How to create an equation for changing friction?

[fi245]

Homework Statement

A piece of butter is sliding down a warm roof. It is known, that on contact, the friction decreases to one tenth per 5 seconds. Not taking into count the fact that the mass reduces due to melting, how can you create the equation for the changing friction, that takes into account both that it is dependent on both time and place?

Homework Equations

The time-depended part could be thought of as µ0exp( − t/ T), where T=5/log(10).

The Attempt at a Solution

 

[fi245]

to clarify, when the piece moves, the front edge
gets in contact with the untouched surface of the roof. The decrease in the friction face starts at the time when the piece
touches the warm roof and begins to melt. And so the friction is dependent on both time and place.

 

[haruspex]

the friction decreases to one tenth per 5 seconds.

Not sure what that means. Does it mean that for any part in sustained contact, the coefficient reduces negatively exponentially, so that after each 5 seconds it is one tenth of what it was at the start of the 5 sconds?
And when does the time start? As it slides down a small distance dx, is the leading dx presumed to be back at its original friction?

 

[fi245]

^^yes, that's how we thought the coefficient would reduce. in this problem, we are not taking account when the movement starts, but when it is already moving, so the time starts when the butter has moved a small distance dx. at this point, the small part of the butter (dx) has the original friction (since the rooftop is "untouched").

 

[jbriggs444]

to clarify, when the piece moves, the front edge
gets in contact with the untouched surface of the roof. The decrease in the friction face starts at the time when the piece
touches the warm roof and begins to melt. And so the friction is dependent on both time and place.

I am not following. A pat of butter on a roof will have all parts of the pat in contact with the roof at all times unless it is falling off the edge.

 

[fi245]

yes of course, but the friction is different on different parts of the block. in the front edge, the butter comes in touch with "new" rooftop surface, where the friciton is at maximum, and on the other hand behind this front edge the butter has already melted a bit ->reduces the friction (lubrication).

 

[jbriggs444]

yes of course, but the friction is different on different parts of the block. in the front edge, the butter comes in touch with "new" rooftop surface, where the friciton is at maximum, and on the other hand behind this front edge the butter has already melted a bit ->reduces the friction (lubrication).

The problem statement says no such thing and does not quantify any such thing.

 

[haruspex]

The problem statement says no such thing and does not quantify any such thing.

Well, that is the only reading I could come up with.
As regards how it work this way, it does seem completely backwards. The leading edge of the butter has already has warmed, and makes contact with fresh, i.e. hottest roof. Meanwhile, at the trailing edge, the roof has been cooled somewhat by the butter.

@fi245, you need to show some attempt. At least set out your thoughts so far.

 

[rude man]

I read it a simply meaning that the coeff of dynamic friction is a function of time: μ_d(t) = µ₀exp( − t/ T).
t=0 is when the butter slab is let go.
Obviously you can't also make it an explicit function of position unless you're given the slope of the roof tanθ.
In any case m d²x/dt² = mg sinθ - u_d(t) mg, x = distance along roof surface would be my guess. x=0 when t=0.
Readily solved linear, const coeff. ODE.

 

[haruspex]

I read it a simply meaning that the coeff of dynamic friction is a function of time: μd(t) = µ0exp( − t/ T).
t=0 is when the butter slab is let go.

Post #1 can be read that way, but not post #2.
@fi245 , can you provide the exact wording?

 

[rude man]

Post #1 can be read that way, but not post #2.

why not?

 

[haruspex]

why not?

Post #2 says that t=0 for a given part of the contact area is when that part of the roof first made contact with the butter.
But I agree this is unlikely to be the original intent of the question.

 

[rude man]

You plop a stick of butter on a virgin sloping roof at t=0. All the butter instantly makes contact with the roof. The stick starts to slide down the roof and as it does so its net dynamic friction coefficient reduces per the given relation. I see nothing complicated here, and I certainly can't make heads or tails of post 6, and I completely agree with post 7.

 

[jbriggs444]

I have a guess at an intended meaning that would be consistent with a distance dependence. What if the coefficient of friction is a function of how long a point on the roof has been in contact with the butter rather than a function of how long a point on the butter has been in contact with the roof?

 

[haruspex]

I have a guess at an intended meaning that would be consistent with a distance dependence. What if the coefficient of friction is a function of how long a point on the roof has been in contact with the butter rather than a function of how long a point on the butter has been in contact with the roof?

yes, that is how I interpret post #2, but the formula provided does seem backwards. At the leading edge, we have already warmed butter contacting roof at its hottest. That should give the least friction.

 

[rude man]

I have a guess at an intended meaning that would be consistent with a distance dependence. What if the coefficient of friction is a function of how long a point on the roof has been in contact with the butter rather than a function of how long a point on the butter has been in contact with the roof?

My post 13 includes distance dependence too. The relationship between time dependence t and distance dependence x is determined by the slope of the roof, since that determines the gravitational force propelling the butter as well as the dynamic friction force:
mx double-dot = mgsinθ(1 - μ₀e^-t/T).