# Plate sliding on ice with friction (Physics competition question)

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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.

jbriggs444
Homework Helper
Gold Member
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
Gold Member
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.

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Lnewqban
Gold Member
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##.

Homework Helper
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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.

jbriggs444, Lnewqban and erobz
Gold Member
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.

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Lnewqban