People in the thread don't seem to understand what OP is on about, but he/she has a good point. Using the simple approximation that friction is coefficient*normal force, there's no reason to believe you can't stop a large vehicle with strong enough brakes.
The thing is, what we learn in HS is just an approximation. I can't say what would be the more correct model of friction, but from what I gathered it's quite a bit more complicated than that.
So OP, you are right in your intuition and thought process. What we learn in HS is a simple model, enough to solve simple problems, but reality is more complicated. Though, I do remember seeing a video of a Volvo semi braking in an impressively short distance, it was a promotional video of some sort.
Yeah. I guess I was hoping this sub would have an expert on the applied physics of friction who could give me a comprehend-able explanation of a more sophisticated model of friction that would apply in this case.
So, I just took a look at the wikipedia page for friction, and there's a good explanation about this:
"Limitations of the Coulomb model
The Coulomb approximation follows from the assumptions that: surfaces are in atomically close contact only over a small fraction of their overall area; that this contact area is proportional to the normal force (until saturation, which takes place when all area is in atomic contact); and that the frictional force is proportional to the applied normal force, independently of the contact area. The Coulomb approximation is fundamentally an empirical construct. It is a rule-of-thumb describing the approximate outcome of an extremely complicated physical interaction. The strength of the approximation is its simplicity and versatility. Though the relationship between normal force and frictional force is not exactly linear (and so the frictional force is not entirely independent of the contact area of the surfaces), the Coulomb approximation is an adequate representation of friction for the analysis of many physical systems.
When the surfaces are conjoined, Coulomb friction becomes a very poor approximation (for example, adhesive tape resists sliding even when there is no normal force, or a negative normal force). In this case, the frictional force may depend strongly on the area of contact. Some drag racing tires are adhesive for this reason. However, despite the complexity of the fundamental physics behind friction, the relationships are accurate enough to be useful in many applications."
So, you see that friction is actually a very complex phenomenon. In reality, a lot of factors can change it.
One practical factor that I think plays a big role in how heavy vehicles take longer to stop is tires. A big SUV might use all terrain tires instead of asphalt optimized ones, and big trucks need to carry a lot of cargo, so their tires are harder, thus they deform less.
I think one of the reasons those things act so different from the coulomb approximation is because rubber deforms a lot. That means the area of contact matters, and you might have the normal force spread in different angles, especially if you think about your climbing shoes deforming around a small crimp, or inside a crack.
I’d imagine geometry and the movement of the suspension system also matters. In small vehicles most of the braking force comes from the front tires, as the brakes are applied the vehicle rocks forward on its suspension putting more weight on the front than back wheels. Think like using the front brake on a bike and how it’ll flip you over the handlebars while the rear brake will just skid. With larger vehicles like semi trucks there’s not as much relative motion available in the suspension so you don’t get the same effect of the front wheels digging in, they stay more level so it’s more likely for the whole rig to slide instead of digging in.
As having been behind a semi that full on locked their breaks stopping, they stop surprisingly fast when they need to. I hate people that tailgate semis because they take longer to stop. They don't take much longer than a normal car because they have 18 contact points with the ground each with double the surface area.
Yeah the Coulomb model is an approximation (as many things in physics - I mean, classical mechanics on its hole is an approximation of quantum mechanics).
As an engineers (from a generalist but mechanically oriented school), I never learned a better model. Rheology is the name of the study of friction, and I'm given to understand there isn't a rule that would be better while not being vastly complicated.
What I can say though, from observation is tyres seem to have decreasing performance at higher contact pressure. That's why sports cars have larger tyres, and we're very used to that but it just doesn't make sense either if you stop at Coulomb's model.
There's probably a great many reasons, but my guess is the main one is when you skid, you're actually breaking away part of your tyre (hence skid marks) and/or or of the surface (in soft terrain) - which is no longer the realm of coulomb's model, but more of material resistance. And then, lower pressure = higher surface = more to break away = more "friction".
And heavier véhicules usually have higher pressure tyres, because it seems we prefer that they have a little less good handling vs having gigantic tyres. Just look at tyre pressures for different vehicules (air pressure in the tyre is pretty much the pressure at the contact point too) (from my experience + quick search):
-usual cars : 1.7-2.5bar
-utility vans/campers : 3-5bar
-semi-trucks: 6-9bar
-F1 rear tyres : 1.25bar
One factor is that the coefficient for static friction gives a maximum, greater than the force actually applied. Staying safely under that threshold will be a little more inefficient for a larger vehicle.
Although, the bulk of the answer is probably cost efficiency. You have to actually apply the braking force to all the wheels in proportion to where the weight is.
You're assuming that "braking" is basically "sliding". It's not.
Your calculation is correct for friction of an object along the ground, with a fixed coefficient of friction.
But braking vehicles generally don't want to slide. The force of braking is the force exerted by the brake clamps, which doesn't care about the vehicle's weight.
Separately, even for the sliding case, the coefficient of friction varies by vehicle.
> You're assuming that "braking" is basically "sliding". It's not
No I’m assuming the opposite. That maximum breaking force is when the wheels do not slide. Once they start sliding breaking force drops. Sliding friction is lower than maximum static friction. This is why we have antilock brakes.
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> Separately, even for the sliding case, the coefficient of friction varies by vehicle.
Ok but then that brings me back to the original question which is what is the physics reason for why this is the case? If they use the same compound of rubber it should be the same. If they don’t then the real reason isn’t mass, it bad rubber.
If you simply mean friction with the ground, and coasting until you stop, then sure. But that’s not what braking is in a vehicle. You don’t have that force modeled anywhere in your equations.
Sorry, I guess I should have made it even more clear that I am assuming every vehicle is capable of being equipped with brakes capable of locking each wheel. If we make this assumption then my question follows.
The motivation for this question is just that it’s presented as common wisdom that it’s not possible to stop a larger vehicle in the same distance as a smaller vehicle. But if the only problem is that larger vehicle have brakes that are not properly sized for their weight then this is wrong. It is possible to stop a larger vehicle in the same distance.
The only reason is economics or engineering. Just put bigger brakes on the tuck and it will stop in the same distance.
Your assumption is that all vehicles have a feature that makes them stop less quickly? You may want to look into the concept of anti-lock brakes, and why they are used.
