Elementary physics question

[Dave Rogers]

As previously mentioned, braking force for a car is generally limited by the maximum friction force between the road and the tires (kinetic if they skid, static if they don't), which depends upon the normal force and hence the weight of the cars, and not by the limitations of the brakes themselves. If the weights are not the same, the braking forces will generally not be either.

That's certainly the case for cars in the real world, where the braking system is designed appropriately to the mass of the car. However, the OP specified the condition that "everything else is identical except the weight of the cars", from which we can deduce that the maximum braking forces applied to the two cars are identical. I say "maximum braking forces" because the braking force is limited to the kinetic friction force if the car skids, whatever theoretical force might be available, and this must vary with the mass of the car.

Dave

 

[Typicallucas]

That's certainly the case for cars in the real world, where the braking system is designed appropriately to the mass of the car. However, the OP specified the condition that "everything else is identical except the weight of the cars", from which we can deduce that the maximum braking forces applied to the two cars are identical. I say "maximum braking forces" because the braking force is limited to the kinetic friction force if the car skids, whatever theoretical force might be available, and this must vary with the mass of the car.

Dave

That's how I read the OP too. If you test a car's stopping distance and then put a heavy load in the backseat and test it again, the car will take longer and farther to stop.

Otherwise if we are talking about "scaling" a car up to be twice as heavy and scale up the brakes as well, I don't think that is an exercise that has any practical basis in reality or any utility. My reasoning follows.

If you want to double the size of the brakes your results will vary depending on how you determine the new size of the brakes. You have to choose a method of scaling up the brakes and all will give you different results:
Increase brake pad area (by how much, double? quadruple?)
Increase the lever arm of the brake disc (by how much, double? 1.4?)
Increase the number of brake pads
Increase the brake pressure available
Modify the brake pad/disc materials to grip better

Although if you are going to actually try this you will also need to modify the suspension to handle the increased weight. Since suspension of a car is a dynamic system especially under braking I don't believe there is a reliable way to scale it without totally mucking up braking characteristics. Are sway bars twice as large, twice as stiff, both?

Also, do you double the width of the tires? Do you double the weight of the tires? Or do you choose tires that have twice the rotational inertia? All would affect the breaking distance.

 

[Dr. Trintignant]

Otherwise if we are talking about "scaling" a car up to be twice as heavy and scale up the brakes as well, I don't think that is an exercise that has any practical basis in reality or any utility.

Perhaps not, but there's an easy way to do it conceptually. The doubled-up car is just that--two of the original cars bolted together (side-by-side, let's say), with the controls slaved together. Clearly, the double car brakes in exactly the same distance as the single car.

- Dr. Trintignant

 

[Soapy Sam]

I'm unsure what this thread tells me about physics- but it confirms my suspicion that Finland is a very odd place.

 

[Typicallucas]

Perhaps not, but there's an easy way to do it conceptually. The doubled-up car is just that--two of the original cars bolted together (side-by-side, let's say), with the controls slaved together. Clearly, the double car brakes in exactly the same distance as the single car.

- Dr. Trintignant

Ok! That would work!

In that case I would also expect the car to brake in exactly the same way.

 

[Ziggurat]

That's certainly the case for cars in the real world, where the braking system is designed appropriately to the mass of the car. However, the OP specified the condition that "everything else is identical except the weight of the cars", from which we can deduce that the maximum braking forces applied to the two cars are identical. I say "maximum braking forces" because the braking force is limited to the kinetic friction force if the car skids, whatever theoretical force might be available, and this must vary with the mass of the car.

Huh? You seem to be saying both that the forces are the same and that they are different.

But it's not the kinetic friction of the tires we care about for a minimum stopping distance, because one never needs to skid. A good driver can prevent skidding even without ABS brakes. And car brakes are over-designed: that is, the brake pads are capable of applying enough force to lock the wheels for expected loads, even though you want to apply less force than that at the brake pads. So without more info on the brakes, I'll assume that both can provide enough to maximize static friction between tires and the road.

 

[RossFW]

Hmm....

Just trying to square this with my real-life experience with aircraft.

We can calculate required stopping distance for take-off and landing. Same aircraft, same brakes, but with a wide range of mass (Boeing 777, empty weight approx 175T, Max takeoff weight, 350T). You ALWAYS need more runway to stop if the mass is greater, and that would have seemed analogous to the OPs senerio.

 

[Skwinty]

Hmm....

