wheel lock and skidding

[tris-]

for anyone in the know, could you possibly tell me how you can determine at what speed a cars wheels will lock, or if thats even possible. as ive been posed the following question -

"1 Determine the minimum stopping distances at all speeds up to the cars maximum when the cars wheels lock and the car skids to a halt on the road for the following conditions"

maybe im reading it wrong. imo its stating the cars wheels will lock at maximum speed, but then again i could be wrong.

 

[Kaun_IA]

for anyone in the know, could you possibly tell me how you can determine at what speed a cars wheels will lock, or if thats even possible. as ive been posed the following question -

"1 Determine the minimum stopping distances at all speeds up to the cars maximum when the cars wheels lock and the car skids to a halt on the road for the following conditions"

maybe im reading it wrong. imo its stating the cars wheels will lock at maximum speed, but then again i could be wrong.

if u dont have ABS then u can lock at quite low speeds... takes about 10 mph or so
the speed is kinda irrelavant.... couse the car so much energy when moveing u can lock easy.... tho i cant tell u the formula for the breaking distance bcouse i dont know it

 

[tris-]

i have the formula for the breaking distance. but my question is how can you determine, by calculation, if the wheels are locked (kinetic friction) or not locked (static friction). or if infact that is even possible. i have a theory that for the wheels to be locked then KE = 0. but rearanging my formulas i can never get a realistic number

 

[Kaun_IA]

i have the formula for the breaking distance. but my question is how can you determine, by calculation, if the wheels are locked (kinetic friction) or not locked (static friction). or if infact that is even possible. i have a theory that for the wheels to be locked then KE = 0. but rearanging my formulas i can never get a realistic number

i dunno that... what i just said was from personal experiance

i think cars breaks can be locked at eny speed... if u go too slow.. they will lock but you just wount skid. just stop right away

 

[Ingafgrinn Macabre]

kinetic friction between asphalt and rubber: μ_k ~~ 0.7
This is equal to the acceleration * 9.81 = 6.87m*s^(-2) (or, deceleration in this case)

(weight m of the car is 2000kg, the friction force is 0.7*20kN = 14kN
F=m*a
a = F/m = 14000N/2000kg = 6.867)

so the deceleration is 6.87 m*s^(-2)

[edit] fucked up there a bit hold on
[edit] there, it's okay now

 

[Neffneff]

wheels on a car can be locked at ANY speed, even when velocity = 0, i.e. you are stopped.

what you need to know is how much deceleration force is applied from the friction of the tyres on the road.

once you have that you should be able to work out at what speed a car of a certain weight will A. just stop on the spot when wheels are locked, or B. skid like a mofo with locked wheels.

im sure there's prolly loads more to it though.

 

[Ingafgrinn Macabre]

Misread your original question a bit, so gave a wrong answer before, or rather, it was correct, but not that which you needed. Therefor here below that which you did require.



A car of mass m [kg] is travelling at an initial speed of v0 [m*s^(-2)]. Its speed after breaking is vt=0 [m*s^(-2)].
The initial position of the car is x0=0 [m] and its position after breaking is xt [m]
The frictioncoefficient between wheels and road is μs for static friction, and μk for kinetic friction.

The maximum force F [N] the wheels can exert on the tarmac without starting to slide is m*μs*g [N].
Exceeding this force will result in the wheels to lose their grip. In case that happens, the exerted force will be m*μk*g [N] which is considerably lower.

The maximum acceleration (deceleration in this case) of the car can be calculated by reforming the formula F=m*a to a=F/m

Substituting F for the formula to calculate it, results in a=(m*μs*g)/m
Eliminating m results in a=μs*g

vt^2 - v0^2 = 2 * a * (xt-x0).
Since vt and x0 are both 0 they can be removed from the calculation, resulting in v0^2=-2*a*xt

a was calculated above, so substituting that in this formula results in:
v0^2=-2*μs*g*xt
However, we don't need v0^2, but would rather have xt singled out.
Therefor:
xt = (v0^2) / (-2*μs*g)

Also, since we're talking about a breaking force here, a will have to be negative, so we can remove the minus infront of the 2 resulting in the necessary formula:

xt = (v0^2) / (2*μs*g)

With that formula you can make a graph with the car's initial speed on one axis, and the travelled distance after maximum breaking on the other.

