# Friction of a sliding door with one of its wheels "frozen"

1. Jul 21, 2015

### J-dizzal

1. The problem statement, all variables and given/known data
The sliding glass door rolls on the two small lower wheels A and B. Under normal conditions the upper wheels do not touch their horizontal guide. (a) Compute the force P required to slide the door at a steady speed if wheel A becomes "frozen" and does not turn in its bearing. (b) Rework the problem if wheel B becomes frozen instead of wheel A. The coefficient of kinetic friction between a frozen wheel and the supporting surface is 0.30, and the center of mass of the 133-lb door is at its geometric center. Neglect the small diameter of the wheels.

2. Relevant equations
FkkFN
where FN is the normal force.

3. The attempt at a solution

I thought symmetry of the door would result in 133/2 lb at each roller.

Last edited: Jul 21, 2015
2. Jul 21, 2015

### Dr. Courtney

I think there is an equilibrium problem hiding in here. The door will rotate up and one of the top wheels may contact if a wheel freezes.

3. Jul 21, 2015

### J-dizzal

I was wondering what all the length dimensions were for...
So then would there be a moment about the non frozen wheel? and then that would cause one of the top wheels to touch? But if a top wheel touches it should roll freely and not add any more friction force.

4. Jul 21, 2015

### Dr. Courtney

It might depend on which wheel is frozen. The frozen wheel becomes the pivot point. You need to consider the equilibrium conditions.

5. Jul 21, 2015

### J-dizzal

For equilibrium conditions, would that be finding the forces at each point knowing that the sum of forces and moments is equal to zero?

edit; ΣMA= -133(14.5) - P(40). P =-48.21 lb Im confused on why this equations solves for P but its not correct.

Last edited: Jul 21, 2015
6. Jul 21, 2015

### J-dizzal

Ok the frozen wheel becomes the pivot point, so i did sum of moments at the frozen wheel and found P=-48.21 but its the wrong answer. I did sum of forces and im going in circles.
My original picture is updated.

7. Jul 21, 2015

### J-dizzal

Im getting stuck at part A. I dont think i need the moment equations because its not tipping its slipping. so then when i do a sum of forces i can only find the normal force which is half of the weight. Then using that to find the kinetic friction force is still giving me the wrong answer for P

8. Jul 21, 2015

### haruspex

Not necessarily. In particular, not if A freezes.
The normal forces will be different at A and B. Put in unknowns for those.
In general, there may also be a normal force at the upper wheel above A.
Do you see why there can never be a normal force at the other top wheel?
Can there be a normal force at the top and bottom wheels on the same side?
How many cases does that give you?

9. Jul 21, 2015

### J-dizzal

For the normal forces, ill have to solve those using moment equations.
there will never be a moment force at the top above B because both the wheels move up top and the moments cancel out, like two hinges.
No, i dont think there will be a moment at the top because of the moments canceling out.
im not sure on how many cases, I only see it one way, two if you include part B of the question.

10. Jul 21, 2015

### haruspex

Sure, so create unknowns and post some equations. Please use only symbols for all the variables, both known and unknown, i.e. don't plug in the numerical distances and masses etc.
I was including both parts of the question. I agree there is only one case to consider in part A (which is that case, and why is it the only one?). There may be two cases to consider for part B.

11. Jul 21, 2015

### J-dizzal

ΣMA=-W(14.5) - P(40)=0. this equation solves for P but its not the correct answer. This appears to be the only moment equations i can create.
There will never be a normal force at B because the moments will rotate the door clockwise.
It is the only case to consider in part A because wheel A is the only point where moments are taken.

edit. my moment equation is missing By(29), im missed this before thinking it was zero because its on the same line of action.

Last edited: Jul 21, 2015
12. Jul 21, 2015

### J-dizzal

At point A if i call the resultant forces in the x and y directions Ax and Ay, is this name convention redundant because the friction force would replace Ax?

13. Jul 21, 2015

### J-dizzal

Here is what i have so far,
ΣMA=-133(14.5) - P(40) + NB=0
ΣFx=P-FkA=0
ΣFy=-133+NA+NB=0
I have 4 unknowns and only 3 equations. I think there must be a way to solve the Normal forces but i just dont see it yet.

14. Jul 21, 2015

### haruspex

That is precisely why the normal force at B will be large!
Anyway, that would give you a negative P.

In general, for equilibrium (non-acceleration) of a rigid body in a 2D problem you have three equations available. Typically one uses two linear force balances at right angles and moments about one reference point. Quite often there is a force which is both unknown and uninteresting in regards to the question, so judicious choice of axis can eliminate that and you then only need a linear equation that also does not involve that force. (I go into more detail in part 4. of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/)
Since part A involves friction at A, it would seem more useful to take moments about B. Which linear force balance equation does not involve the normal force at B?

15. Jul 22, 2015

### J-dizzal

This is what i dont understand, if B is free to rotate then how can a moment equation be made? because it only resists movement in the y axis. This is what i learned form previous equilibrium materials, why does it change now?

sum of forces in the x direction, does not involve the normal force at B.

I only put in the given dimensions from the problem no calculations were made.

16. Jul 22, 2015

### haruspex

I don't think I am going to understand your difficulty here unless you can quote a prior example.
It might help if you would answer a question I asked in post #10:
That still constitutes plugging in numbers. The consequence is that I have to go back to the OP and reverse engineer your equation to figure out what those numbers represent. Sometimes that's not clear because the same number might crop up in different variables.

17. Jul 22, 2015

### J-dizzal

there is only one case here, this case is when wheel A freezes and there is a moment about A. None of the top wheels would contact their frame because the door is not tipping. It is not tipping because the location of the normal force and friction force are not at the bottom right corner. Additionally, wheel B stops tipping by contributing a normal force.

ΣMB im getting the same result at before with the sum of moments at A.
ΣMB=W(Length of half the distance from A to B)-P(Length from P to bottom right corner)=0
If im wrong and there is a normal force because the door is tipping then add into the moment equation: (Top A)(vertical length of door)

18. Jul 22, 2015

### haruspex

As I wrote, there is a normal force at bottom right, but yes, the reason it won't tip is that there is no friction at bottom right. If it started to tip there'd be no friction at A either, so no horizontal force to counter P.
Don't assume there's no normal force at A.
What linear force equation does not involve the normal force at B?

19. Jul 22, 2015

### J-dizzal

sum of forces in the x direction; ΣFx=P-(force of friction at A)=0.
unknowns;P,NB,NA,and friction force at A.
But only 3 equations, moment and two linear force equations.

20. Jul 22, 2015

### haruspex

You know the relationship between the frictional force and NA.
If you take moments about B, NB does not feature.