So, I asked a pretty innocuous question. What's more slippery? Smooth Ice, or bumpy Ice? Well, ice is slippery not because it's smooth, though its reasonably so. The exact reason ice is slippery is apparently poorly understood. Back when I learnt my elementary physics, I was told a fairly simple explanation, one that'd been repeated through many text books over the years. This explanation, that ice's melting point drops when pressure is increased, is partially true, (though not the whole story), but its what's relevant to this discussion. As pressure on the ice increases, the melting temperature of the ice decreases (and becomes more slippery for our purposes).
When we're walking on smooth ice, the whole of our mass is distributed across the surface of the ice. However, when you walk across bumpy ice, your foot only comes in contact with a smaller percent of it. Let's say you you weigh 150 lbs. I'll use English units, even though it makes me feel scientifically dirty. Using a quick approximation, the surface area under a men's size ten boot is around 34 square inches, for a total of 68 inches total in contact with the ground. Your feet exert a total of 2.2 pounds per square inch on the ground. However, in reality, the whole of the bottom of the boot isn't in contact with the ground. For every 50% reduction in area, the pressure increases by 50% - 4.4 psi for half the surface area, 8.8 for 25%, etc.
Now imagine a complex boot sole coming into contact with either of our ice surfaces. One is bumpy, and one is smooth. Which one has the greater contact between the boot and the ground? Obviously, the smooth one. The bumpy surface, you'll only come in contact with the tops of the crests. The surface area is then very small, and so the pressure quite large. At this point, it is easier to slip and fall. Bumpy ice is more slippery, C.P.
This is part of the reason that old style boot soles are much better. A old style kamguk (mukluk in English) has a soft sole, meaning the foot conforms to the ground. This has two immediate effects. First, it increases the surface area, so pressure is more evenly distributed. Secondly, that increased surface area means the friction between the ice and the boot is greater. I'd actually be interested in demonstrating this numerically, which means getting a force meter. An idea for another day!
Here's another fun one from the book! And a bit easier, since I think a lot of people intuitively know the answer. Let's say you want to cool a completely full pot, and you only have one block of ice. Should you put the ice below it, above it, or to the side of it? Why?
3 comments:
I respectfully disagree. If you look at the phase diagram of water, you'll find that you need a very high pressure, much higher than a human's weight, to appreciably cause melting. Also, if you work out the calculations, pressure melting could no-way-no-how cause slipperiness below -20F, but clearly we can still slip on ice (and whang ourselves while crossing the street, which makes us feel really awesome, and I ought to know) at -20F.
The reason that smooth ice is less slippery than bumpy ice is not that the applied pressure is lower. It's simply that there is more contact area. Ice is slippery indeed, but it still provides more traction, after all, than air. :)
And in case you're wondering, the current theory on why ice is slippery is that the dangling bonds at the edges of ice crystals form a liquid-like-layer, which thins with decreasing temperature. That's why ice ceases to be slippery at -40. Ice only *seems* to have strange behaviors because we live near its melting point. If we were to observe, say, iron near its melting point, we'd see it have a slippery surface, also.
I was hoping you'd post! Could you elaborate on the calculations? I'd be interested in seeing the actual math. My source (the Russian book) says `The math clearly shows blah...` but doesn't show the numbers at all. And I know it was wrong about other stuff. Or, not wrong, per-se, but not entirely correct after ~90 years of science.
Haha! The ol' I.C.B.S. ("It can be shown that...")
I actually recommend you just print out the phase diagram of water, and use your ruler to interpolate the amount of pressure you'd need to bring the melting point of ice to the current air temperature. Then figure your weight, and the area of your shoes, and you'll see it just doesn't work.
Maybe it'll work if I stand only on the heel of a stiletto-heeled shoe. Then my pressure is something like 20x my regular foot pressure, eh?
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