 

[tris-]

kinetic friction between asphalt and rubber: μ_k ~~ 0.7
This is equal to the acceleration * 9.81 = 6.87m*s^(-2) (or, deceleration in this case)

(weight m of the car is 2000kg, the friction force is 0.7*20kN = 14kN
F=m*a
a = F/m = 14000N/2000kg = 6.867)

so the deceleration is 6.87 m*s^(-2)

[edit] fucked up there a bit hold on
[edit] there, it's okay now

ive been given these coefficients.

Surfaces µ (static) µ (kinetic)
Tire on dry asphalt road 0.95 0.80
Tire on wet asphalt road 0.60 0.40
Tire on ice 0.30 0.15

sorry but it seems i may not have explained my self in good enough detail. what im asking is your oppinion of how the question reads. do you think its saying "work out the stopping distance, the µ and then determine the wheels have locked. or is it saying use the µ(kinetic) coefficients to work out the stopping distances.

although im sure the calculations you done there will come in handy, thanks

infact the brief states all the equations we need to use, and only those ones -

ke = 1/2mv^2

d = v^2(initial)/2µg

(d is stopping distance)

stopping distance with reaction time -

d = 1.5v initial + v^2 initial / 2µg

bear in mind i dont want anyone to actually work things out for me, the whole point of this work is i can gain skills in getting a problem and creating the right method to solve it. i am merely asking how you interpret the question and also possibly how youd know which coefficient to use.

 

[Ingafgrinn Macabre]

ive been given these coefficients.

Surfaces µ (static) µ (kinetic)
Tire on dry asphalt road 0.95 0.80
Tire on wet asphalt road 0.60 0.40
Tire on ice 0.30 0.15

sorry but it seems i may not have explained my self in good enough detail. what im asking is your oppinion of how the question reads. do you think its saying "work out the stopping distance, the µ and then determine the wheels have locked. or is it saying use the µ(kinetic) coefficients to work out the stopping distances.

although im sure the calculations you done there will come in handy, thanks

infact the brief states all the equations we need to use, and only those ones -

ke = 1/2mv^2

d = v^2(initial)/2µg

(d is stopping distance)

stopping distance with reaction time -

d = 1.5v initial + v^2 initial / 2µg

bear in mind i dont want anyone to actually work things out for me, the whole point of this work is i can gain skills in getting a problem and creating the right method to solve it. i am merely asking how you interpret the question and also possibly how youd know which coefficient to use.

"Determine the minimum stopping distances at all speeds up to the cars maximum" This to me says that they want a formula, or graph that sets out the speed of the car on one axis, and the resulting stopping distance on the other.

"when the cars wheels lock and the car skids to a halt on the road for the following conditions" so use μk apparently... hmm... did that wrong in above post ahwell, just replace μs with μk in all the formulae.

So, to me it says you'll have to determine the distance the car travelled from an ungiven initial speed with all the wheels locked. that locked wheels or not is an assumption you'll have to make at the start, because the frictioncoefficient is a given fixed factor. Cannot calculate that in this case.


"d = 1.5v initial + v^2 initial / 2µg"
This formula doesn't make any sense.... [m] = [m/s] + [m^2/s^2] / [m/s^2]
It cannot be resolved...

 

[tris-]

ah ha i see, this is what i thought and im glad someone agrees lol.

but here is another question, why is it that µk must be used? this is why im confused. for some reason i thought it needed to be used, but i dont know. is it because only µk will cause the wheels to lock? where as µs basically causes a 'wheel spin'. like if you pull away too quickly the wheels just spin and the revs shoot up?

 

[Ingafgrinn Macabre]

ah ha i see, this is what i thought and im glad someone agrees lol.

but here is another question, why is it that µk must be used? this is why im confused. for some reason i thought it needed to be used, but i dont know. is it because only µk will cause the wheels to lock? where as µs basically causes a 'wheel spin'. like if you pull away too quickly the wheels just spin and the revs shoot up?

The force of friction is dependent upon the weight that rests on the frictionate surfaces. the frictioncoefficient μ is actually something like a percentage number. The weight resting on the surfaces times μ results in the force of friction. Up to μs the surfaces will stay in fixed contact and will not slide. However, if you exceed that force, the static friction will transform into kinetic friction, which is much lower. That's the reason why locking up your wheels will make your breakingdistance a lot longer. Because the force available to break then isn't dependent on the higher μs but on the lower μk
μs doesn't mean wheel spin, it actually means that the contact surfaces are not moving in respect of eachother.

