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

In summary, the problem presents a pirate walking a plank with a weight of 500 N at the center. With the dimensions in meters, the question asks how far the 780 N pirate can walk before the plank tips and he falls. By using the sum of forces and moment equations, it is possible to solve for the distance x. However, since there are 4 unknowns in the problem, there is a limitation in solving for only 3 unknowns in 2D. The far left support must have a normal force of 0 when the plank is about to tip, assuming it is not anchored. Ultimately, the solution is 0.769 m.
  • #1
Oblivion77
113
0

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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  • #2
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.).
 
  • #3
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
 

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

What is 2D equilibrium of rigid body?

2D equilibrium of rigid body is a concept in physics that describes the state of a rigid body when it is at rest and the sum of all the forces acting on it is equal to zero. In other words, the body is not moving or rotating and all the forces acting on it are balanced.

What are the conditions for 2D equilibrium of rigid body?

There are two main conditions for 2D equilibrium of rigid body: first, the net force acting on the body must be equal to zero, and second, the net torque (or rotational force) acting on the body must also be equal to zero. These conditions ensure that the body remains at rest and does not start to move or rotate.

How is 2D equilibrium of rigid body different from 3D equilibrium?

2D equilibrium of rigid body only considers forces and torques acting in two dimensions, while 3D equilibrium takes into account forces and torques acting in all three dimensions. This means that in 2D equilibrium, the body is constrained to move or rotate along a plane, while in 3D equilibrium, it can move and rotate freely in all directions.

What are some real-life examples of 2D equilibrium of rigid body?

Some common examples of 2D equilibrium of rigid body include a ladder leaning against a wall, a book resting on a table, and a seesaw with two people of equal weight on either side. In all of these cases, the forces acting on the body are balanced, and it remains at rest without any movement or rotation.

How is 2D equilibrium of rigid body used in engineering and design?

Understanding 2D equilibrium of rigid body is essential in engineering and design to ensure that structures and machines are stable and do not collapse or fail. Engineers use this concept to calculate the forces and torques acting on different parts of a structure and make sure that they are balanced, thus ensuring the safety and stability of the design.

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