Plate sliding on ice with friction (Physics competition question)

In summary: What are the forces acting on the plate when it is treated as a free body and the motorcycle is on top of it?In summary, the problem involves finding the frictional force and work done on a plate by a motorcycle moving on top of it and then off of it. The first work done is equal to the first frictional force multiplied by a distance of 3 meters. The second work done is equal to the second frictional force multiplied by a distance of x meters. The sum of these two energies is equal to the kinetic and frictional energy of the motorcycle. However, there may be other forces at play and more information is needed to accurately solve the problem.
  • #1
Aristarchus_
95
7
Homework Statement
A person is riding a motorbike on an icy lake. The speed is constant at 70km/h relative to the ice. Air and rolling resistance together are constant equal to 400N. The mass of the motorbike with the rider is 140kg.
On the ice, there is a 10 kg metal late of a length of 3 m in the direction of motion. The motorbike crosses the plate. The upper side of the plate has sufficient friction so that the wheels will not slip. The friction coefficient between the plate and the ice is 0.18.
What distance will the plate move due to the passing of the bike (until it is again at rest)
Relevant Equations
Kinetic energy, friction, work
I reason the frictional force on the plate from the ice is doing work first 3 meters (while the motorbike is moving on top) and then an "x" distance after the motorbike has left it. Does anybody have an idea of how one might solve this problem?
 
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  • #2
What is 0.18? Some words are missing, maybe?
 
  • #3
nasu said:
What is 0.18? Some words are missing, maybe?
Fixed it. Thanks
 
  • #4
nasu said:
What is 0.18? Some words are missing, maybe?
Most likely a coefficient of kinetic friction but between what and what? The upper side of the plate is in contact with the motorcycle wheels, not the ice. It would make sense to assume that the wheels roll on the plate without slipping and that the coefficient of kinetic friction between the lower face of the plate and the ice is 0.18. @Aristarchus_ please confirm.
 
  • #5
Aristarchus_ said:
Fixed it. Thanks
Yes, as I thought. Thank you.

There are two entities that can exert forces on the plate: motorcycle and ice. Can you find expressions for each one of these when the motorcycle is (a) moving on top of the plate and (b) when the motorcycle is off the plate?
 
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  • #6
kuruman said:
Yes, as I thought. Thank you.

There are two entities that can exert forces on the plate: motorcycle and ice. Can you find expressions for each one of these when the motorcycle is (a) moving on top of the plate and (b) when the motorcycle is off the plate?
I wrote that the first frictional force when the motorbike is moving on top of the plate is ##R_1 = u*(m+m_p)*g##, where ##u=0.18## times the sum of the masses (m_p -mass of the plate) times the acceleration of gravity. Thus the first work done is R_1*(3m)
Then for the second situation, when the motorcycle is off the plate, the friction exerted is just R_2 = u*m_p*g. The work done here is R_2*x, where x is the distance the plate moves. These two energies combined equal the motorbikes' kinetic + frictional energy. However, I do get a big number, which made me suspicious...
 
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  • #7
Aristarchus_ said:
I wrote that the first frictional force when the motorbike is moving on top of the plate is #R_1 = u*(m+m_p)*g#, where #u=0.18# times the sum of the masses (m_p -mass of the plate) times the acceleration of gravity. Thus the first work done is R_1*(3m)
Then for the second situation, when the motorcycle is off the plate, the friction exerted is just R_2 = u*m_p*g. The work done here is R_2*x, where x is the distance the plate moves. These two energies combined equal the motorbikes' kinetic + frictional energy. However, I do get a big number, which made me suspicious...
Shouldn't you be aiming to calculate the kinetic energy of the plate just as the bike leaves it for the first step?
 
  • #8
erobz said:
Shouldn't you be aiming to calculate the kinetic energy of the plate just as the bike leaves it for the first step?
What answer do you get?
 
  • #9
Aristarchus_ said:
What answer do you get?
You said the work done was ##R_1 \cdot ( 3 \rm{m}) ##, I would say that isn't the whole picture as far as work done on the plate goes?
 
