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Braking distances, and weight

by phil4321
Tags: braking, distances, weight
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phil4321
#1
May10-06, 12:47 PM
P: 2
Hi im new to the forums and i'll try and keep this short and sweet.

A couple weeks ago i started going to driving school, its boring but neccessary. On the first day she said (the teacher) that a fully loaded transport truck will take LESS time to stop than an empty one. her reasoning for this is the weight of the truck pulls it down and slows it faster. i know this cannot be right becuse the weight of that cargo is going to carry momentum and is going to be harder to stop. My dad has been driving truck for 30+ years and he agrees with me, so im pretty confident with my side of the argument but i need a way to prove it. ive been looking on the net for the past while but the math involved with it is way over my head and i cant find one that involves weight either. Do i need to find the kinetic energy first before i can solve this. i found an equation to find the KE that take speed and weight into factor but i couldnt find an equation that finds braking distance when you know the kinetic energy. if some one knows an equation i could use for this, or some way i can find this out it would be greatly appretiated.

ps: a site supporting my side would work too but if i had the math to back it up she cant really argue with that. and that would really be supperior

thanks alot
phil
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rbj
#2
May10-06, 11:54 PM
P: 2,251
it's an issue of momentum and friction force.

friction force is proportional to the "normal" force, [itex] F_n [/itex],, the component of all resulting forces acting on an object that is "normal" or perpendicular to the surface. often the normal force is simply the weight of the object.

there are, in classical mechanics, two constants of proportionality. 1. the coefficient of static friction, [itex] k_s [/itex], which governs the maximum force (parallel to the surface) needed to dislodge the object from sticking to the surface and skidding. 2. the coefficient of kinetic friction, [itex] k_k [/itex], which governs the force of friction when the object is skidding. these coefficients of friction are dependent mostly on the materials of the two surfaces rubbing together. (so there is some coefficients of static and kinetic friction for rubber against road.) usually the coefficient of static friction is greater than the coefficient of kinetic friction. (it takes more push to get your refridgerator to start sliding than what you need to keep it sliding.)

assuming that the brakes are applied to the maximum degree so that slippage between rubber and road (or skidding) does not occur, that's where the maximum braking can occur.

the bracking force is

[tex] F_b = k_s F_n [/tex]

the normal force is the weight which is proportional to the mass.

[tex] F_n = m g [/tex]

where [itex] g [/itex] is the acceleration of gravity. so the braking force is

[tex] F_b = k_s m g [/tex]

and that braking force is proportional to the deceleration [itex]a[/itex] of the vehicle

[tex] F_b = k_s m g = m a [/tex]

so the deceleration due to braking is

[tex] a = k_s g [/tex]

independent of the mass of the vehicle. the increased mass increases the braking force, but it also increases the momentum by the same factor. if the vehicle was on a slippery surface, [itex] k_s [/itex] would be less and the vehicle would decelerate more slowly. if the vehicle was on the moon, [itex]g[/itex] would be less and the vehicle would decelerate more slowly. also if the vehicle is pulling a trailer which does not have braking on its wheels, that increased mass from the trailer and load contributes to the momentum, but not to the braking force because only the pulling vehicle weight is pushing down on its wheels (that have brakes), so it decelerates more slowly.

if you know the velocity when the brakes are applied, the braking distance is:

[tex] d = v t - \frac{1}{2} a t^2 [/tex]

and since the final velocity is zero

[tex] 0 = v - a t [/tex]

then you can calculate the braking time

[tex] t = v/a [/tex]

and

[tex] d = v (v/a) - \frac{1}{2} a (v/a)^2 = \frac{v^2}{2 a} = \frac{v^2}{2 k_s g} [/tex]


i think your teacher is wet. it is possible that the increased weight and normal force somehow increases this coefficient of friction (which would increase the deceleration), but she would need some kind of experimental evidence to support that.
Stingray
#3
May11-06, 07:41 AM
Sci Advisor
P: 674
A real vehicle is much more complicated than that. Tires have an effective friction coefficient that is not really constant. It decreases slightly with increasing normal force (weight). That usually implies that braking times are worse when more mass is added, which is one reason that race cars are kept as light as possible.

