Stopping distance for car, should increase with increase of mass... by how much?

[inquisitive]

Hello .

Have a question and am interested to find a simple equation to include mass as a variable regarding stopping distances of a car

In the UK highway code..... a braking distance = 14m for a car at 30 mph (48 kph)

my questions are..... if you load the car up fully (increase mass)..... then (for same v) that distance will increase????
Q1: by how much ?
Q2: what's a simple equation ?

NB
1. [The car and the road conditions are the same for each stop]
2. [the original car is say 1000kg]
3. [the fully loaded car is say 1500kg]

thanks for time / thoughts
inq

[swansont]

s = vt + 1/2 at^2 where s is distance, v is the original speed, a is the acceleration and t is time. You can find the acceleration the brakes supply

 

Assuming the brakes are giving the same force, a=F/m, so if you double the mass, the acceleration drops in half

[ewmon]

You're talking about a panic stop where the tires apply the greatest traction to the road without breaking free. Increasing the mass of the car increases the car's normal force on the road and, thus, the traction and the greatest traction. So, the heavier car could decelerate at the same rate and stop in the same distance.

 

HOWEVER, I'm not remembering my mechanics enough to recognize the limitations of a car's braking system. Obviously if the car weighs 1,000,000 kg, the braking system will not stop it in the same distance. The braking system transforms the car's kinetic energy into thermal energy absorbed and dissipated by the disks/calipers and drums/shoes. I think the failure occurs when these components fail due to this heat.

[inquisitive]

s = vt + 1/2 at^2 where s is distance, v is the original speed, a is the acceleration and t is time. You can find the acceleration the brakes supply

 

Assuming the brakes are giving the same force, a=F/m, so if you double the mass, the acceleration drops in half

very helpful thanks...... having the brakes assumed as giving same force, makes for simple relation of mass and acceleration.... therefore can figure that stopping distances vary also.

 

...... I can move on to study brakes mechanisms themselves in a little more detail now.... such as affects of ABS - i know it switches brakes on/off according to traction on road surface (anti skid).... but perhaps it is also governed by max force loading (and release) or max heat tolerence (and release) type of thing.

 

thanks for equations and time to answer

 

 

You're talking about a panic stop where the tires apply the greatest traction to the road without breaking free. Increasing the mass of the car increases the car's normal force on the road and, thus, the traction and the greatest traction. So, the heavier car could decelerate at the same rate and stop in the same distance.

 

HOWEVER, I'm not remembering my mechanics enough to recognize the limitations of a car's braking system. Obviously if the car weighs 1,000,000 kg, the braking system will not stop it in the same distance. The braking system transforms the car's kinetic energy into thermal energy absorbed and dissipated by the disks/calipers and drums/shoes. I think the failure occurs when these components fail due to this heat.

Interesting that heavy car could come to rest in same distance as force on road is greater.... understand principle of this.... and presume that the work done by the braking system must be greater to accomplish this......

 

If there were a max limit that a braking system were allowed to work at..... then (brakes working at that limit), I guess an extra massed car would travel further to stop..... but I would think that is where the study of braking systems would be interesting.

 

Anyway, thanks for info

I think I could go onto study car brake systems to see if such things as ABS, work at optimising max load (and release) and or max heat (and release)..... as well as max traction (and release if lock).

 

cheers for now

 

 

 

 

NOTE TO SELF:

experiment to do.....

 

 

1. jump into a small car..... travel west at 30 mph..... EMERGENCY STOP.... [stamp fully on ABS brakes].... measure distance

 

(load up same car with four extra persons, dog, camping gear, food including 10 bags of spuds to drop off at grandads allotment)

 

2. Do same as 1. now fully loaded.... measure distance.

 

 

distance in 1. = x meters

distance in 2. = same or less or more

 

 

any bets regarding the outcome?

Edited April 20, 2013 by inquisitive

[Enthalpy]

...Increasing the mass of the car increases the car's normal force on the road and, thus, the traction and the greatest traction. So, the heavier car could decelerate at the same rate and stop in the same distance....

Approved.

