Friction: A non-uniform surface with a uniform weight

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Hello Physics Forums! Longtime lurker, first time poster here.

I've got a deceptively simple question that had my friends and me occupied (read: bickering) for quite some time during a road trip, and I thought I might pose it to you. The situation is this: we have a 2x4 piece of wood, and we've covered half of one of the long sides with sandpaper and the other half with teflon. Let's forget for now that the sandpaper would increase the height of that side compared to the teflon, and imagine that both halves would contact the ground equally. Now we push the piece of wood (textured side down) perpendicularly to the ground and calculate the coefficient of friction to be [itex]\mu[/itex][itex]_{1}[/itex].

The question is this: if we change the proportion of sandpaper/teflon and push the wood as before, will the coefficient change from [itex]\mu[/itex][itex]_{1}[/itex]? In other words, if we make 90% of the surface covered in sandpaper, would it have a greater frictional force? It seems that, as per Amontons' Second Law, the surface area of either side would make it such that the frictional force (and thus the coefficient) wouldn't change with the surface area of the sandpaper or the teflon. However, it was also argued that given a large sandpaper/teflon value, there would be more weight on the sandpaper covered end, giving a larger frictional force. This implies that the force of friction would be something like this:

Given S = dimension of sandpaper covered wood, T = dimension of teflon covered wood, and W = total surface area in contact with the ground,

F[itex]_{f}[/itex] = F[itex]_{N}[/itex]*([itex]\frac{S}{W}[/itex]*[itex]\mu[/itex][itex]_{sandpaper}[/itex] + [itex]\frac{T}{W}[/itex]*[itex]\mu[/itex][itex]_{teflon}[/itex])

After a good amount of time spent googling for an answer, it became clear that this situation just doesn't come up in PHY 101 classes, and thus isn't on the internet! What do you think?
 

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