But there are many other factors to worry about on a large truck. A loaded semi is very very different from an unloaded one. The weight is obviously completely different. The tire pressures would also have to be changed. The center-of-mass location changes, which affects the optimal split in brake pressure between the different axles. There are probably details in the braking system which are changed to account for this. So comparing loaded and unloaded trucks to each other isn't very simple at all.

Gelsamel Epsilon
#4
May12-06, 12:10 AM
P: 316
Braking distances, and weight

Assuming same speed, same brake friction, same road friction on a perfectly flat surface a heavier truck will always take longer to stop due to higher inertia. But it is possible that brake friction and road friction is increased due to more force pushing down, but I seriously doubt that it is enough to counter the change in inertia.

~Gelsamel

(And yes I'm too lazy to do any math, you'll just have to take my word for it).
Stingray
#5
May12-06, 03:41 AM
Sci Advisor
P: 674
No. As rbj showed, braking distances do not depend on weight when the friction coefficient is constant. More force can be generated by the tires with more weight on them, which exactly makes up for the extra inertia. But the case of identical friction isn't all that great of an idealization.
al_201314
#6
May12-06, 08:42 AM
P: 116
If you wouldn't mind me bringing this up as well, I had difficulty understanding this.

For a given weight of a truck for example, why does increasing the number of wheels not increase the friction acting on the truck? I would have thought more tires would spread out its weight but also offer more resistance since there is more contact with the ground?
phil4321
#7
May12-06, 10:16 AM
P: 2
alright picture this, instead of to semi's we have a car and a motorcycle.

which is going to stop sooner? obviously the bike. why? becuse it weighs next to nothing compared to the car. to me its not wether a heavier vehicle takes longer to stop, i know it does.

i just wanted to prove this drving instructor wrong, cause i know she is wrong. i just dont know how to do it
berkeman
#8
May12-06, 10:34 AM
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P: 41,090
Quote Quote by phil4321
alright picture this, instead of to semi's we have a car and a motorcycle.

which is going to stop sooner? obviously the bike. why? becuse it weighs next to nothing compared to the car.
Actually, that's not true. A high performance sportscar and a sportbike stop in about the same distance. And given the big difference in mass like you pointed out, that kind of shoots down the idea that braking distance depends on mass (given equal friction coefficients and equal brakes). Both the sportbike and sportscar are able to brake at the edge of traction, which gives you the shortest stopping distance. The sportbike can out accelerate most sportscars, because it has a better engine power to overall weight ratio. That's where a sportbike has the edge, not necessarily in braking.

BTW, your driving instructor is definitely wrong in saying that a loaded truck will stop faster than an empty one. It will generally be the other way around, because a fully loaded semi does not have good enough brakes to stay at the limit of traction effectively on all 18 wheels during full-on braking.
Stingray
#9
May12-06, 06:32 PM
Sci Advisor
P: 674
Quote Quote by al_201314
For a given weight of a truck for example, why does increasing the number of wheels not increase the friction acting on the truck? I would have thought more tires would spread out its weight but also offer more resistance since there is more contact with the ground?
Add together the weight supported by each tire, and you'll get the total weight of the truck. In the simplest approximation, the maximum frictional force a tire can generate grows linearly with how much weight is on it. So if each tire has the same friction coefficient, the total force will not depend on how many wheels there are. Reality is a bit more complicated, though.

Quote Quote by phil4321
which is going to stop sooner? obviously the bike. why? becuse it weighs next to nothing compared to the car. to me its not wether a heavier vehicle takes longer to stop, i know it does.
You are changing so many variables that the example is pointless. By far the most important variable in stopping distances is the tires. But do a semi and bike have tires that are in any way comparable? Of course not.

