Plate sliding on ice with friction (Physics competition question)

[jbriggs444]

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.

 

[erobz]

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?

 

[jbriggs444]

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.

 

[jack action]

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.

 

[jbriggs444]

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]

 

[erobz]

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]

It seems like rolling resistance would tend to be negligible for these conditions for both its very low magnitude, and this tendency toward canceling.

 

[Lnewqban]

It seems like rolling resistance would tend to be negligible for these conditions for both its very low magnitude, and this tendency toward canceling.

Not if the plate is very thick (or slightly concave) and light and the tire hits it and climbs over it, …perhaps.

 

[erobz]

Not if the plate is very thick (or slightly concave) and light and the tire hits it and climbs over it, …perhaps.

Fair enough, but we probably aren't talking about a metal plate then. A 10 kg metal plate of length 3m is probably not going to be particularly thick.

Lets say it's an aluminum strip that's about 1/3 m wide. It's about 3 mm thick.

 

[jack action]

Front wheel force is equal to half of the 400 N total rolling resistance. So that is 200 N.

No, 400 N is total rolling resistance + drag force. The drag force component would be entirely on the rear wheel.

Assuming a rolling resistance coefficient $C_{RR}$, a friction coefficient $\mu$, a plate of weight $W_p$, and a front wheel normal force $N_f$, the horizontal forces acting on the top of the plate is $C_{RR}N_f$ and the static friction at the bottom of the plate is $\mu (N_f + W_p)$.

Even if the plate weight was very small compared to the front wheel's normal force, for the plate to move, $C_{RR} \geqslant \mu$. This is a very unreasonable assumption, especially when $\mu = 0.18$.

 

[haruspex]

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?

Not by adding drag, by substituting it for rolling resistance. In general, drag is an unknown fraction of the 400N. To solve it you need to know exactly how much of each.

 

[erobz]

Not by adding drag, by substituting it for rolling resistance. In general, drag is an unknown fraction of the 400N. To solve it you need to know exactly how much of each.

Ok, the gripe is we don't know the magnitude of the drag force, which is what would (indirectly) be accelerating the plate. I think I get it now.

The author, trying to be overly specific, actually introduced an issue that they could have simply avoided by stating rolling resistance is negligible...if it's not negligible absolutely (in theory) it almost certainly is negligible relative to the drag force acting on the motorcycle/rider traveling at 70 kph for this scenario.

 

[jbriggs444]

No, 400 N is total rolling resistance + drag force.

In the case that I was considering, drag force was explicitly set to zero.