Free body diagram drawing for a two force members on machine

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SUMMARY

The discussion focuses on analyzing a free body diagram for a two-force member system involving a brad squeezer. When a force of 2 lb is applied, the calculations reveal that the force P exerted on the smooth brads at points C and D results in a value of -2 lb, indicating a direction opposite to the initial assumption. The moment equation around point E confirms that the force F_AB equals 4 lb. The participants emphasize the importance of maintaining consistent units throughout the calculations.

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  • Understanding of free body diagrams and their application in mechanics
  • Knowledge of static equilibrium equations: ∑F_x=0, ∑F_y=0, ∑M=0
  • Familiarity with two-force member principles in structural analysis
  • Proficiency in unit conversion and management, particularly between English and Imperial units
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  • Study the principles of static equilibrium in mechanical systems
  • Learn how to accurately draw and analyze free body diagrams
  • Explore the concept of two-force members and their applications in engineering
  • Review unit conversion techniques to ensure consistency in calculations
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Engineering students, mechanical engineers, and anyone involved in structural analysis or mechanics who seeks to understand the application of free body diagrams in two-force member systems.

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Homework Statement


When a force of 2 lb is applied to the handles of the brad squeezer, it pulls in the smooth rod AB. Determine the force P exerted on each of the smooth brads at C and D

media%2F760%2F7609e26d-7481-4d34-b4f6-eb8d308bd379%2FphpsT6miC.png


Homework Equations


##\sum F_x=0##
##\sum M=0##
##\sum F_y=0##

The Attempt at a Solution


I drew a free body diagram of Members EA with the initial force applied. Since A and B are two force members, ##F_{AB}## is drawn opposite direction in free body diagram BCD. However when I solve the the equations I get that P is opposite to the Free body diagram drawn. Why is that is there some mistake?[/B]

Moments about ##E## yields ##M_E=2\cdot 2-F_{AB}=0## so ##F_{AB}=4##
In second FBD, I get that ##P=-2##there must be some mistake in this?
 

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Wow, I thought I was the only one who ever used English/Imperial units any more.

The sum of the torques about a given axis must be zero.

Taks care, and carry your units along.
 

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