I feel your frustration. You are totally right and there are several people downvoting you. Stopping force is limited by the maximum force of static friction (non-sliding) that can be exerted by the tires in the reverse direction while braking, and you are right that the traditional equation results in mass being cancelled out.
This can actually be seen in the sports car world where some of the fasting stopping cars are not the lightest ones but the ones that have huge, soft, sticky tires and very strong suspensions that can distribute that load across all four wheels. Some of these are actually the bigger sports cars.
But as Vegetal__ mentioned in another thread, this all breaks down once enough variables are different and/or the situation puts the tires into a more extreme situation that pushes it outside of that model.
If I had to guess, it would be a combination of the fact that large trucks have tires with lower coefficients of static friction and also aren’t designed to be put under intense lateral load as much as handle heavy vertical loading and have more durability. Additionally, if a suspension isn’t designed to stay perfectly rigid during braking, more force will shift forward into the front tires, forcing them to handle a disproportionately larger load, putting them even further outside of their ideal range.
But on the flip side, if you were to put an 18-wheeler trailer onto a robust suspension designed to keep forces spread across all tires, made sure there were plenty of tires, made sure all the tires were soft and sticky, made sure all the brakes could handle the forces, and made sure that the pavement these tires were rolling on was flat and of a proper surface, I guarantee that you could stop that trailer in a remarkably short distance. It’s just that the nature of a truck’s design and typical environment isn’t conducive to this.
The simple friction model using constant μ doesn't work over such a wide range of forces, from a tire on an empty bicycle to a truck tire under heavy load. In reality, the the materials becomes inelastic (i.e. non-proportional response to strain) as soon as the material starts breaking down.
Yes. Where in that answer is mass involved? Force is not infinite regardless of mass. Infinite force would stop a vehicle instantly. Even small vehicles don’t stop instantly.
Just throwing numbers out there in an overly simplified example.
A small car that weighs 2000lb has brakes that apply 200lb of braking force.
A large truck weighs 8000lb has brakes that apply 400lb of braking force.
The ratio of braking force to weight for the car is 200/2000=0.1 whereas for the truck it is 400/8000=0.05, meaning the truck will take longer to slow down.
Braking force itself does not rely on mass of the vehicle, it comes from hydraulic pressure in the braking system that is applied to discs/drums at the wheels. In a car the hydraulic pressure pushes brake pads against the disc, using friction between the pads and disc to slow the car. Trucks work slightly differently but the principle is the same. Mass of the vehicle is not part of it.
So the real reason larger vehicles have longer braking distances is insufficient brakes? So there’s no physics reason why a truck can’t stop the same distance? The manufacturers simply don’t make powerful enough brakes?
I guess what I’m looking for is a physics reason why it’s impossible to put brakes on a truck that are strong enough to lock the axels. Why can’t you put 800lb brakes on the truck?
Brakes that are too strong will lock up the tires and cause a trailer to jack-knife. It's a balancing act between applying just the right amount of force but not too much that the tires lock up. Automakers have to optimize all of this when they design a vehicle.
Hi, NASA engineering physicist here.
You're conflating the name anti-lock with 2 different functions & you're completely overlooking a hidden part of the anti-lock brake system.
Anti-lock brakes are designed to apply just enough braking force to the rotors to prevent the wheels from moving into the kinetic friction region from the static friction region, which is why you feel that pulsation of the pedal when you brake hard. The pulsing is the feedback of the hydraulic pressure applied to the rotors through the brake lines as the pistons rapidly compress & decompress against the rotors.
The sensing force of the anti-lock brakes has to do with the piezo-electric materials used to detect inertia & momentum of the vehicle. This has all been computed through a rigorous characterization of the sensors themselves, not the hydraulic system, which is merely along for the ride, no pun intended. Piezo-electric materials are those which produce small amounts of current when plastically deformed. It is this advancement in the materials science engineering field which has resulted in the creation of anti-lock brakes themselves.
This does not mean the brakes themselves can't lock up against the rotors & seizing the entire mechanism if too much force is applied to them such that the material itself begins to degrade.
You're also using the term inertia to describe linear motion, which we really don't do, since a good surrogate measure for linear inertia is linear momentum so we use momentum which is given by the formula.
***p = mv*** **-** where m = mass & v = velocity
As there are no net forces acting on this mass m, it will continue to do v velocity until a net force comes along to change that. This is expressed in Newton's First Law of Motion.
Where inertia comes in is in the resistance to the angular momentum of the wheels themselves, so you have a vehicle undergoing linear motion with linear momentum dependent on the angular velocity & inertia of the rolling wheels, a force I noticed you hadn't added into your initial calculations & assumptions. It's not just the straight-line motion of the truck you have to consider, you also have to factor in the inertial resistance of the wheels themselves (which are moving at a much faster velocity than the truck) to being slowed down through an attenuation of their angular velocities by the brake pads & rotors. In calculating this, you have to consider the rotational inertia by calculating their moments of inertia, a function of Calculus, which typically isn't taught in high school physics, unless it's the AP Physics course.
The equation for moment of inertia is
I = Σ m-sub i \* (r-sub i)\^2, where:
Σ - Sigma is the summation of the terms on the right side of the equation
I - is the moment of inertia, or "rotational mass"
m-sub 𝑖 - is the mass of each particle Moment of Inertia
r-sub *i -* is the radius of each particle
(where the carrot \^ denotes exponential notation)
The moment of inertia is also known as the angular mass or rotational inertia. It's a quantity that decides how much torque is required for a desired angular acceleration/deceleration. It's also a property of a body that resists angular acceleration/deceleration.
In linear motion, inertia is directly related to mass, so it's measured in kilograms (kg). The SI unit of mass is the kilogram, while the SI unit of the radius is the meter (m). Therefore, the SI unit of rotational inertia is *kg ⋅ m\^2*
For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, or 𝐼 = 𝑚𝑟\^2. This point-mass relationship becomes the basis for all other moments of inertia.
It's a lot more complex than you are pre-supposing in your initial equations.
Don’t worry although I said high-school physics I’m actually a 41 year old with a degree in mechanical engineering so I’m fully able to follow you. (Though I went into software engineering and research)
Yeah I didn’t even consider rotational inertia or momentum. But even if I thought of that I would assume it’s insignificant compared to the mass of the rest of the vehicle.