Just trying to square this with my real-life experience with aircraft.

Hi RossFW
With aircraft of that size you would have auto braking at various setting.
If you used the brakes to maximum effect, there is the possibility of fire and tyre blow outs etc. Even if you had no fire or blowouts your tyre replacement costs would sky rocket. Just a thought.

 

[RossFW]

Skwinty,

Auto-brakes give you a set rate of retardation up to an energy limit.

In the case of a high-speed rejected take-off, we use maximum available breaking. In a high-speed, high-weight case, it will almost certainly mean losing tyres in the process, but we would do it just the same as the alternative would be runing off the end of the runway.

But, as I said, in calculating a limiting case, it always comes out that a heavier aircraft needs more runway to stop.

 

[Ziggurat]

We can calculate required stopping distance for take-off and landing. Same aircraft, same brakes, but with a wide range of mass (Boeing 777, empty weight approx 175T, Max takeoff weight, 350T). You ALWAYS need more runway to stop if the mass is greater, and that would have seemed analogous to the OPs senerio.

No. First off, landing speed tends to vary with mass. Secondly, there are significant forces aside from tire/road friction, including air resistance and reverse thrust, which won't scale directly with mass. And third, I don't think you can generally apply enough braking force to max out the frictional forces, because the heat load in the brakes for something that heavy and fast would be too great.

 

[RossFW]

Ziggy,

Yes, approach speed is higher at higher weights as it is a function of stall speed, but not so much ( I would have thought) that it would be an over-ridding factor. As an example a difference in landing weight of 70T yields a difference in Vref (Basically landing speed) of 19kts (132 as opposed to 151), and a difference in max-braking landing roll of 350m (830 vs 1180). Reverse thrust is not included in these calculations as an added safety factor, and in any case would be around equal in both cases, as would aerodynamic drag which is minimal once the wing is no longer producing significant lift.

Crunching the speed difference, I would make a 23% increase in KE assuming two aircraft of the same weight landing at different speeds, whilst we have a 40ish% increase in the landing distance required for the heavier aircraft.

Our brakes are quite capable of locking the wheels even at these speeds and weights (we have two MASSIVE brake shoes on EACH of the twelve main wheels) but have an Anti-skid system (basically a really good ABS system) to pull them up just short of skidding, so we are putting down as much friction as is available.

I am very interested in this, as it is challenging a well rooted pre-supposition on my part. If someone can definitively show this one way or the other, that would be great.

My own feeling is that, while I see that the heavier vehicle would have greater friction due to the higher pressure on the contact patch, I can't see this as being directly proportional to the mass, and therefore the retarding force needed to stop.

 

[Dave Rogers]

Huh? You seem to be saying both that the forces are the same and that they are different.

No, not at all. I'm saying that the OP specifies that the braking systems are identical, and assuming that the force applied to the brake pads is the same in both cases. If that force is too small to cause either car to skid, then the car that is twice as heavy stops in twice the distance. If the force is sufficiently great to cause both to skid, then both cars stop in the same distance. However, there's an interesting regime between the two where the same force applied to the brake pads can cause only the lighter car to skid, and in that regime the stopping distance of the heavier car may be less.

But it's not the kinetic friction of the tires we care about for a minimum stopping distance, because one never needs to skid. A good driver can prevent skidding even without ABS brakes.

If the cars are being driven optimally for their weights and braking systems and both braking systems are powerful enough to lock the wheels, then I agree that the heavier car will require a longer stopping distance. However, neither of those conditions was specified in the OP, so I was giving a more general analysis.

Dave

 

[X]

Assuming the two cars have identical brakes both acting at maximum efficiency, this is not correct.

Brakes dissipate energy over time. More energy (which is so with a more massive car) takes more time to dissipate. Light cars stop quicker than heavy cars- a fact borne out in everyday life.

"A fact borne out in everyday life" indeed.

I experience this whenever I drive my Lincoln. It weighs 5,200 lbs. And its brakes were state-of-the-art, 40 years ago.
For what should be obvious reasons, I drive it a little slower than the limit, especially in heavy traffic. And leave a lot of room ahead of me.
Yet I still get people in compact cars pulling in front of me and hitting their brakes. Which can be terrifying, because I physically can not stop in the distance they can.



Also, doesn't the friction generate by tires on the road depends not only on the normal force, but also on the contact area?
Perhaps not directly on the contact area. Rubber compound and tire strength would be the main factors, and bigger tires can take more force. As well, there is the roughness of the road.