Sportbike tires are very sticky. The tradeoff is that they last only a couple of thousand miles. They're designed for performance at the cost of almost everything else. In contrast, the tires on a semi are meant to last as long as possible and to minimize fuel consumption. It's no surprise that they perform poorly.

As berkeman mentioned, sports cars (which don't come with economy tires) stop as well or better than sportbikes. And the car tires are still less compromised than the bike tires. Stick some R-compounds on the car (which are probably comparable to sportbike tires), and no bike on factory tires could come close.

Anyway, as I mentioned before, the main reason that weight does influence braking distances is because tire friction does not grow linearly with the normal force. It is linear at lower weights, but it eventually starts to deviate a bit when you load the tire enough. Where that deviation starts to become significant is completely different for different tires.

Quote Quote by berkeman
It will generally be the other way around, because a fully loaded semi does not have good enough brakes to stay at the limit of traction effectively on all 18 wheels during full-on braking.
Are you sure about that? I'd be very surprised if trucks couldn't generate enough line pressure to be on the edge of lock-up. Or do you mean that there isn't enough control over all of the axles to distribute the pressure optimally? Do big trucks have ABS?
berkeman
#10
May12-06, 06:39 PM
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P: 41,090
Quote Quote by Stingray
Are you sure about that? I'd be very surprised if trucks couldn't generate enough line pressure to be on the edge of lock-up. Or do you mean that there isn't enough control over all of the axles to distribute the pressure optimally? Do big trucks have ABS?
Most big trucks do not have ABS, not even on the rears. That's why they often lock the rears at the end of hard stops when they are empty. And ABS would just lengthen the stopping distance anyway for full-on braking.

Yeah, I was referring mostly to how the different axles of a fully-loaded semi have different weights on them, so there's no way you could get maximum braking out of all of them at once. Heck, even the dual tire assemblies work against you. And the tractor's 6 tires are going to be loaded differently than the other 12 tires. Plus, I think that for a fully-loaded semi, it would be pretty hard to gnerate enough brake force to get to max braking. I don't think I've ever seen the back 4 tractor tires of a loaded semi lock up under hard braking....
willyadventur
#11
Aug16-11, 09:37 AM
P: 17
I know this is an old post but

1) When braking the front of the car nose dives causing the car body to pitch forward (till at an angel), does this alter the normal force of gravity (of the vehicles body mass) acting on the all of the tyres Fnorm = M*g*Cosine Angel (Inclined plane, like)?

I think not; because the road wheels are still horizontal with the road, (and presumably the direction of motion hasn’t changed), and all that has changed is only a proportional weight transfer from the rear to front wheels, but I’m not certain.

2) Towing a ‘non-braked trailer’. How do we factor in the Force (m*a) of the un-braked trailer? (The trailer has no brakes of its own)

----------------Force Braking = Ks_g* Mass vehicle = (Mass-vehicle + Mass trailer) * Accel
----------------Accel = (Ks_g) / Mass of the trailer-------------------

At first this looks right, but I’m not sure if we should consider some torque affect due to the diameter of the wheels, but I don't think so?

Willy
PeterO
#12
Aug16-11, 09:55 AM
HW Helper
P: 2,318
Quote Quote by phil4321 View Post
Hi im new to the forums and i'll try and keep this short and sweet.