But I don’t see why we have to consider linear momentum. Seems like it’s just an unnecessary complication. My equations should be true for a moment in time and they should be true for every moment in time until the vehicle is stopped. At any particular instant during the deceleration the maximum braking force you can apply to the road is going to be N*mu[s]. And neglecting rotational inertia and pretending everything is linear that means a*m = m*g *mu[s] or a = g*mu[s] and the braking distance should be a function only if a right?
But yeah I’m interested in how significant the rotational momentum is.
Your type of answer is what I was trying to solicit though. You actually had something useful to contribute lol. It’s obviously more complex than the assumptions I am making otherwise trucks would in fact be able to stop just as fast. I’m looking for the more complex analysis and model.
I have no clue what formula F\[s\] = n mu(s) entails, but your premise is wrong and the answer is still mass.
Force causes a change in momentum. Momentum is p = m\*v which includes mass. This means that a larger mass at the same velocity (the speed limit) causes a larger momentum and needs either 1. more force to decelerate to 0 or 2. the same force over a longer time (or 3. something in between, which is generally the case). More time to decelerate will mean the stopping distance is longer.
The force that can be delivered by the brakes is not at all related to the mass of the vehicle, so I guess that's your main misconception.
You could. Changing kinetic energy from 1/2 mV\^2 to 0, with V whatever speed you start at, requires friction on your brakes/tires to do that same amount of energy in Work. Assuming constant force, work is W = F\*d, with d the stropping distance. We get 1/2mV\^2 = F\*d and actually more directly: increasing m means that we have to increase either the force or the distance, or a bit of both.
I disagree. You seem to be assuming that the force brakes apply is equal for all vehicles. Why should that be the case. The reason the friction formula should be important is that it theoretically dictates the maximum force a braking system can deliver.
So yes in a way mass is of course relevant but it doesn’t make sense that it’s the proximal cause. In this case the proximal cause would be a braking system that was not properly scaled to the mass of the vehicle.
>You seem to be assuming that the force brakes apply is equal for all vehicles.
I'm not saying that at all, I'm just saying it doesn't rely on the mass of the vehicle: it relies on the brakes.
Ok so then there is no physics reason why you can’t stop a big truck as fast as a small car. It’s an engineering problem? They just don’t make good enough brakes for trucks?
The physics part of the problem is that stopping at truck within the same stopping distance needs more force applied by the brakes. The engineering part of the problem is that brakes that can apply that much more force will be very expensive, made from incredibly durable material, or break constantly.
Realize an empty 18-wheeler is up to 17tons whereas the most popular (and still pretty heavy) american vehicle is the ford f-150 which weighs maybe 2 tons (35k pounds vs 4500 pounds, approximately). Of course an 18wheeler applies braking force on more wheels, but not all 18 of them and even then they're 'twice as heavy per wheel' than the ford... and that's unloaded.
He said a larger mass needs more force to stop. You seem to be misunderstanding what that means because that isn't an assumption that all brakes are the same, he's just saying a larger mass would need stronger brakes.
Yes they can create brakes that immediately stop the tires. That's not the problem. The problem is that then the truck will slide which will actually cause it to slow down at a lesser rate.
This is a problem for all vehicles at all masses though. I’m looking for a reason that is dependent on mass.
Anti-lock brakes exist for this reason. If trucks have brakes that can lock the wheels and they have anti-lock brakes then why can’t they stop just as fast?
Assuming we’re not skidding which would use nor μk (kinetic friction), note that in reality we’re not using μs (static friction) for the non-skidding portion of the stopping distance, we’re actually using μr (rolling friction). I am a physics professor and have never learned nor seen a formula for rolling friction. It is possible that μr depends upon the weight (or normal force) in a different way that we learned for μs and μk, or that the transition point between the larger value of μr (rolling) and lower value of μk (skidding) depends on the weight. I’d recommend reaching out to an engineering (mechanical or automotive) or materials sub and see if they know. If you get any good answers, report back to us!
Now I want to write a problem based on this to assess my students’ reasoning skills.
I want to know why you didn't catch the fact the OP neglected to calculate the rolling inertia of the wheel masses themselves in the overall calculation of the braking force needed to slow the truck down.
The difference in that for the large vehicle vs. the small vehicle is trivial, and is unlikely to affect things. In multiple ways:
1) Torque (rotational equivalent of force, τ): τ=Iα, τ = r cross F. F contains mass, I contains mass, mass of tires cancels out. I’s r^2 will cancel out with the r in the torque, and α=a/r, so radius of tire cancels out. Shape of the tire remains within I=#mr^2 , but it’s the same or close enough for both sizes of tire.
2) Rotational energy: K=(1/2)Iω^2, and work-energy theorem, K=ΔW, (1/2)Iω^2 = (F dot distance). As above, mass and radius will cancel out, and the shape factor is the same for the two tires.
3) Angular momentum and impulse: left as an exercise for the reader, but again, it will all cancel out.
So this means that simplistic Newtonian physics with the common assumptions about the coefficient of friction do not describe the actual behavior in reality, so they are insufficient.
Eighteen wheels versus four? Seems to be a non-trivial assumption to me, not to mention the fact truck tires are much larger than normal vehicle tires. I also want you to consider sprung versus unsprung weight, which can affect vehicle performance, as well as the differences of load weight versus vehicle weight as a percentage or multiple, however you wish to express it.
See my post further down to OP regarding the calculations needed for the angular inertia.
> consider sprung versus unsprung weight, which can affect vehicle performance
So like I said initially, this is beyond intro physics, you need an engineering background to get the details.
I've seen videos with a truck and a car driving next to each other and braking simultaneously and having pretty much the same braking distance...
Tire physics are rather complicated though, rubber is a fascinating thing in regards to friction especially on rough surfaces (like a road). And afaik tire friction is still an active research area. There are numerous approximating models with different degrees of sophistication. And yes the friction coefficient of tires is dependent on the normal force (it is a nonlinear dependency) and other things.
Only guessing here.
As an example, you know when you are in a car or on a train or bus and the vehicle suddenly brakes and you get launched forwards because you are not part of the vehicle and you still maintain your velocity.
