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

[J-dizzal]

Homework Statement

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.

Homework Equations

F_k=μ_kF_N
where F_N is the normal force.

The Attempt at a Solution

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

 

[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.

 

[J-dizzal]

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.

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.

 

[Dr. Courtney]

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

 

[J-dizzal]

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

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; ΣM_A= -133(14.5) - P(40). P =-48.21 lb I am confused on why this equations solves for P but its not correct.

 

[J-dizzal]

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

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 I am going in circles.
My original picture is updated.

 

[J-dizzal]

Im getting stuck at part A. I don't 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

 

[haruspex]

The door will rotate up and one of the top wheels may contact if a wheel freezes.
...
The frozen wheel becomes the pivot point.

Not necessarily. In particular, not if A freezes.

i can only find the normal force which is half of the weight.

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?

 

[J-dizzal]

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?

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 don't 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.

 

[haruspex]

For the normal forces, ill have to solve those using moment equations.

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.

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.

I asked about forces, not their moments.

im not sure on how many cases, I only see it one way, two if you include part B of the question.

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.

 

[J-dizzal]

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 asked about forces, not their moments.
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.

ΣM_A=-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 B_y(29), I am missed this before thinking it was zero because its on the same line of action.

 

[J-dizzal]

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

 

[J-dizzal]

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.

Here is what i have so far,
ΣMA=-133(14.5) - P(40) + N_B=0
ΣF_x=P-F_kA=0
ΣF_y=-133+N_A+N_B=0
I have 4 unknowns and only 3 equations. I think there must be a way to solve the Normal forces but i just don't see it yet.

 

[haruspex]

There will never be a normal force at B because the moments will rotate the door clockwise.

That is precisely why the normal force at B will be large!

ΣM_A=-W(14.5) - P(40)=0.

What part of "please don't plug in numbers" was unclear? Please read part 1. of https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/
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?

 

[J-dizzal]

it would seem more useful to take moments about B. Which linear force balance equation does not involve the normal force at B?

This is what i don't 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.

please don't plug in numbers" was unclear?

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

 

[haruspex]

This is what i don't 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?

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:

there is only one case to consider in part A (which is that case, and why is it the only one?).

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

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.

 

[J-dizzal]

which is that case, and why is it the only one?

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.

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

 

[haruspex]

this case is when wheel A freezes ... It is not tipping because the location of the normal force and friction force are not at the bottom right corner.

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.

ΣM_B=W(Length of half the distance from A to B)-P(Length from P to bottom right corner)=0

Don't assume there's no normal force at A.
What linear force equation does not involve the normal force at B?

 

[J-dizzal]

What linear force equation does not involve the normal force at B?

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

 

[haruspex]

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

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