complex structures.
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Students and professionals in civil and mechanical engineering, particularly those studying structural analysis and mechanics of materials.

thegreengineer
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<<Mentor note: Missing template due to being moved from other forum.>>

I took this problem from Vector Mechanics for Engineers by Beer et al. The reason why I am writing this is because I don't understand where I am wrong in this problem, yet I know I am wrong.

This is problem 6.123 and it's a problem concerning analysis of structures (in this case a machine).
https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13590368_1744245399193330_1531717957213866948_n.jpg?oh=b73b768c39128c189fcca2f3a5d3e3f4&oe=5835700D
So the first thing I did was calculating the reactions the system has. In this case there is a reaction in node A which consists of a rough surface. Therefore there are two forces which is in fact one single force but decomposed into its horizontal and vertical components which I called Ax and Ay respectively.
https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13567002_1744248252526378_7606622242639327423_n.jpg?oh=23ec4d5bd163fcbba959ba52711e8d41&oe=57F96698
This is where my problem begins, let's suppose I just considered the reactions Ax and Ay in node A and the 250 N load acting in node C. Then I work with equilibrium equations:
\sum F_{x}=0
\sum F_{y}=0
For the equation \sum F_{x}=0 we find out that the only force acting upon the x-axis is the horizontal component Ax that becomes zero. Then for the other equation \sum F_{y}=0 I see that there are two forces: the 250 N that goes downwards and the Ay that I supposed that goes upwards. When I perform the operations required I get that the Ay force is 250 N that goes upwards. Since the vertical component is the only component for the reaction in node A then I conclude that the reaction in node A is 250 N upwards.
So far we have answered part b).
When I consult the answers to selected problems section just to verify I am right I see that the answer doesn't match:
https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13511060_1744249799192890_8621689707186190488_n.jpg?oh=cd46cd6c0101d44c8d8a4d65b6dd0a2b&oe=57FB23BC
It is not that the answer doesn't match in magnitude, but it also says that the reaction has an angle of 61.3°.
I don't know where I am wrong.
 
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MarcusAu314 said:
For the equation ∑Fx=0∑Fx=0\sum F_{x}=0 we find out that the only force acting upon the x-axis is the horizontal component Ax that becomes zero. Then for the other equation ∑Fy=0∑Fy=0\sum F_{y}=0 I see that there are two forces: the 250 N that goes downwards and the Ay that I supposed that goes upwards.

You've ignored what's going on at B (and/or D). They contribute forces in both x and y.

Complete question a) before proceeding to b).
Taking moments about A is a good first step.
 

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