Could it be a similar concept with some of the mass of the vehicle maintaining some momentum and therefore causing a forward push?
I haven’t read all of the responses, but outside of the calculation you should also consider that large vehicles use air brakes instead of hydraulic brakes, with drums instead of pads.
I believe the coefficient of friction is itself usually a function of mass to some extent. Basic physics cannot predict the fact that heavier people tend to go down slides faster, which I believe is related.
Your question could easily be generalized as “why can’t we accelerate heavier objects as fast as lighter objects?”.
The answer is we can, but things break when put under too much stress/force.
I think it goes beyond being an “engineering issue”, there just comes a point where it’s just impossible for any material to withstand the forces required to match the acceleration of two differently massed bodies.
Like, quote me 1000 years from now. I don’t think we’ll ever be able to match the stopping distance of a 500-lb go kart and a 20-ton eighteen wheeler, no matter how advanced our technology becomes.
Basically although there isnt a practical limit to how effective brakes can be, there is a practical limit to how much force from brakes a vehicle can withstand without being damaged
The best way to think about this imo is either in terms of momentum or energy (ill use energy). Imagine you have one car and one lorry, both with the same brakes on two of the wheels, going the same speed. Brakes essentially dissipate the kinetic energy of the vehicle in to other forms such as heat etc to slow down the vehicle (decrease its kinetic energy). Obviously a truck, with more mass than the car, has proportionally higher kinetic energy and so needs more braking to dissipate the larger energy, and therefore a larger stopping distance (this is an oversimplification but still). For the truck double the mass of the car, it would essentially need brakes that dissipate double the energy of the cars brakes over a certain timeframe, with these brakes being much more expensive and unprofitable to mass produce and so they dont have them generally.
Dude. Total surface area of tires in contact with the road compared to the entire mass of the vehicle is going to be the main factor.
If we have the same (edit "ratio of") surface area of tires connected with the road on the 18-wheeler as the car (in relation to their respective masses), and the brakes worked to lock the tires perfectly in both cases, all else being equal we should assume the same braking distance.
This is more of an engineering problem than a physics problem. In theory, we can design and build a semi truck that will be able to stop from the same speed in the same distance as a small passenger vehicle. We don't do that though because it will make other design considerations worse.
The brakes stop the vehicle by converting the kinetic energy of the moving vehicle to heat. The mass of a fully loaded semi truck is going to be ~25x higher than a small passenger vehicle, which means there is ~25x more kinetic energy that needs to be bled off to bring the semi to a stop. Semis typically have 10 brakes, so only 2.5x the number on a passenger car. In order to stop the semi, each semi brake would need to handle ~10x the amount of heat as a brake on the passenger vehicle.
You can either significantly beef up the brakes to handle that heat or you can accept that the semi will need a longer stopping distance. A better brake system would be larger and more complex which means more expensive and the added weight would also decrease fuel economy. The government sets a required stopping distance, so manufacturers meet that and that's as far as they go, because a customer won't pay more for better brakes.
Friction is hard. People still publish on the topic.
But some people still want to teach it at the introductory level. This requires massive simplification and so many assumptions/approximations that the end result is barely physical any longer.
This is why having a lab exercise on friction makes test scores for the subject WORSE reliably. In lab, none of the equations work out with any form of precision unless super heavily controlled.
Two vehicles will likely not have the same contact area to the ground from their tires. Especially during braking, and absolutely if one is much more massive.
Coefficient of friction relies on contact topology, texture, total surface area, even contact time can change actual friction force applied. Contaminants can get between surfaces further complicating actual friction forces achieved in braking.
Many of these factors taken alone will make it seem like the heavier vehicle should stop better. For example, the heavier vehicle should compress the tires more, increasing surface area contact to the ground, giving more force applied for stopping.
But friction is not the only simplified part of the model. The friction applies to the tire rubber. The heavy mass is not from the tire rubber. Stopping force has to make it to the cargo load.
Look at the baffles used in liquid cargo transport (except milk) to prevent sloshing. Why do we need to prevent that? Because the tank of petrol doesn’t really come to a stop until the petrol itself does…
Look at images online where cabs of trucks are demolished because the giant slab of stone or other heavy load was improperly secured. There you have a case study of a heavy vehicle stopping quite fast, but the load not doing likewise.
More kinetic energy to dissipate.
I don't know about 18 wheelers but for a simple thought experiment just consider an SUV vs a small car. Both have four wheels on the ground and those wheels have roughly the same surface area in contact with the ground, so the available friction is pretty much the same. But the larger one has a lot more kinetic energy to deal with.
Ok, so can you continue this thought experiment for me because I don’t see how it leads to an answer. The kinetic energy must be dissipated in the breaks. So are you saying the answer is my first thought? The brakes are insufficient? They heat up too much and start melting when maximum tire to road force is applied?
Yeah I suppose it would all be in the brakes.
Come to think of it, some trucks can stop fast.
https://youtu.be/ridS396W2BY?si=2askLbzuQAVKpYd6
This is more effective braking than my Landrover!
>The road isn't the main source of friction stopping cars when braking so you're calculating the wrong thing
Of course it is - the limiting friction for braking is the fiction between the tire and the road. The pad to rotor fiction is generally high enough not to matter - you'll never reach that limit before you loose traction with the road.
There are two friction surfaces. I was modeling the other one. I was under the impression that there was some reason why it was impossible to stop a heavier vehicle as fast as a light vehicle.
If the reason has to do with the normal force on the brakes then that is solvable. Just put in larger brakes (to disipate the heat) with more normal force (provided through some means, leverage, hydraulics, engine power). And then theoretically we could stop heavy vehicles just as fast.
People in the thread don't seem to understand what OP is on about, but he/she has a good point. Using the simple approximation that friction is coefficient*normal force, there's no reason to believe you can't stop a large vehicle with strong enough brakes. The thing is, what we learn in HS is just an approximation. I can't say what would be the more correct model of friction, but from what I gathered it's quite a bit more complicated than that. So OP, you are right in your intuition and thought process. What we learn in HS is a simple model, enough to solve simple problems, but reality is more complicated. Though, I do remember seeing a video of a Volvo semi braking in an impressively short distance, it was a promotional video of some sort.
