I was glancing through an old textbook for a course called “Principles of Technology” when I came across something that has been puzzling me. The book states that the force of friction on an object is the product of two components: (1) the force caused by the weight of the object, and (2) the types of material that are in contact with each other. This makes sense, but I had expected that the surface area that is in contact would also play a role. Isn’t that why wider tires are used on race cars?
Yes and no. You’re talking about the total force of friction being applied to the entire object.
However, friction need not be measured that way. The textbook is talking about the force of friction applied per unit size of surface area. If you know that, the former is trivially easy.
No, the textbook is talking about the total force. The idea is that there is a “coefficient of friction” and the area drops out of a ratio.
I think the idea of the normal and tangent forces being in constant ratio given the two materials is a pretty big oversimplification that often doesn’t work well enough. Race car tires, and the dynamics of their use, involve cutting things pretty close sometimes, and fine tuning all kinds of details. I bet there are all kinds of major deviations away from the “coefficient of friction” concept, so much so that it might not be useful at all in the analysis of race car tire design.
Think of a brick w/ a piece of felt on a big side and a piece on a small side. If you put the brick down in the flat-ish position, so that the big-side felt is in contact w/ the ground (or whatever), there is more surface area in contact. However, the weight gets distributed over a wider area, so there is less weight over top of each individual square inch. If you turn it so that the small felt side is down, there is less area in contact, but each inch has more weight behind it.
Turns out, it’s all proportional. Say the brick weighs 10 pounds. You can either have 10 pounds spread over 10 in @ 1lb/sqin or have 1 sqin @ 10lbs/sqin. It all comes out the same in the equations.
Does it make mathematical sense now? Does it make logical sense now?
As others have pointed out, in the idealized world of textbooks with microscopic friction the area is not important, just the force and the coefficient of friction.
In the real world things are not as simple. There can be macroscopic components to friction that do not scale linearly with pressure and hence the area of contact becomes important.
Yes, I had worked out what you have described here (very well) after I initially puzzled over why the surface area component was missing. That much makes sense – large area, weight is more spread out; small area, it is concentrated. But what is really puzzling me is why there are many examples in the real world of improving friction by creating more surface area. For example, why do dragsters have very wide – and even somewhat deflated for more contact – tires in the rear (where friction would matter) and very skinny tires in the front (where it wouldn’t)?
The classic friction calcualtion, where the frictional force is a function of only the normal force (and not the area) is a simplification. As Chessic Sense says, there’s a rough and fairly reasonable linear relationship between pressure and sustainable shear stress: ten square inches pressed together at 1 pound per square inch will support roughly the same shear force as one square inch pressed together at 10 pounds per square inch.
In real life, that’s a simplification. The sustainable frictional shear is a complex relationship dependant on exactly what the contact patch looks like and how the materials interact and any other materials (say, stones or water, in the case of a tire) in the interface.
However, in real life that’s often a fairly reasonable simplification. “Friction coefficients,” which relate the sustainable frictional force to the normal force for particular materials, are known only within a wide range, and can be off by10, 20, even 50%. The additional complexity required to increase the accuracy of the calculation often isn’t worthwhile, and an approximate calculation of friction is good enough. In this case, ignoring any effect of area is reasonable, and doesn’t introduce much additional error.
If anyone is knowledgeable about such things, I would be interested in hearing about what macroscopic components of friction or other factors might come into play with my wide tire example that are unaccounted for in the friction equation.
Here’s a quick reference on tire friction, and why larger tires might be useful. In a nutshell, the pressure underneath a rolling tire varies over the contact patch, and thus so does the sustainable shear. A wider tire distributes the load better.
With dragster tires in particular, I think there’s an adhesive term in addition to what might normally be called the frictional term. The frictional term is dependent on pressure, but the adhesive term is not. So when you make a large contact area, you keep the same adhesive force per area, but increase the total area, for more total adhesive force.
I was just thinking:
Could it also be that when driven over a real world surface that is not perfectly flat, a wider tire with more surface area is more likely to have at least some of its surface in contact with the road (as opposed to in the air)?
Yes, absolutely. The surface a tire rolls over is unsmooth even at a macro scale.
Consider what hapens when a car tire rolls over a rock about the size of a golf ball. Now imagine the car had (very strong) motorcycle-sized tires & rolled over the same rock. Now (very very strong) bicycle-sized tires. There is a big difference in the amount of disruption the same stone causes to the three contact patches.
A lot of the real-world variation in tire design has to do with performance on wet roads. On dry roads free of ball-bearing-like contaminants such as sand, a slick will always outperform a treaded tire of the same size and material. Add some loose sand or standing water, and the slick’s performance plummets, while the treaded tire degrades only slightly, at least for small amounts of contaminants.
As the contamination load gets bigger, eventually the tread is overwhelmed and the treaded tire’s performance plummets to something akin to the slick’s. Which is why snow tires are more aggressively treaded than all-weather tires, and off road mud tires more aggressively treaded yet. They can “absorb” more contaminants before their performance fails.
He set the brick on a board and measured the force to move it to show that it didn’t matter which side the brick was on.
It failed miserably; the larger surface area required more force. So the mathematical model lead us to think it wouldn’t make a difference. But the mathematical model and the real world model did not agree.
He was a crappy teacher anyway;
Bricks are bad examples. In fact, any rough surfaces are bad examples because the roughness of the surface is not really the same as friction. A better example uses smoother surfaces, like a brick with felt on it over a smooth surface like glass or a counter top. Otherwise, you’re just measuring the force necessary to get over and around rough obstructions on the brick and board.
Going back to tires, another factor is heat. Especially in race cars, the tires have the potential to melt. A bigger tire means more heat capacity and more area to dissipate the heat before it melts.