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

Click For Summary
SUMMARY

The discussion centers on calculating the force required to move a 200-pound door hanging from two rollers, A and B, which are 5 feet apart, with coefficients of friction of 0.15 and 0.25, respectively. The user initially calculated the force P as 40 pounds, while the correct answer is 45.5 pounds, as confirmed by the book. The solution involves resolving forces in both the x and y directions and recognizing that the normal forces at rollers A and B are not equal due to the applied moment. The final values for the normal forces are NA = 45.5 pounds and NB = 154.5 pounds.

PREREQUISITES
  • Understanding of statics and force equilibrium
  • Knowledge of friction coefficients and their application
  • Ability to set up and solve systems of equations
  • Familiarity with free body diagrams (FBD)
NEXT STEPS
  • Study the principles of static equilibrium in mechanical systems
  • Learn about the calculation of moments and their effects on forces
  • Explore the role of friction in mechanical systems and its implications
  • Practice solving similar problems involving multiple forces and friction
USEFUL FOR

Students of engineering mechanics, particularly those focusing on statics, as well as educators and professionals seeking to reinforce their understanding of force analysis in mechanical systems.

SoylentBlue
Messages
50
Reaction score
8

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]
 

Attachments

  • Prob816FBD.jpg
    Prob816FBD.jpg
    6 KB · Views: 531
  • Prob816.jpg
    Prob816.jpg
    49.1 KB · Views: 619
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
3
Views
4K
Replies
3
Views
8K
Replies
5
Views
9K
  • · Replies 3 ·
Replies
3
Views
13K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 14 ·
Replies
14
Views
2K