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  • #10
Aristarchus_ said:
I wrote that the first frictional force when the motorbike is moving on top of the plate is #R_1 = u*(m+m_p)*g#, where #u=0.18# times the sum of the masses (m_p -mass of the plate) times the acceleration of gravity. Thus the first work done is R_1*(3m)
Is friction the only force that does work on the plate at all times and only its magnitude changes when the motorcycle is off the plate? The plate is sitting there at rest minding its own business with no frictional force exerted by the ice. Does the ice suddenly change its mind and decides to exert a frictional force when the motorcycle gets to the plate?
 
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  • #11
kuruman said:
Is friction the only force that does work on the plate at all times and only its magnitude changes when the motorcycle is off the plate? The plate is sitting there at rest minding its own business with no frictional force exerted by the ice. Does the ice suddenly change its mind and decides to exert a frictional force when the motorcycle gets to the plate?
What is your method of solving this problem?
 
  • #12
Aristarchus_ said:
What is your method of solving this problem?
It seems like you are fishing for the answer, I don't think you are going to get a bite on that...

What are the forces acting on the plate when it is treated as a free body and the motorcycle is on top of it?
 
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  • #13
erobz said:
It seems like you are fishing for the answer, I don't think you are going to get a bite on that...

What are the forces acting on the plate when it is treated as a free body and the motorcycle is on top of it?
Friction, and the normal force from the motorbike, and friction from the motorbike
 
  • #14
Aristarchus_ said:
Friction, and the normal force from the motorbike, and friction from the motorbike
ok, you haven't represented the last of those forces in your calculation?
 
  • #15
Aristarchus_ said:
Friction, and the normal force from the motorbike, and friction from the motorbike
What is the direction of friction from the motorcycle?
 
  • #16
Do a Free-Body Diagram (FBD) for the motorbike to find the force at the rear tire. The key information is that speed is constant.

Then do an FBD for the plate. Here, the key information is that the plate may be accelerating (but not necessarily).

If the plate moves, you can find how much energy went to it from the motorbike, and how much kinetic energy it translates to.

You should figure out how to get the result you need afterward.

Lesson to learn: Always start with an FBD.
 
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  • #17
Aristarchus_ said:
I wrote that the first frictional force when the motorbike is moving on top of the plate is ##R_1 = u*(m+m_p)*g##, where ##u=0.18## times the sum of the masses (##m_p## -mass of the plate) times the acceleration of gravity. Thus the first work done is ##R_1*(3m)##
Then for the second situation, when the motorcycle is off the plate, the friction exerted is just ##R_2 = u*m_p*g##. The work done here is ##R_2*x##, where x is the distance the plate moves. These two energies combined equal the motorbikes' kinetic + frictional energy. However, I do get a big number, which made me suspicious...
To make the LaTeX work you need to put double hashes, as I have done above.

I would guess you are expected to ignore the facts that there will be two periods when only one wheel is on the plate, that the weight will not be distributed evenly between them, that the drags they exert will be in opposite directions, and that the bike will slow down a little.
 
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  • #18
With enough low friction between plate and ice (specially if a film of water beteen both is present), the drag of the front wheel may induce a forward movement on the plate.

Using front brake, even slightly, on wet leaves or street markings or a piece of sheetmetal can quickly demonstrate the above.

Irrelevant, I know... :smile:
 
  • #19
Aristarchus_ said:
What answer do you get?
erobz said:
It seems like you are fishing for the answer, I don't think you are going to get a bite on that...
Agreed. Please stop doing that.

haruspex said:
I would guess you are expected to ignore the facts that there will be two periods when only one wheel is on the plate, that the weight will not be distributed evenly between them, that the drags they exert will be in opposite directions, and that the bike will slow down a little.
There are actually multiple different interaction periods to this problem, if it is a competition problem and not an intro uni physics problem with assumed simplifications.

You are given the rolling resistance, which translates to friction pushing the plate forward while the front wheel is in contact with it. Is that enough to push the plate forward against that coefficient of friction while only the front wheel is rolling over it?

You are given the thrust of the rear wheel to propel the bike forward. Is that enough to propel the plate backward when only the rear wheel is in contact with the plate? How about when both wheels are in contact with the plate?

