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

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The problem involves a pirate weighing 780 N walking on a 500 N plank, with the goal of determining how far he can walk before the plank tips. The solution requires applying equilibrium equations, specifically the sum of forces and moments. The key insight is that the normal force at the left support becomes zero when tipping occurs. After working through the calculations, the distance the pirate can walk before tipping is found to be 0.769 m. The discussion highlights the importance of understanding support reactions in 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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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