 

The only remote limit is that the friction "coefficient" isn't really constant... Mankind is still in need of a proper theory of friction, and a proper theory wouldn't have just a "coefficient of friction".

 

The brakes are designed to stop the car even if heavily loaded and at maximum speed.

[PureGenius]

It seems as if the heavier weight would slow the accelleration due to increased mass thus gravity comes into play reducing the ability of the car to accelerate. Also when the car reached the point of breaking its overall mass would b increased therefore its kinetic energy would increase its breaking distance I agree the frictional pressure would b higher and therefore add to its initial breaking power ,would this completely offset the increased mass I can't say. Also i agree there should be a friction eqaution that is exact in order to deal with this qestion. Also we need a eqaution to deal with mass acceleration and kinetic energy increase due to velocity ...

[krash661]

there's to much to type for explanation so,

try this link..

it should help.

 

http://www.gcsescience.com/pfm30.htm

 

weird, same exact question here,

 

http://www.physicsforums.com/showthread.php?t=686694

Edited April 26, 2013 by krash661

[ewmon]

Did you guys take into consideration the doubling of the normal force between the car and the road (and possibly the braking force) when the car's mass is doubled? So, the standard equation, a = F/m, would produce the same deceleration —

(2F_B)/(2m) = F_B/m = a

 

[krash661]

yeah, that link i provided address that .

[ewmon]

Well, it's just that the first link says

 

the braking distance is changed as the mass of a car changes

...

 

The car with twice the mass will take twice as long to stop

 

and the second link is a forum whose members give widely differing opinions.

[krash661]

and the second link is a forum whose members give widely differing opinions.

it's another physics forum, like this one.

but i understand what you are saying,

I have been on that one for some months now ,

it appears to be filled with mathematicians.with some physicist.

but i would not disregard the knowledge there.they and it is very usefully and very qualified.

and also,

that exact same question on that forum is what i posted.as in inquisitive asked it there also and they answered it.

Well, it's just that the first link says

not at all,

it explains the whole proccess, did you notice you can click on words and it will explain.

for example,

click on the underlined words.

Edited April 26, 2013 by krash661

[ewmon]

The GCSE Science site uses fixed braking values and does not consider the increase in the normal force.

 

What is your take away from the varying opinions on the Physics Forum, and why?

[krash661]

The GCSE Science site uses fixed braking values and does not consider the increase in the normal force.

 

What is your take away from the varying opinions on the Physics Forum, and why?

Umm ok..

 

and also i'm not sure what you are asking on that last line.

[ewmon]

 

What is your take away from the varying opinions on the Physics Forum, and why?

 

i'm not sure what you are asking .

 

Because the thread that you linked to has varying opinions and you posted it for a reason. It's like answering a question "Yes and no", so I'm asking about the point you were trying to make. Which opinion in that thread do you agree with and why? That's all.

[krash661]

hmm, maybe refer to post 11 again.

[ewmon]

Are you saying that you agree with the last post in that thread?

[krash661]

Are you saying that you agree with the last post in that thread?

when i stated ,

" hmm, maybe refer to post 11 again. ",

i meant this link below.

 

http://www.scienceforums.net/topic/74421-stopping-distance-for-car-should-increase-with-increase-of-mass-by-how-much/?p=740790

 

i already told you that link i provided is the answer.

 

as for the other forum.

i have moved on already.

i do not remember, have to read all that, which i have no interest to do so.

 

Edited May 3, 2013 by krash661

[dumbbloke]

An increase in car mass does not greatly increase grip before the brakes lockup. Tyre grip is very temperature dependent. Once the tyres slip too much {~8%?}, decelleration is reduced. It is safe to say a increase in vehicle mass always increases braking distance.

[ewmon]

As I have told others here, the increase in the normal force must be considered.

 

Thus, the braking force remains directly proportional to the mass and the deceleration (a=F/m) remains constant, and the car stops in the same distance as long as the brakes are powerful enough to almost lockup.

 

If the car is heavily weighed down and the driver stands on the brakes but cannot bring them to the verge of locking up, then braking distance will increase.

 

Once the tires begin slipping at all, then µ_dynamic < µ_static, and deceleration is reduced, thus braking distance is increased.