Time to get busy with FBDs and calculations... :wink:

Edit/Add -- and if the plate can slip during any of the n time periods that you divide this problem (and FBDs) up into, how does that affect the time duration of that period and the position changes for the plate? For example, if the rolling resistance of the front wheel is enough to push the plate forward before the rear wheel contacts it how does that change things? :smile:
 
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  • #20
berkeman said:
rolling resistance, which translates to friction pushing the plate forward while the front wheel is in contact with it
Yes, but the details are complex. Mostly, rolling resistance comes from the imperfect elasticity of the tyre (in this case) or of the substrate (caster on carpet) or both.
That shifts the average location of the normal force forwards, ahead of the lowest point of the wheel. This exerts a torque tending to slow the wheel, leading to a forward frictional force exerted on the plate.
 
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  • #21
Aristarchus_ said:
Air and rolling resistance together are constant equal to 400N.
Not solvable. The question setter does not know what "rolling resistance" is. It is not the same thing as an external retarding force.
 
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  • #22
jbriggs444 said:
Not solvable. The question setter does not know what "rolling resistance" is. It is not the same thing as an external retarding force.
Yes, it would be an interesting exercise to solve for the two extreme cases, all drag or all rolling resistance.
 
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  • #23
jbriggs444 said:
Not solvable. The question setter does not know what "rolling resistance" is. It is not the same thing as an external retarding force.
So if there's only rolling resistance, you are saying that a wheel-driven vehicle rolling without accelerating requires no friction force from the ground?
 
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  • #24
They are in good company... Apparently, I still don't understand how rolling resistance is to be utilized in Newtons Laws.

If it was all drag force surely its solvable?
 
  • #25
jack action said:
So if there's only rolling resistance, you are saying that a wheel-driven vehicle rolling without accelerating requires no friction force from the ground?
It will require more careful analysis, but my guess is it will turn out the same. Will check when I get time.

Edit: I was wrong. See post #29.
 
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  • #26
jack action said:
So if there's only rolling resistance, you are saying that a wheel-driven vehicle rolling without accelerating requires no friction force from the ground?
Yes, that is what I am saying.

Rolling resistance manifests as an external torque, not an ordinary external force. The engine's forward torque on the wheel goes into cancelling the external torque from rolling resistance. Under pure rolling resistance, there is no external linear force to resist and no static friction required to resist it.
 
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  • #27
jbriggs444 said:
The question setter does not know what "rolling resistance" is. It is not the same thing as an external retarding force.
FWIW I'd guess the intention is that
“Air and rolling resistance together are constant equal to 400N”​
is saying that air and rolling resistances can be treated as a single, constant retarding force of 400N.

To maintain the constant speed of 70mph on ice, it follows that there must be a 400N forwards force. This is the forwards frictional force exerted by the ice on the tyres; so the tyres exert a 400N backwards frictional force on the ice. When the tyres reach the plate they will exert a backwards frictional force on the plate.
 
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  • #28
jbriggs444 said:
Yes, that is what I am saying.

Rolling resistance manifests as an external torque, not an ordinary external force. The engine's forward torque on the wheel goes into cancelling the external torque from rolling resistance. Under pure rolling resistance, there is no external linear force to resist and no static friction required to resist it.
For the rolling resistance to exist, there must be rolling; for rolling to exist, there must be friction. Otherwise, the wheel would only spin freely, without any deformation, thus no rolling resistance.

The wheel torque must be applied to the ground first (thus the ground is pushing on the tire), then it fights back the rolling resistance it created by deforming the tire. If there is not enough friction force to compensate for the rolling resistance, the tire should only deform without rolling.

If that wasn't the case, it would mean that the available force would be:
$$F_{avail} = \min\{\frac{T}{r} - C_{rr}N; \mu N\}$$
But I always saw:
$$F_{avail} = \min\{\frac{T}{r}; \mu N\} - C_{rr}N$$

Edit: Changed ##\max## for ##\min## in equations.
 
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  • #29
Consider the pure rolling resistance scenario, no drag.
The tyre has a greater elastic constant during deformation than during relaxation. That displaces the normal force ##N## some distance ##s## ahead of the axle. Combined with the bike's load on the axle, that produces a retrograde torque ##Ns## on the wheel.

In the steady motion of the bike, the motor provides a balancing torque to the driving wheel. ##\tau_{motor}=Ns##. The reaction torque on the bike is matched by a slight shift in the distribution of weight between the two axles. At this point, no horizontal forces are involved, so there is no frictional force in the direction of travel (but some lateral to stay upright, no doubt).
To overcome an external retarding force ##F## instead, the motor would need to provide axle torque ##Fr##, where r is the wheel radius. Thus the external force equivalent, as far as the motor is concerned, is ##Ns/r##.