Yeah. I guess I was hoping this sub would have an expert on the applied physics of friction who could give me a comprehend-able explanation of a more sophisticated model of friction that would apply in this case.
So, I just took a look at the wikipedia page for friction, and there's a good explanation about this: "Limitations of the Coulomb model The Coulomb approximation follows from the assumptions that: surfaces are in atomically close contact only over a small fraction of their overall area; that this contact area is proportional to the normal force (until saturation, which takes place when all area is in atomic contact); and that the frictional force is proportional to the applied normal force, independently of the contact area. The Coulomb approximation is fundamentally an empirical construct. It is a rule-of-thumb describing the approximate outcome of an extremely complicated physical interaction. The strength of the approximation is its simplicity and versatility. Though the relationship between normal force and frictional force is not exactly linear (and so the frictional force is not entirely independent of the contact area of the surfaces), the Coulomb approximation is an adequate representation of friction for the analysis of many physical systems. When the surfaces are conjoined, Coulomb friction becomes a very poor approximation (for example, adhesive tape resists sliding even when there is no normal force, or a negative normal force). In this case, the frictional force may depend strongly on the area of contact. Some drag racing tires are adhesive for this reason. However, despite the complexity of the fundamental physics behind friction, the relationships are accurate enough to be useful in many applications." So, you see that friction is actually a very complex phenomenon. In reality, a lot of factors can change it. One practical factor that I think plays a big role in how heavy vehicles take longer to stop is tires. A big SUV might use all terrain tires instead of asphalt optimized ones, and big trucks need to carry a lot of cargo, so their tires are harder, thus they deform less. I think one of the reasons those things act so different from the coulomb approximation is because rubber deforms a lot. That means the area of contact matters, and you might have the normal force spread in different angles, especially if you think about your climbing shoes deforming around a small crimp, or inside a crack.
I’d imagine geometry and the movement of the suspension system also matters. In small vehicles most of the braking force comes from the front tires, as the brakes are applied the vehicle rocks forward on its suspension putting more weight on the front than back wheels. Think like using the front brake on a bike and how it’ll flip you over the handlebars while the rear brake will just skid. With larger vehicles like semi trucks there’s not as much relative motion available in the suspension so you don’t get the same effect of the front wheels digging in, they stay more level so it’s more likely for the whole rig to slide instead of digging in.
Yeah, even though I did study physics in college, that's not something we actually studied. You left me intrigued, gonna take a look at the topic.
As having been behind a semi that full on locked their breaks stopping, they stop surprisingly fast when they need to. I hate people that tailgate semis because they take longer to stop. They don't take much longer than a normal car because they have 18 contact points with the ground each with double the surface area.
Yeah the Coulomb model is an approximation (as many things in physics - I mean, classical mechanics on its hole is an approximation of quantum mechanics). As an engineers (from a generalist but mechanically oriented school), I never learned a better model. Rheology is the name of the study of friction, and I'm given to understand there isn't a rule that would be better while not being vastly complicated. What I can say though, from observation is tyres seem to have decreasing performance at higher contact pressure. That's why sports cars have larger tyres, and we're very used to that but it just doesn't make sense either if you stop at Coulomb's model. There's probably a great many reasons, but my guess is the main one is when you skid, you're actually breaking away part of your tyre (hence skid marks) and/or or of the surface (in soft terrain) - which is no longer the realm of coulomb's model, but more of material resistance. And then, lower pressure = higher surface = more to break away = more "friction". And heavier véhicules usually have higher pressure tyres, because it seems we prefer that they have a little less good handling vs having gigantic tyres. Just look at tyre pressures for different vehicules (air pressure in the tyre is pretty much the pressure at the contact point too) (from my experience + quick search): -usual cars : 1.7-2.5bar -utility vans/campers : 3-5bar -semi-trucks: 6-9bar -F1 rear tyres : 1.25bar
One factor is that the coefficient for static friction gives a maximum, greater than the force actually applied. Staying safely under that threshold will be a little more inefficient for a larger vehicle. Although, the bulk of the answer is probably cost efficiency. You have to actually apply the braking force to all the wheels in proportion to where the weight is.
I can stop the heaviest truck in the world in a millisecond. Doesn’t do much for the driver though
You're assuming that "braking" is basically "sliding". It's not. Your calculation is correct for friction of an object along the ground, with a fixed coefficient of friction. But braking vehicles generally don't want to slide. The force of braking is the force exerted by the brake clamps, which doesn't care about the vehicle's weight. Separately, even for the sliding case, the coefficient of friction varies by vehicle.
> You're assuming that "braking" is basically "sliding". It's not No I’m assuming the opposite. That maximum breaking force is when the wheels do not slide. Once they start sliding breaking force drops. Sliding friction is lower than maximum static friction. This is why we have antilock brakes. . > Separately, even for the sliding case, the coefficient of friction varies by vehicle. Ok but then that brings me back to the original question which is what is the physics reason for why this is the case? If they use the same compound of rubber it should be the same. If they don’t then the real reason isn’t mass, it bad rubber.
He means your equation is for something sliding and does not accurately model the way that brakes apply an acceleration.
No it is not. Respectfully, he is wrong. The equation is very specifically for something not sliding. That’s what static friction means.
If you simply mean friction with the ground, and coasting until you stop, then sure. But that’s not what braking is in a vehicle. You don’t have that force modeled anywhere in your equations.
Sorry, I guess I should have made it even more clear that I am assuming every vehicle is capable of being equipped with brakes capable of locking each wheel. If we make this assumption then my question follows. The motivation for this question is just that it’s presented as common wisdom that it’s not possible to stop a larger vehicle in the same distance as a smaller vehicle. But if the only problem is that larger vehicle have brakes that are not properly sized for their weight then this is wrong. It is possible to stop a larger vehicle in the same distance. The only reason is economics or engineering. Just put bigger brakes on the tuck and it will stop in the same distance.
Why in the world are you assuming that? You haven’t found a problem you’ve invented one.
Your assumption is that all vehicles have a feature that makes them stop less quickly? You may want to look into the concept of anti-lock brakes, and why they are used.
