How Does Equilibrium Apply in the Classic Ladder Problem?

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Homework Help Overview

The discussion revolves around a classic physics problem involving a ladder in equilibrium, where a man stands on the ladder at a specified height. The problem includes parameters such as the ladder's length, weight, the man's weight, and the angle of the ladder against a wall.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of equilibrium and torque in the context of the ladder problem. There are attempts to calculate forces acting on the ladder, including the force exerted by the wall and the normal force from the ground. Questions arise regarding the correct interpretation of the man's position on the ladder and the implications for calculations.

Discussion Status

The discussion is ongoing, with some participants questioning the assumptions made about the man's position and the forces involved. There is a suggestion to clarify what is being solved for, indicating a productive direction for the conversation.

Contextual Notes

There is mention of a previous thread related to the same problem, suggesting that participants may be referencing earlier discussions or solutions. Additionally, the problem includes a scenario with a frictionless floor, which raises questions about the setup and assumptions of the problem.

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Homework Statement


http://i.imgur.com/5iA1G4i.jpg
A ladder that is 10ft long, weighs 30lbs, and has a 180 lb man standing 5ft up it, resting at 60 degrees on a wall

Homework Equations


equilibrium, and torque

The Attempt at a Solution


180lbs cos60° * 2.5ft + 30lbs cos60° * 10ft = Fwall sin60° * 10ft (all trig functions used to find component of force that is perpendicular to the ladder)
and when I solve for Fwall i get ≈43.3lbs

then I would simply set the force of friction (or as he has in his problem, the person standing at the bottom of the ladder on a frictionless floor, god know how that works, lol) to that as they are the only horizontal forces, then the normal force from the ground would be the two weights added together. So Ffriction is also 43.3lbs and Fnormal is 210lbs.
 
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What is the problem??
 
Let's start all over...

The self-weight of a uniform ladder acts through the ladder's midpoint.

The question says the man is half way up the ladder, your working implies he is one quarter of the way up the ladder. Which shall we use?

What are you solving for? I'm guessing the reaction at the floor?
 

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