A couple weeks ago i started going to driving school, its boring but neccessary. On the first day she said (the teacher) that a fully loaded transport truck will take LESS time to stop than an empty one. her reasoning for this is the weight of the truck pulls it down and slows it faster. i know this cannot be right becuse the weight of that cargo is going to carry momentum and is going to be harder to stop. My dad has been driving truck for 30+ years and he agrees with me, so im pretty confident with my side of the argument but i need a way to prove it. ive been looking on the net for the past while but the math involved with it is way over my head and i cant find one that involves weight either. Do i need to find the kinetic energy first before i can solve this. i found an equation to find the KE that take speed and weight into factor but i couldnt find an equation that finds braking distance when you know the kinetic energy. if some one knows an equation i could use for this, or some way i can find this out it would be greatly appretiated.

ps: a site supporting my side would work too but if i had the math to back it up she cant really argue with that. and that would really be supperior

thanks alot
phil
Of course the only really logical answer for that is :
"Wow, thanks ma'am; I never would have thought of it that way - I will certainly keep my eyes open for trucks, whether they are empty or loaded!"

And if your instructor was male you would say:
"Wow, thanks sir; I never would have thought of it that way - I will certainly keep my eyes open for trucks, whether they are empty or loaded!"



And I do believe a loaded truck will take longer to stop.

If you were to accelerate a truck for 10 seconds, and then apply the brakes to stop - the fully laden truck may certainly stop quicker, but only because it will only have made it to 10 - 15 mph, where as the empty truck may have already reached 30 mph or more!!
But if you let both trucks wind up to 40 mph, the empty one is going to stop more quickly, though is more prone to having the trailer wheels lock-up in an emergency stop.
willyadventur
#13
Aug16-11, 02:00 PM
P: 17
Quote Quote by PeterO View Post
Of course the only really logical answer for that is :
"Wow, thanks ma'am; I never would have thought of it that way - I will certainly keep my eyes open for trucks, whether they are empty or loaded!"

And if your instructor was male you would say:
"Wow, thanks sir; I never would have thought of it that way - I will certainly keep my eyes open for trucks, whether they are empty or loaded!"



And I do believe a loaded truck will take longer to stop.

If you were to accelerate a truck for 10 seconds, and then apply the brakes to stop - the fully laden truck may certainly stop quicker, but only because it will only have made it to 10 - 15 mph, where as the empty truck may have already reached 30 mph or more!!
But if you let both trucks wind up to 40 mph, the empty one is going to stop more quickly, though is more prone to having the trailer wheels lock-up in an emergency stop.

Momentum = M*V and the units are Kg*m/s,This can also be considered as measure of work required to alter your vehicles speed (or Ke (energy required), which is not exactly right, but close enough for the now)

F= m*a, and the units are Newton’s, and this is a measure of Force required (don't confuse your units). If your brakes (that is the brake drum and shoe, not tyres) are working well, then they will always provide the necessary force. But! Ke (momentum) requires work to stop, OR MORE PERSISLY you have to use up the energy, and (on a motor vechile) this is done in the form of heat. Ask your dad what happens when your brakes get to HOT!

So all, that you are left with is the maximum friction that your tyres can exert on the ground Vs change in momentum. As the weight goes up, so dose the friction, but momentum also increase and your tyres only can generate so much friction before they lose traction (blow/disintegrate) and your brakes can only get rid of so much heat!

I think Your teacher is confusing force and acceleration (F=mass*acell and Accel=F/m) with Velocity (Momentum = Mass*velocity)

Thus a loaded truck can take longer to stop, simply because the tyres and braking systems requierd, aren't good enough and/or you can't afford. (Tell your techer there is a maximum speed and weight limit for a reason!).

Note: A very good truck can exert a breaking force (including brakes, tyres, and load) 0f about 1.2 G,’s. A formula 1 car is about 5 G's and cost about $1,000,000. (G’s = G force) and at 5 g’s your chest is pressing so hard against your rib cage it’s hard to breath, also, the average Air-Force fighter pilot will pass out at about 8 G’s, so the limiting factor is the driver).
PeterO
#14
Aug16-11, 08:55 PM
HW Helper
P: 2,318
Quote Quote by willyadventur View Post
Momentum = M*V and the units are Kg*m/s,This can also be considered as measure of work required to alter your vehicles speed .....
A posting that starts out like this is a real worry.