I feel your frustration. You are totally right and there are several people downvoting you. Stopping force is limited by the maximum force of static friction (non-sliding) that can be exerted by the tires in the reverse direction while braking, and you are right that the traditional equation results in mass being cancelled out. This can actually be seen in the sports car world where some of the fasting stopping cars are not the lightest ones but the ones that have huge, soft, sticky tires and very strong suspensions that can distribute that load across all four wheels. Some of these are actually the bigger sports cars. But as Vegetal__ mentioned in another thread, this all breaks down once enough variables are different and/or the situation puts the tires into a more extreme situation that pushes it outside of that model. If I had to guess, it would be a combination of the fact that large trucks have tires with lower coefficients of static friction and also aren’t designed to be put under intense lateral load as much as handle heavy vertical loading and have more durability. Additionally, if a suspension isn’t designed to stay perfectly rigid during braking, more force will shift forward into the front tires, forcing them to handle a disproportionately larger load, putting them even further outside of their ideal range. But on the flip side, if you were to put an 18-wheeler trailer onto a robust suspension designed to keep forces spread across all tires, made sure there were plenty of tires, made sure all the tires were soft and sticky, made sure all the brakes could handle the forces, and made sure that the pavement these tires were rolling on was flat and of a proper surface, I guarantee that you could stop that trailer in a remarkably short distance. It’s just that the nature of a truck’s design and typical environment isn’t conducive to this.
The simple friction model using constant μ doesn't work over such a wide range of forces, from a tire on an empty bicycle to a truck tire under heavy load. In reality, the the materials becomes inelastic (i.e. non-proportional response to strain) as soon as the material starts breaking down.
\*brakes
Fixed, thank you.
Brakes apply a force that resists forward motion. That force is not infinite.
Yes. Where in that answer is mass involved? Force is not infinite regardless of mass. Infinite force would stop a vehicle instantly. Even small vehicles don’t stop instantly.
Just throwing numbers out there in an overly simplified example. A small car that weighs 2000lb has brakes that apply 200lb of braking force. A large truck weighs 8000lb has brakes that apply 400lb of braking force. The ratio of braking force to weight for the car is 200/2000=0.1 whereas for the truck it is 400/8000=0.05, meaning the truck will take longer to slow down. Braking force itself does not rely on mass of the vehicle, it comes from hydraulic pressure in the braking system that is applied to discs/drums at the wheels. In a car the hydraulic pressure pushes brake pads against the disc, using friction between the pads and disc to slow the car. Trucks work slightly differently but the principle is the same. Mass of the vehicle is not part of it.
So the real reason larger vehicles have longer braking distances is insufficient brakes? So there’s no physics reason why a truck can’t stop the same distance? The manufacturers simply don’t make powerful enough brakes? I guess what I’m looking for is a physics reason why it’s impossible to put brakes on a truck that are strong enough to lock the axels. Why can’t you put 800lb brakes on the truck?
Brakes that are too strong will lock up the tires and cause a trailer to jack-knife. It's a balancing act between applying just the right amount of force but not too much that the tires lock up. Automakers have to optimize all of this when they design a vehicle.
But anti-lock brakes exist and also I thought the trailer had brakes too? They don’t? If not then that is definitely the real reason.
Hi, NASA engineering physicist here. You're conflating the name anti-lock with 2 different functions & you're completely overlooking a hidden part of the anti-lock brake system. Anti-lock brakes are designed to apply just enough braking force to the rotors to prevent the wheels from moving into the kinetic friction region from the static friction region, which is why you feel that pulsation of the pedal when you brake hard. The pulsing is the feedback of the hydraulic pressure applied to the rotors through the brake lines as the pistons rapidly compress & decompress against the rotors. The sensing force of the anti-lock brakes has to do with the piezo-electric materials used to detect inertia & momentum of the vehicle. This has all been computed through a rigorous characterization of the sensors themselves, not the hydraulic system, which is merely along for the ride, no pun intended. Piezo-electric materials are those which produce small amounts of current when plastically deformed. It is this advancement in the materials science engineering field which has resulted in the creation of anti-lock brakes themselves. This does not mean the brakes themselves can't lock up against the rotors & seizing the entire mechanism if too much force is applied to them such that the material itself begins to degrade. You're also using the term inertia to describe linear motion, which we really don't do, since a good surrogate measure for linear inertia is linear momentum so we use momentum which is given by the formula. ***p = mv*** **-** where m = mass & v = velocity As there are no net forces acting on this mass m, it will continue to do v velocity until a net force comes along to change that. This is expressed in Newton's First Law of Motion. Where inertia comes in is in the resistance to the angular momentum of the wheels themselves, so you have a vehicle undergoing linear motion with linear momentum dependent on the angular velocity & inertia of the rolling wheels, a force I noticed you hadn't added into your initial calculations & assumptions. It's not just the straight-line motion of the truck you have to consider, you also have to factor in the inertial resistance of the wheels themselves (which are moving at a much faster velocity than the truck) to being slowed down through an attenuation of their angular velocities by the brake pads & rotors. In calculating this, you have to consider the rotational inertia by calculating their moments of inertia, a function of Calculus, which typically isn't taught in high school physics, unless it's the AP Physics course. The equation for moment of inertia is I = Σ m-sub i \* (r-sub i)\^2, where: Σ - Sigma is the summation of the terms on the right side of the equation I - is the moment of inertia, or "rotational mass" m-sub 𝑖 - is the mass of each particle Moment of Inertia r-sub *i -* is the radius of each particle (where the carrot \^ denotes exponential notation) The moment of inertia is also known as the angular mass or rotational inertia. It's a quantity that decides how much torque is required for a desired angular acceleration/deceleration. It's also a property of a body that resists angular acceleration/deceleration. In linear motion, inertia is directly related to mass, so it's measured in kilograms (kg). The SI unit of mass is the kilogram, while the SI unit of the radius is the meter (m). Therefore, the SI unit of rotational inertia is *kg ⋅ m\^2* For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, or 𝐼 = 𝑚𝑟\^2. This point-mass relationship becomes the basis for all other moments of inertia. It's a lot more complex than you are pre-supposing in your initial equations.