Now consider the idling wheel. With no motor to provide torque, the rolling resistance torque would tend to slow the wheel down. ##I_{wheel}\alpha=Ns##. To keep the wheel turning at the rolling speed, there will be a backwards frictional force from the ground. ##Ns=rF_{fric}##. That is transmitted through the axle to the bike and to the driving wheel. As the motor works to maintain speed, the driving wheel therefore experiences an equal forward frictional force.

Adding these up, the torque the motor has to supply is ##2Ns##, whence ##2Ns/r=400N##. The frictional forces the wheels will exert on the plate are ##Ns/r## forwards at the front wheel and ##Ns/r## backwards at the rear wheel.
 
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  • #30
haruspex said:
Adding these up, the torque the motor has to supply is ##2Ns##, whence ##2Ns/r=400N##. The frictional forces the wheels will exert on the plate are ##Ns/r## forwards at the front wheel and ##Ns/r## backwards at the rear wheel.
Just making sure I ineterpret correctly: If it was pure rolling resistance the plate wouldn’t move?
 
  • #31
erobz said:
Just making sure I ineterpret correctly: If it was pure rolling resistance the plate wouldn’t move?
I agree that as a first approximation the plate would not move. This ignores the distinction between front and back wheels and considers the motorcycle as a unicycle.

As a second approximation, the 200 N forward force on the plate from the free-wheeling front tire is adequate to cause the plate to shift forward at the beginning of the event when the plate is under only half load. Then it should shift rearward in a similar fashion at the end of the event. As I follow the sequence of accelerations, the net should be a slight forward shift. I believe that wheelbase information would be required to quantify it.

As a third approximation, we would need to contemplate mass distribution on the motorcycle and how rolling resistance is assigned to the two wheels (proportional to load?) and, perhaps, the behavior of the engine when the drive wheels are running against a plate that is sliding out from under. Not really worth pushing this far in my opinion.
 
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  • #32
Ok, so if there is a first order approximation net neutral effect from rolling resistance, then by adding drag by we should expect the plate to go backwards as if rolling resistance were irrelevant and solve the problem as stated?
 
  • #33
erobz said:
Ok, so if there is a first order approximation net neutral effect from rolling resistance, then by adding drag by we should expect the plate to go backwards as if rolling resistance was irrelevant and solve the problem as stated?
Yes, I would agree with that. In the unicycle model, rolling resistance is irrelevant.
 
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  • #34
jbriggs444 said:
As a second approximation, the 200 N forward force on the plate from the free-wheeling front tire is adequate to cause the plate to shift forward at the beginning of the event when the plate is under only half load.
But, if I understood correctly:

The static friction force between the plate and the ice would be much greater than the front wheel rolling resistance (and its equivalent friction force acting on top of the plate) and thus the plate wouldn't move with only the front wheel on the plate.

When both wheels will be on the plate, then the net horizontal force on top of the plate will be equivalent to the drag force, most likely pushing the plate backward. (Assuming the drag force is greater than the plate-ice static friction force.)

When only the rear wheel is on the plate, the net force increases to the sum of the drag force and the front wheel rolling resistance.
 
  • #35
jack action said:
But, if I understood correctly:

The static friction force between the plate and the ice would be much greater than the front wheel rolling resistance (and its equivalent friction force acting on top of the plate) and thus the plate wouldn't move with only the front wheel on the plate.
I did a quick and dirty calculation. Front wheel supports half of the cycle+rider mass: 70 kg or 700 N.
Add to that the 10 kg of the plate and we are up to 800 N.
Multiply by the coefficient of friction of 0.18 to yield 144 N.
Front wheel force is equal to half of the 400 N total rolling resistance. So that is 200 N.
200 N > 144 N, so this seems to be sufficient for the plate to slip on the ice.

Mind you, 200 N rolling resistance for a 700 N load is indeed ludicrously high. Thirty percent would be about right for a car on sand. A motorcycle on ice should be more like one percent. Reference here

[The calculation is for the case in which air resistance is negligible and the 400 N retarding effect is entirely from rolling resistance]
 
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