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The classical model of friction has a coefficient of friction depend only on the materials, but not area, and the force proportional to the normal force and coefficient of friction. So a given object on the same surface has the same friction whether it is supported by full bottom area or small legs as long as the materials are the same.

However every child knows that on a slide one goes faster if one lays down on their back compared to sitting on their butt. The slide is obviously still the same and since jackets usually extend below butt, the other material is also the same. So the friction should be the same as well, but it clearly isn't. So what is going on here?

Note: I mean typical stainless steel or fibreglass laminate slide, not ice, which is soft enough to complicate the matter further.

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When the child lies down there is less deformation of the slide's surface than when he is sitting.   Whether the deformation is relatively deep and short in length, or more shallow and extends far beyond the child, it is always nonzero.  The force required to move the trough of this deformaton down the slide opposes the component of the child's weight directed down the slide.

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My guess would be that if you're sitting, the surface of contact between you and the slide is smaller, thus the pressure is higher than if you were to lay on your back. Your jacket is a deformable material, its friction coefficient with steel might vary with applied pressure.

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In laboratory experiments with bricks, the sliding angle comes to be the same irrespective of the orientation of the brick. However, when it comes to a child sliding there are some extra factors. The child usually is not sitting at normal to the plane especially at large angles of sliding due to the fact that he can topple more easily than slide. He wants his COM nearer to the slide in order to prevent the toppling and so lays on his back. If he stands to the normal, then the free body diagram states that the acceleration of his centre of mass is dependent on the frictional force. So along with acceleration, he also topples.

If a standing child starts to topple, his feet won't move much, and therefore he won't 'slide', although his COM definitely would and therefore would land on his head. Thus in order for his fun and security, the child prefers to lay on back instead of doing anything else. The jacket may not play a big role here. The coefficient of friction should remain the same and not dependent on the surface area. I encourage you to try the same thing too. :P

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  • $\begingroup$ I am comparing laying with sitting on the butt, not standing. Standing is completely irrelevant, for the reasons you mention and also because shoes have much higher friction than jacket (by design). When sitting, the CoM is slightly higher then laying, but low enough to slide without toppling. $\endgroup$ – Jan Hudec Feb 1 '16 at 17:56
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Going faster on a dry playground slide if you lie down seems not to be well-documented, so I shall discount the effect of posture. The amount of friction depends on your weight, which is the same lying down as it is sitting up.

If there is any difference on a dry slide it is probably due to the reduced coefficient of friction provided by clothing, especially wool, nylon, polyester and polythene. Skin and rubber have relatively high coefficients, so lifting shoes and bare legs and arms off the slide makes a big difference. It makes no difference whether you lie down, sit up or stand (surf- or skate-board style) if the material beneath you is the same.

(I once almost slid off a steep rock face while mountain climbing by lying down on nylon clothing to avoid losing my balance, instead of standing upright on rubber-soled boots as my experienced companions did.)

On a water slide the key difference is the lubrication provided by the stream of water which flows down with you. Lying down spreads your weight over the greatest area, reducing the pressure beneath you, which allows a relatively thick layer of water to lubricate your motion over the plastic slide. Sitting up squeezes the water beneath you into a much thinner layer which offers less lubrication.

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See this answer: What happens to the coefficient of friction as the normal force increases?

The "classical model of friction" - the expression $f_k=\mu_k n$ - holds for small normal forces $n$ only. Small means that the normal force per area (the normal pressure $q=n/A$) is much smaller than the material's shear strength $k$ (which is how much sideways force the material can withstand per area).

Firstly, friction will certainly change while the snow is being compressed. In new powdered snow I would expect $k$ to be much smaller, and even after making it compact it is still not fully compressed and stone-hard. The normal pressure $q$ is higher when sitting than lying, because lying increases surface area for the same normal force. So sitting might bring $q$ close to $k$, and so the expression wouldn't hold and friction would be depending on area.

On a well-used sliding hill the snow would assumably be much more compact. Here I would expect the expression to hold.

With all that being said, remember the not insignificant change in air-resistance when lying down compared to sitting.

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  • $\begingroup$ I had a playground slide—made of glass laminate or polished metal—in mind rather than snow. Those should have quite a bit of shear strength and still lying makes a big difference on them compared to sitting in practice (it is worth noting that shear strength of the clothing and body is much smaller). $\endgroup$ – Jan Hudec Dec 4 '16 at 22:08
  • $\begingroup$ @JanHudec Hmm okay. Did you transit from sitting to lying during the sliding? That would have a huge effect on the felt effect, I would pressume. $\endgroup$ – Steeven Dec 4 '16 at 22:11
  • $\begingroup$ This is the right physics, but you haven't given he OP a way to quantify the smallness of a force. I usually go with 'neither material is significant deformed, neither is permanently deformed, and no abrasion occurs' as a rule of thumb for when you can apply the naive model. $\endgroup$ – dmckee Jan 6 '17 at 18:43

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