I did an experiment the other day in my Grade 11 Physics class. We were measuring the force required to move a few different objects. One object was this block; one side was two rubber hockey pucks, and the other side, was a piece of lumber. We used a dynamometer and we recorded the force just before the object moved. These are the results/dimensions of the objects:

(in this case, the force reading is right before movement, so the force exerted by me = force of friction)

rubber side : area = 88,36cm^2, mass = 890,2g, friction = 7,22N

wood side : area = 91,14cm^2, mass = 890,2g, friction = 3,32N

I am trying to figure out what caused the rubber side's friction to be so much higher. I'm thinking either the rubber is better at gripping to the surface (laminated table) or the rubber side is smaller, so there is more mass per square cm, causing more friction, or both, or neither of those hypotheses.

The problem with that second hypothesis is that we measured the friction for a block of wood, about the same size as the other chunk of lumber, but with a smaller mass. Yet, the lighter block had about the same amount of friction. Here are the results/dimensions for the lighter block:

small wood: area = 85,68cm^2, mass = 197,6g, friction = 3,10N

I can't, for the life of me, figure out what is going on with the experiment. I saw a few suggestions like the wood is more lubricated than rubber, the rubber is more malleable (at a microscopic level, anyway) that it "fits" into the other surface better, etc. I don't know which is right, or if some kind of human error is involved.

P.S. Sorry, not all my units are SI, I figured that having "cleaner" numbers makes it a bit easier to visualize. These are all significant digits.


2 Answers 2


Friction is a complicated matter that is still not clear in many aspects. The degree of friction crucially depends, among other things, on the surface details of the contacting materials. This suggests that your first scenario is more likely (rubber is "grippier" that wood). The second hypothesis is less likely in my opinion. Moreover, there is this "Amonton's law" of (dry) friction that states that the only factors that determine the friction force $f_{fr}$ is the type of materials in contact quantified by the "coefficient of friction" $\mu$ and the force with which the one is pressed against the other $N$ (typically the weight of the sliding body): $f_{fr}=\mu N$. You sure remember this expression from your textbook, but if you think over it, it is almost mysterious: it tells you that contact area ${\it does\, not}$ matter. So if you take say a steel block with various facets, it will not matter for the friction force which side it is lying on. Weird as it is, it is an empirical law valid for what is called "dry" friction. If we assume your hockey puck does not stick to the table, nor do other parts of your sample make dents in the table top, Amonton's law should be valid for your experiment too, hence the contact area should not matter and it should be the difference in $\mu$ (different "grippiness"), which as a matter of fact can be very different for different sorts of wood; it can also strongly depend on the condition of the wood (e.g. how dry or polished it is) which can explain your results with different blocks of wood which seemingly violate Amonton's law.


My instinct would be to attribute it to the material properties of the rubber being "grippy."

A potential way to test this would be to see if one of the following tests increases the starting friction

  • Mash the rubber and the wood into the table before you try tugging them.
  • Increase the temperature in the room somewhat before you try tugging them.

If either of these changes the result, then likely what you are observing is due to material properties of the objects.

It's also worth noting that this is quantified as the coefficient of static friction, and in general $\mu_s$ varies with material properties -- consider that brake pads in an auto aren't typically made out of wood. Under common assumptions, and in the experiment you're performing with a level table the friction force is proportional to the coefficient and the weight, $f=w\mu_s=mg\mu_s$


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