I must say, for such deceptively simple questions, this thread has become interesting.

 

[Ziggurat]

No, not at all. I'm saying that the OP specifies that the braking systems are identical, and assuming that the force applied to the brake pads is the same in both cases.

If the brake pads apply equal force to the brake discs/drums for both cars, there's really no way to tell what happens without more information. If the brake pads apply enough force to lock the wheels on both cars, then they should stop at the same time. If the brake pads don't apply enough force to lock wheels on either car, then the lighter car will stop first. But if they apply enough force to lock the wheels on the lighter car but NOT the heavier car, then a curious thing can (but doesn't necessarily) happen: the light car could take longer to stop than the heavier car.

 

[Ziggurat]

Our brakes are quite capable of locking the wheels even at these speeds and weights (we have two MASSIVE brake shoes on EACH of the twelve main wheels) but have an Anti-skid system (basically a really good ABS system) to pull them up just short of skidding, so we are putting down as much friction as is available.

Hmmm... Well, I guess I was wrong about that aspect. Though given that we're talking about airplanes, where the normal force against the ground is a function of aerodynamics and not simply mass, we should probably be careful about assuming too much about how braking power scales with mass. But it still leaves the question of other forces (reverse thrust and drag from those flaps they stick up).

 

[Ziggurat]

Also, doesn't the friction generate by tires on the road depends not only on the normal force, but also on the contact area?

Not on dry pavement, at least not directly.

Rubber compound and tire strength would be the main factors, and bigger tires can take more force.

Yes, this is the main reason racecars use fat tires. They push the rubber to the limits of its material strength, and if you start "erasing" the rubber onto the pavement, you lose traction. But until that starts to happen, the friction force is basically independent of contact area.

 

[jasonpatterson]

Since friction depends (at least in its most commonly used, ideal incarnation) only on the normal force and the coefficient of friction between two surfaces, changing the surface area does affect the force.

Try thinking of it this way:
Suppose I have a box that is 2x3x4 cm and has a weight of 10N. On the 2x3 face, there are 6cm² of surface area that support 10N. We could imagine that instead of pulling a single box with a 2x3 face, we are instead pulling 6 smaller boxes, each with a 1x1cm face and a weight of 1.67N. The total force required to move all of these smaller boxes would be the same as the force required to move the single larger box.

OK, now flip the box onto its 2x4 side. It now has 8cm² that are supporting 10N. Again we could imagine this as 8 1x1cm boxes each with a weight of 1.25N.

Suppose the coefficient of friction (the ratio of the frictional force to the normal force) is 0.5. Each of the 6 small boxes from the 2x3 situation would then experience a 0.83N frictional force, for a total of 5N. In the case of the 2x4 situation, each of the 8 small boxes would experience 0.625N of frictional force, for a total of 5N. In either case the total frictional force is the same because although the area over which the friction occurs increases going from the first to the second example, the pressure that the object applies to the surface decreases by the exact same factor. The two effects cancel and, in general (and again, ideally) the same frictional force is experienced.

 

[MattusMaximus]

I think we need to construct an experiment. Who wants to burn some rubber?

 

[dlorde]

No, I think your reading of what I wrote was correct, but it's the same as Dave's reading. The gravitational force on the large ball is ~981N and on the small one ~98.1N. The acceleration due to gravity is the same for both balls, 9.81 ms^-2
In a vacuum the balls would accelerate equally, but when there is a drag force dependent only on velocity (since the shape and surface area are the same for both balls), it reduces the total downward force on the lighter ball much more rapidly..

OK, that clarifies it nicely. Thanks.

 

[Dave Rogers]

If the brake pads apply equal force to the brake discs/drums for both cars, there's really no way to tell what happens without more information. If the brake pads apply enough force to lock the wheels on both cars, then they should stop at the same time. If the brake pads don't apply enough force to lock wheels on either car, then the lighter car will stop first. But if they apply enough force to lock the wheels on the lighter car but NOT the heavier car, then a curious thing can (but doesn't necessarily) happen: the light car could take longer to stop than the heavier car.

Agreed, as that's exactly what I said in post #4.

Since friction depends (at least in its most commonly used, ideal incarnation) only on the normal force and the coefficient of friction between two surfaces, changing the surface area does affect the force.

Did you mean "does not affect the force", given that that's what your example shows?

Dave