If you got a heavy book of 2kg on your desk, which you want to slide a bit to the side, you'll first have to overcome the static friction force (m*μs) untill it "comes loose" and then you'll be able to continue sliding it around with a much lighter force, the kinetic friction (m*μk)

 

[tris-]

AH HA. now its starting to make sense. there is two pictures to demonstrate µs and µk. it shows for µk that all positions of the tyre move at the same velocity. so to put that in my own words -
the wheel its self is not turning but being dragged by the car, as it where, like the book being dragged. so the wheel is not turning BUT its still moving along the surface so there is kinetic friction trying to stop it?

for µs it shows the same picture but at contact with the road the velocity is zero but the top of the wheel has a velocity of 2v.
now from youve explained, im assuming that static friction would cause the wheel to stop but also not allow the wheel to slid across the surface, therfore its static.

but how come the top of the tyre has a velocity of 2v and the bottom is zero v?

 

[Ingafgrinn Macabre]

AH HA. now its starting to make sense. there is two pictures to demonstrate µs and µk. it shows for µk that all positions of the tyre move at the same velocity. so to put that in my own words -
the wheel its self is not turning but being dragged by the car, as it where, like the book being dragged. so the wheel is not turning BUT its still moving along the surface so there is kinetic friction trying to stop it?

Exactly

for µs it shows the same picture but at contact with the road the velocity is zero but the top of the wheel has a velocity of 2v.
now from youve explained, im assuming that static friction would cause the wheel to stop but also not allow the wheel to slid across the surface, therfore its static.

Well, I guess you try to say it correct, but it's not causing the wheel to stop, it's causing it to decelerate. And yes, it's called static if the touching surfaces do not glide over eachother

but how come the top of the tyre has a velocity of 2v and the bottom is zero v?

Speed is relative to a reference frame. In this case the frame is the road, therefor when a round object with a dot on one point at the perimiter is rolling over this road, at a certain point the dot touches the road. At this moment the speed relative to the road is 0.
Rotate the object 90 degrees, and the speed of the dot will be upwards with 1v and sidewards with 1v. rotate it again, and it'll go only sidewards again, with 2v. At the last quarter the dot will be moving sidewards with 1v like in the first quarter, but now also downwards with 1v untill it'll stand still again at the moment it touches the ground. It's a sine function with the lower tips on the x-axis, and the highest tips at 2v

 

[GimmlyTheDorf]

Some serious physics going on this thread

 

[Ingafgrinn Macabre]

Some serious physics going on this thread

Well, kinda, but it's dynamics, and it's still pretty rudimentary (yeah, I know it can be tricky tris- but it can get a whole lot more difficult )

serious physics to me is more like stuff to do with particles and charges etc

 

[tris-]

ive done some a level physics last year but we never coverd friction. i only need to know the physics for this for one assignment. the rest of the physics im gonna learn is to do with physics in forensic applications (optics etc)


cheers for the help. youve explained it far better than the teacher did in the brief.

 

[Ingafgrinn Macabre]

You're welcome

And tell that teacher of yours that his formulae are wrong, making it nigh impossible for anyone to correctly solve the thingy
(really, teachers love to be corrected by students )

[edit] hmm... reading it over again, the formula ½*m*v² totally isn't necessary here. The last formula he gave is utterly wrong, so I guess those three formulae given are just to check if the students know what they're doing...

 

[Bugz]

Not taking a-level physics thx god xD

 

[Job]

Have you allowed for the change in the friction of a locked wheel as the rubber melts, of course temperature plays a big part for the road surface and also weight transfer to the front wheels...you could go on for ever

 

[Bahumat]

All i know is the wheels on the bus go round and round.

Not sure if thats any help?

 

[Ingafgrinn Macabre]

All i know is the wheels on the bus go round and round.

Not sure if thats any help?

Well, it is, because if they didn't, they wouldn't at all make their schedule and people would complain about not getting to work etc.
I also hope they go round and round and round and round and round and round and round and.....
cos if they only go round twice, you need reaaaallllly big wheels to get anywhere