Statics: force to open a door on 2 rollers given friction

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Discussion Overview

The discussion revolves around a statics problem involving a 200-LB door suspended from two rollers, A and B, with given coefficients of friction. Participants explore how to calculate the force required to move the door to the left, considering the effects of friction and the distribution of forces at the rollers.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states that the normal force at each roller is 100 LB, assuming symmetry, but acknowledges uncertainty in their calculations.
  • Another participant suggests that the pull force creates a moment about roller B, indicating that the downward force at roller A will be greater than 100 LB.
  • A later reply outlines the equations needed to solve the problem, noting that the assumption of equal normal forces at A and B is incorrect.
  • The participant who initially struggled expresses gratitude for the guidance received, indicating that they have arrived at a solution consistent with the book's answer.
  • Final values presented include NA = 45.5 and NB = 154.5, leading to a calculated force P of 45.5 LB.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial assumptions regarding the normal forces at the rollers, as one participant corrects the assumption of symmetry. The discussion reflects a progression from confusion to clarity, but initial disagreements on the approach remain evident.

Contextual Notes

Participants note the importance of considering moments and the distribution of forces, highlighting that assumptions about equal normal forces can lead to errors in calculations. The discussion includes multiple equations and variables that are interconnected, which may complicate the resolution of the problem.

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Homework Statement


We have a 200-LB door hanging from 2 rollers, A & B, which are 5 feet apart. Coeffs of friction for A and B are .15 and .25, respectively. A force P is applied to the door. How big must it be to move the door to the left?[/B]

Homework Equations

The Attempt at a Solution


I have included a FBD. The downward force is the 200LBs, so the normal force would be 200, or 100 at each roller. That's reasonable, since the whole thing is symmetrical. The forces in x direction are P (to the left) and the total of muAN and muBN to the right.
So here we go:
Sum of forces in x: -P + (.15)(100) + (.25)(100) = 0
This yields a value of 40 for P. The book's answer is 45.5

Hmmm...been struggling with this for a week. Not sure where I made a wrong turn. Any help appreciated. Thank you.:biggrin:

[/B]
 

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SoylentBlue said:
The downward force is the 200LBs, so the normal force would be 200, or 100 at each roller. That's reasonable, since the whole thing is symmetrical

The pull force creates a moment about B so the downward force at A will be greater than 100.
You'll have to assume C is directly below B.
 
Thank you thank you thank you for pointing me in the right direction. Your answer was perfect--just a hint, without giving the answer away.
As soon as I saw your reply, I actually said out loud "D'Oh!" I am embarrassed that I did not see something so obvious.:H Ok, in my defense, I am still a student and still learning.

Anyway, for anyone else struggling with this, here is how it is solved. Note that we are looking at problem 8.16, in which the force pushes to the left, not to the right.Sum of forces about B: -6P + 2.5(200) – 5NA=0

Sum of forces in x: -P + .15NA + .25NB=0

Sum of forces in y: -200 + NA + NBEasy mistake to make: assuming NA and NB are equal. They are not!

OK, 3 equations with 3 unknowns, which is solvable. I won’t bore you with grunt-work.

NA=45.5

NB=154.5

Then P is 45.5, which agrees with the book’s answer.

Thank you again:smile:
 
SoylentBlue said:
Thank you again:smile:
You're welcome.
 

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