MOMENTUM and Work don't even have the same unit - so to think if you think that Momentum is a measure of the work needed you are sorely mistaken.

If you double the speed of a truck, you double its momentum, but the work needed to stop it is 4 times as great.

triple the speed, triple the momentum, but 9 times the work needed.
willyadventur
#15
Aug17-11, 10:03 AM
P: 17
Dear PeterO

Please consider your statement very carefully (which is correct and is supposed to rebut mine)
“If you double the speed of a truck, you double its momentum, but the work needed to stop it is 4 times as great. triple the speed, triple the momentum, but 9 times the work needed”.

Please read my posting carefully “Momentum can be considered as a measure of 'work required' to 'alter your vehicles speed' (or Ke (energy required), which is not exactly right, but close enough for the now). I also said “be careful about confusing your units”, because confuses the hell out of others

Can you see the correlation in your own statement?

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2


Background information

Momentum = M*v --- (mass times velocity)
v= m/s ----- (m=meters)
Therefore Momentum = M * (m/s)

And the momentum impulse exchanged from one object to another is
N*s = (M*9.8)*s (this means applying a force of 1n for 1 second)

Are you with me so far??

KE= 1/2mv^2 = __(1/2)(kg*m^2/s^2) = __1/2 Mass * V^2 =____= Mass * V^2 / 2_____= joules
Energy = (joules) __ = M*(m/s)^2 __ = M * M^2 / s^2___ = 1 N•m =___1 Joule

(In this case 1n.m is not torque : It is equal to the energy expended (or work done) in applying a force of 1 Newton through a distance of 1 meter (or 1 Newton meter = 1 Newton force applied over a 1 meter distance?).

The formula for Ke is 1/2M*v^2 = 1 joule__ = M*(m/s)^2 = M*v^2----------and Momentum = M*v = M*(m/s) ------ (M=Mass and m=meters)

Are you still with me and starting to see the correlation here??

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2

And Kenetic energy = Momentum^2 / 2*mass

SO you still don't think that the MOMENTUM of a truck driving down the road IS not RELATED to the ENERGY that is required to stop it? OF COURSE IT IS! It all depends on how you want to go about Calculating it and in What Units. But for basic, stick to the basics and don’t interchange your calculations, unless you really know them well.

‘O’ Dear
willyadventur
#16
Aug17-11, 10:04 AM
P: 17
Dear PeterO

Please consider your statement very carefully (which is correct and is supposed to rebut mine)
“If you double the speed of a truck, you double its momentum, but the work needed to stop it is 4 times as great. triple the speed, triple the momentum, but 9 times the work needed”.

Please read my posting carefully “Momentum can be considered as a measure of 'work required' to 'alter your vehicles speed' (or Ke (energy required), which is not exactly right, but close enough for the now). I also said “be careful about confusing your units”, because confuses the hell out of others

Can you see the correlation in your own statement?

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2


Background information

Momentum = m*v (mass times velocity)
v= m/s
Therefore Momentum = m * (m/s)

And the momentum impulse exchanged from one object to another is
N*s = (M*9.8)*s (this means applying a force of 1n for 1 second)

Are you with me so far??

Energy
Ke = m^2/s^s
1 Joule = 1 N•m = 1 kg•m^2/s^2

(In this case 1n.m is not torque : It is equal to the energy expended (or work done) in applying a force of 1 Newton through a distance of 1 meter (or 1 Newton meter = 1 Newton force applied over a 1 meter distance?).

The formula for Ke is 1/2m*v^2 = m^2/s^s----------and Momentum = m*(m/s)

Are you still with me and starting to see the correlation here??

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2


SO you still don't think that the MOMENTUM of a truck driving down the road IS not RELATED to the ENERGY that is required to stop it? OF COURSE IT IS! It all depends on how you want to go about Calculating it and in What Units. But for basic, stick to the basics and don’t interchange your calculations, unless you really know them well.

