Can You Solve This 2D Equilibrium Problem with a Pirate on a Plank?

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SUMMARY

The problem involves a pirate weighing 780 N walking on a plank that weighs 500 N, with the plank tipping at a critical distance. The solution requires applying static equilibrium principles, specifically the sum of forces in both the X and Y directions, and the sum of moments about a point. The calculated tipping point for the pirate is 0.769 meters from the pivot point, which is the far left support of the plank. Understanding the conditions under which the normal force at the left support becomes zero is crucial for solving this equilibrium problem.

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  • Static equilibrium principles
  • Understanding of forces and moments in 2D
  • Knowledge of normal force concepts
  • Ability to solve equations with multiple unknowns
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  • Study the principles of static equilibrium in greater detail
  • Learn about calculating moments and forces in 2D systems
  • Explore problems involving multiple unknowns in physics
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Students in physics or engineering courses, particularly those focusing on mechanics and statics, as well as educators seeking to enhance their teaching of equilibrium problems.

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Homework Statement



The pirate is walking the plank. If the plank weighs 500 N and the weight acts at
the centre of the plank, how far (distance x) can our pirate, who weighs 780 N, walk
before the plank tips and he falls. All dimensions are in meters. ANS: 0.769 m

(Look at picture)

Homework Equations



Sum of forces in X =0
Sum of forces in Y=0
Sum of moment(about a point) = 0

The Attempt at a Solution



I can get pretty far in the solving, but since there are 4 unknowns in the problem(and you can only solve for 3 unknowns in 2D I ran into a problem). The Pin support has 2 reactions and the other support is 1 normal force. Thanks for any help.

2hplwzs.jpg
 
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Hint: The Normal force (reaction) at the far left support must be 0 when the plank is about to tip (it is assumed that the left support is not anchored, such that it can take a downward load, but not an upward load.).
 
PhanthomJay said:
Hint: The Normal force (reaction) at the far left support must be 0 when the plank is about to tip (it is assumed that the left support is not anchored, such that it can take a downward load, but not an upward load.).

Thank you so much! I got the answer :D
 

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