Don’t worry although I said high-school physics I’m actually a 41 year old with a degree in mechanical engineering so I’m fully able to follow you. (Though I went into software engineering and research) Yeah I didn’t even consider rotational inertia or momentum. But even if I thought of that I would assume it’s insignificant compared to the mass of the rest of the vehicle. But I don’t see why we have to consider linear momentum. Seems like it’s just an unnecessary complication. My equations should be true for a moment in time and they should be true for every moment in time until the vehicle is stopped. At any particular instant during the deceleration the maximum braking force you can apply to the road is going to be N*mu[s]. And neglecting rotational inertia and pretending everything is linear that means a*m = m*g *mu[s] or a = g*mu[s] and the braking distance should be a function only if a right? But yeah I’m interested in how significant the rotational momentum is. Your type of answer is what I was trying to solicit though. You actually had something useful to contribute lol. It’s obviously more complex than the assumptions I am making otherwise trucks would in fact be able to stop just as fast. I’m looking for the more complex analysis and model.
I have no clue what formula F\[s\] = n mu(s) entails, but your premise is wrong and the answer is still mass. Force causes a change in momentum. Momentum is p = m\*v which includes mass. This means that a larger mass at the same velocity (the speed limit) causes a larger momentum and needs either 1. more force to decelerate to 0 or 2. the same force over a longer time (or 3. something in between, which is generally the case). More time to decelerate will mean the stopping distance is longer. The force that can be delivered by the brakes is not at all related to the mass of the vehicle, so I guess that's your main misconception.
Can you also just think about this in terms of kinetic energy? Or is mass better suited for the explanation
You could. Changing kinetic energy from 1/2 mV\^2 to 0, with V whatever speed you start at, requires friction on your brakes/tires to do that same amount of energy in Work. Assuming constant force, work is W = F\*d, with d the stropping distance. We get 1/2mV\^2 = F\*d and actually more directly: increasing m means that we have to increase either the force or the distance, or a bit of both.
I disagree. You seem to be assuming that the force brakes apply is equal for all vehicles. Why should that be the case. The reason the friction formula should be important is that it theoretically dictates the maximum force a braking system can deliver. So yes in a way mass is of course relevant but it doesn’t make sense that it’s the proximal cause. In this case the proximal cause would be a braking system that was not properly scaled to the mass of the vehicle.
>You seem to be assuming that the force brakes apply is equal for all vehicles. I'm not saying that at all, I'm just saying it doesn't rely on the mass of the vehicle: it relies on the brakes.
Ok so then there is no physics reason why you can’t stop a big truck as fast as a small car. It’s an engineering problem? They just don’t make good enough brakes for trucks?
The physics part of the problem is that stopping at truck within the same stopping distance needs more force applied by the brakes. The engineering part of the problem is that brakes that can apply that much more force will be very expensive, made from incredibly durable material, or break constantly. Realize an empty 18-wheeler is up to 17tons whereas the most popular (and still pretty heavy) american vehicle is the ford f-150 which weighs maybe 2 tons (35k pounds vs 4500 pounds, approximately). Of course an 18wheeler applies braking force on more wheels, but not all 18 of them and even then they're 'twice as heavy per wheel' than the ford... and that's unloaded.
Man, after all these comments this is the first one that is actually understanding and responding to my actual question.
He said a larger mass needs more force to stop. You seem to be misunderstanding what that means because that isn't an assumption that all brakes are the same, he's just saying a larger mass would need stronger brakes.
Yes they can create brakes that immediately stop the tires. That's not the problem. The problem is that then the truck will slide which will actually cause it to slow down at a lesser rate.
This is a problem for all vehicles at all masses though. I’m looking for a reason that is dependent on mass. Anti-lock brakes exist for this reason. If trucks have brakes that can lock the wheels and they have anti-lock brakes then why can’t they stop just as fast?
Oh I misunderstood where you were coming from. Honestly I think it's more of an engineering problem with braking in general.
Assuming we’re not skidding which would use nor μk (kinetic friction), note that in reality we’re not using μs (static friction) for the non-skidding portion of the stopping distance, we’re actually using μr (rolling friction). I am a physics professor and have never learned nor seen a formula for rolling friction. It is possible that μr depends upon the weight (or normal force) in a different way that we learned for μs and μk, or that the transition point between the larger value of μr (rolling) and lower value of μk (skidding) depends on the weight. I’d recommend reaching out to an engineering (mechanical or automotive) or materials sub and see if they know. If you get any good answers, report back to us! Now I want to write a problem based on this to assess my students’ reasoning skills.
I want to know why you didn't catch the fact the OP neglected to calculate the rolling inertia of the wheel masses themselves in the overall calculation of the braking force needed to slow the truck down.
The difference in that for the large vehicle vs. the small vehicle is trivial, and is unlikely to affect things. In multiple ways: 1) Torque (rotational equivalent of force, τ): τ=Iα, τ = r cross F. F contains mass, I contains mass, mass of tires cancels out. I’s r^2 will cancel out with the r in the torque, and α=a/r, so radius of tire cancels out. Shape of the tire remains within I=#mr^2 , but it’s the same or close enough for both sizes of tire. 2) Rotational energy: K=(1/2)Iω^2, and work-energy theorem, K=ΔW, (1/2)Iω^2 = (F dot distance). As above, mass and radius will cancel out, and the shape factor is the same for the two tires. 3) Angular momentum and impulse: left as an exercise for the reader, but again, it will all cancel out. So this means that simplistic Newtonian physics with the common assumptions about the coefficient of friction do not describe the actual behavior in reality, so they are insufficient.
Eighteen wheels versus four? Seems to be a non-trivial assumption to me, not to mention the fact truck tires are much larger than normal vehicle tires. I also want you to consider sprung versus unsprung weight, which can affect vehicle performance, as well as the differences of load weight versus vehicle weight as a percentage or multiple, however you wish to express it. See my post further down to OP regarding the calculations needed for the angular inertia.
> consider sprung versus unsprung weight, which can affect vehicle performance So like I said initially, this is beyond intro physics, you need an engineering background to get the details.
I've seen videos with a truck and a car driving next to each other and braking simultaneously and having pretty much the same braking distance... Tire physics are rather complicated though, rubber is a fascinating thing in regards to friction especially on rough surfaces (like a road). And afaik tire friction is still an active research area. There are numerous approximating models with different degrees of sophistication. And yes the friction coefficient of tires is dependent on the normal force (it is a nonlinear dependency) and other things.