‘O’ Dear
PeterO
#17
Aug17-11, 11:02 AM
HW Helper
P: 2,318
Quote Quote by willyadventur View Post
Dear PeterO

Please consider your statement very carefully (which is correct and is supposed to rebut mine)
“If you double the speed of a truck, you double its momentum, but the work needed to stop it is 4 times as great. triple the speed, triple the momentum, but 9 times the work needed”.

Please read my posting carefully “Momentum can be considered as a measure of 'work required' to 'alter your vehicles speed' (or Ke (energy required), which is not exactly right, but close enough for the now). I also said “be careful about confusing your units”, because confuses the hell out of others

Can you see the correlation in your own statement?

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2


Background information

Momentum = m*v (mass times velocity)
v= m/s
Therefore Momentum = m * (m/s)

And the momentum impulse exchanged from one object to another is
N*s = (M*9.8)*s (this means applying a force of 1n for 1 second)

Are you with me so far??

Energy
Ke = m^2/s^s
1 Joule = 1 N•m = 1 kg•m^2/s^2

(In this case 1n.m is not torque : It is equal to the energy expended (or work done) in applying a force of 1 Newton through a distance of 1 meter (or 1 Newton meter = 1 Newton force applied over a 1 meter distance?).

The formula for Ke is 1/2m*v^2 = m^2/s^s----------and Momentum = m*(m/s)

Are you still with me and starting to see the correlation here??

I’ll give you a hint it is something about
speed^2 or maybe M^2
Or maybe the MOMENTUM^2


SO you still don't think that the MOMENTUM of a truck driving down the road IS not RELATED to the ENERGY that is required to stop it? OF COURSE IT IS! It all depends on how you want to go about Calculating it and in What Units. But for basic, stick to the basics and don’t interchange your calculations, unless you really know them well.

‘O’ Dear
I am perfectly aware of what momentum is, and I am perfectly aware of what Kinetic energy is. I also know that a change in momentum can be measured by the impulse of a force - Force*time, and that a change in KE can be measured by the Work done by a force - Force* distance.
I also know that you should never ue momentum and Energy in the same sentence other than to say that momentum and energy are completely different quantities.
Energy can be transformed from one type to another - with a braking system the Kinetic energy is converted to heat energy.
Momentum can not be transformed into anything, since there is only one form of momentum. Instead it can only be transferred to another body. Hopefully that other body is only the Earth.

I also stand by my opening statement that any explanation that starts off “Momentum can be considered as a measure of 'work required' to 'alter your vehicles speed'" should be instantly dismissed since it will contain a mish-mash of conflicting concepts and units and have only a passing resemblance, if any, to the Physics of the situation.
You also said " “be careful about confusing your units”, because (it) confuses the hell out of others." Your explanations certainly showed the units "confuse the hell out of you" - not sure about others
willyadventur
#18
Aug17-11, 11:24 PM
P: 17
Dear PeterO

Does

Kinetic energy = Momentum^2 / (2*mass),
um, Sure it does ,
(but I wouldn't want to confuse Ke with momentum)


MomentumF • t = m • Δ v.
a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N•s. This impulse would cause a momentum change of 720 kg•m/s. In a collision, the impulse experienced by an object is always equal to the momentum change.

Can we correlate m/s to Distance, Um yes

Work = Force X Distance, but I'm no expert

Can we correlate m/s and distance to Acceleration, Um I think so

Force = Mass X Accl

In this case the colliding object is the tyres which provide the force (mass) to change momentum. And the brakes do work on the Ke in the form of friction and heat

In my original post (two posts ago) I was merely trying to explain an easy way to correlate the change in momentum in the vehicle to brakes to Heat (Ke) to force to tyres without having to go through many sets of various equations for each component. Which I pretty sure I have done. And if I was a computer programmer, who's model would they like best, the long version or the shorter algorithm?


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