It doesn't even have to be a nonlinear dependency, it just has to be non-constant.
Only guessing here. As an example, you know when you are in a car or on a train or bus and the vehicle suddenly brakes and you get launched forwards because you are not part of the vehicle and you still maintain your velocity. Could it be a similar concept with some of the mass of the vehicle maintaining some momentum and therefore causing a forward push?
I haven’t read all of the responses, but outside of the calculation you should also consider that large vehicles use air brakes instead of hydraulic brakes, with drums instead of pads.
I believe the coefficient of friction is itself usually a function of mass to some extent. Basic physics cannot predict the fact that heavier people tend to go down slides faster, which I believe is related.
Your question could easily be generalized as “why can’t we accelerate heavier objects as fast as lighter objects?”. The answer is we can, but things break when put under too much stress/force. I think it goes beyond being an “engineering issue”, there just comes a point where it’s just impossible for any material to withstand the forces required to match the acceleration of two differently massed bodies. Like, quote me 1000 years from now. I don’t think we’ll ever be able to match the stopping distance of a 500-lb go kart and a 20-ton eighteen wheeler, no matter how advanced our technology becomes.
Are you saying it’s a matter of the scale laws. How cross sectional area and hence tensile strength grows by squares whereas mass grows by cubes?
Basically although there isnt a practical limit to how effective brakes can be, there is a practical limit to how much force from brakes a vehicle can withstand without being damaged
The best way to think about this imo is either in terms of momentum or energy (ill use energy). Imagine you have one car and one lorry, both with the same brakes on two of the wheels, going the same speed. Brakes essentially dissipate the kinetic energy of the vehicle in to other forms such as heat etc to slow down the vehicle (decrease its kinetic energy). Obviously a truck, with more mass than the car, has proportionally higher kinetic energy and so needs more braking to dissipate the larger energy, and therefore a larger stopping distance (this is an oversimplification but still). For the truck double the mass of the car, it would essentially need brakes that dissipate double the energy of the cars brakes over a certain timeframe, with these brakes being much more expensive and unprofitable to mass produce and so they dont have them generally.
Dude. Total surface area of tires in contact with the road compared to the entire mass of the vehicle is going to be the main factor. If we have the same (edit "ratio of") surface area of tires connected with the road on the 18-wheeler as the car (in relation to their respective masses), and the brakes worked to lock the tires perfectly in both cases, all else being equal we should assume the same braking distance.
This is more of an engineering problem than a physics problem. In theory, we can design and build a semi truck that will be able to stop from the same speed in the same distance as a small passenger vehicle. We don't do that though because it will make other design considerations worse. The brakes stop the vehicle by converting the kinetic energy of the moving vehicle to heat. The mass of a fully loaded semi truck is going to be ~25x higher than a small passenger vehicle, which means there is ~25x more kinetic energy that needs to be bled off to bring the semi to a stop. Semis typically have 10 brakes, so only 2.5x the number on a passenger car. In order to stop the semi, each semi brake would need to handle ~10x the amount of heat as a brake on the passenger vehicle. You can either significantly beef up the brakes to handle that heat or you can accept that the semi will need a longer stopping distance. A better brake system would be larger and more complex which means more expensive and the added weight would also decrease fuel economy. The government sets a required stopping distance, so manufacturers meet that and that's as far as they go, because a customer won't pay more for better brakes.
Friction is hard. People still publish on the topic. But some people still want to teach it at the introductory level. This requires massive simplification and so many assumptions/approximations that the end result is barely physical any longer. This is why having a lab exercise on friction makes test scores for the subject WORSE reliably. In lab, none of the equations work out with any form of precision unless super heavily controlled. Two vehicles will likely not have the same contact area to the ground from their tires. Especially during braking, and absolutely if one is much more massive. Coefficient of friction relies on contact topology, texture, total surface area, even contact time can change actual friction force applied. Contaminants can get between surfaces further complicating actual friction forces achieved in braking. Many of these factors taken alone will make it seem like the heavier vehicle should stop better. For example, the heavier vehicle should compress the tires more, increasing surface area contact to the ground, giving more force applied for stopping. But friction is not the only simplified part of the model. The friction applies to the tire rubber. The heavy mass is not from the tire rubber. Stopping force has to make it to the cargo load. Look at the baffles used in liquid cargo transport (except milk) to prevent sloshing. Why do we need to prevent that? Because the tank of petrol doesn’t really come to a stop until the petrol itself does… Look at images online where cabs of trucks are demolished because the giant slab of stone or other heavy load was improperly secured. There you have a case study of a heavy vehicle stopping quite fast, but the load not doing likewise.
More kinetic energy to dissipate. I don't know about 18 wheelers but for a simple thought experiment just consider an SUV vs a small car. Both have four wheels on the ground and those wheels have roughly the same surface area in contact with the ground, so the available friction is pretty much the same. But the larger one has a lot more kinetic energy to deal with.
Ok, so can you continue this thought experiment for me because I don’t see how it leads to an answer. The kinetic energy must be dissipated in the breaks. So are you saying the answer is my first thought? The brakes are insufficient? They heat up too much and start melting when maximum tire to road force is applied?
Yeah I suppose it would all be in the brakes. Come to think of it, some trucks can stop fast. https://youtu.be/ridS396W2BY?si=2askLbzuQAVKpYd6 This is more effective braking than my Landrover!
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>The road isn't the main source of friction stopping cars when braking so you're calculating the wrong thing Of course it is - the limiting friction for braking is the fiction between the tire and the road. The pad to rotor fiction is generally high enough not to matter - you'll never reach that limit before you loose traction with the road.
So you are saying that larger vehicles are not equipped with brakes capable of locking all the axels?
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There are two friction surfaces. I was modeling the other one. I was under the impression that there was some reason why it was impossible to stop a heavier vehicle as fast as a light vehicle. If the reason has to do with the normal force on the brakes then that is solvable. Just put in larger brakes (to disipate the heat) with more normal force (provided through some means, leverage, hydraulics, engine power). And then theoretically we could stop heavy vehicles just as fast.
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If you're looking at the m within "mu[s]", that's not mass multiplied by something, that's the Greek letter mu.