another academic question

[tris-]

afaik the stopping distance of a vehicle should increase with mass of the car.

i.e. a car of 1500kg should take longer to stop than a car of 1000kg.

problem is, on paper the mass seems to make no difference.

using the formula

umgd = 1/2mv^2

(u = friction coefficient)

d = 1/2mv^2 / umg
d = 1/2v^2 / ug

which shows mass will cancel out. except using only those formulas im supposed to show increasing mass will make a difference . is it even possible?

 

[noblok]

It's been a while since I've done any real maths, but aren't you supposed to get: d = 1/2gu(m^2)(v^2) ?

Since if 2x = 1/2, then x = 1/(2x2) and NOT x = (1/2)/2

Or am I looking at this all wrong?

 

[tris-]

not quite sure why you squared the mass there

1/2 mv^2 = umgd
Transpose:
2* (1/2mv^2)/ 2*(umg) = d
Cancel out:
(v^2) / 2*(ug) = d

this is what i get anyway. sorry forgot to cancel out the half before. but still, the mass is canceling out.

 

[Bugz]

The maths behind it is true. They cancel out & it's a bit of a 'wtf' as to why they do. However, I think when you use that formula, you take into mind the idea that the friction stopping the car is caused by the interaction between the car & the ground.

In effect, mass does not work this way and does affect the braking time.

As for finding a formula which takes into account mass - I have no fuckin clue.

I'll google it and see what I come up with.

 

[noblok]

As English isn't my mother tongue I don't know the English terms, so I'll try to explain it with an image .

Edit: I just noticed: in my first post the 'x' is supposed to be the letter X, the second 'x' (the one in 2x2) should be the mathematic symbol for multiplication.

 

[Bugz]

Okey, googled it a bit and came up with this formula (not sure if it'll be useful to you though):

s= 0.5
-------
-u^2m / F


In which case, s = stopping distance, u = initial velocity, m = mass & f = force.

That should give you a variable on the idea that mass affects the stopping distance, which will hopefully eventually end up as a proportional line in the sense that 's' and 'm' are directly proportional.

Unfortunately, that's all I could find in reference to equations involving the mass. The kinematics one (I think that's the one you're using) doesn't take mass into account.

Goodluck!

 

[crispy]

Well looking physically on the problem a you will get larger friction energy if your object has a larger mass, but also larger kinetic energy. so i guess for certain speeds and friction constants the friction energy would 'cancel' out the extra kinetic energy no matter what your mass is.

Might make it easier not to think of a car since i guess its friction constant is pretty low. Think if a box sliding on sandpaper instead

but i dont quite get what formula it is you got there :O

umgd = 1/2mv^2 ?

Seem something in the lines of friction energy = kinetic energy

so in thew special case when ugd = 1/2v^2 (which is also umgd = 1/2mv^2) then your mass doesnt matter (haah get it?).

 

[tris-]

@bugz : is that some variation of the suvat equation?

sadly it doesnt take into consideration the friction coefficient

thanks anyway. and noblok, i think there is some discrepancy with your math there. in that im not sure how you managed to square the mass

 

[tris-]

but i dont quite get what formula it is you got there :O

umgd = 1/2mv^2 ?

Seem something in the lines of friction energy = kinetic energy

so in thew special case when ugd = 1/2v^2 (which is also umgd = 1/2mv^2) then your mass doesnt matter (haah get it?).

not sure how it works either, but plug some numbers in and it works out.

m = 1000
u = 0.80
d = 57.33944954
v = 30^2

should get 450,000 for each formula.

 

[noblok]

Because you've got one m in the first fraction and one in the second. As you have to multiply them m*m gives m^2.

 

[tris-]

Because you've got one m in the first fraction and one in the second. As you have to multiply them m*m gives m^2.

ah ha. you dont multiply them. the formula need transposing. so you gotta move umg over to the left (so you gotta divide both sides by umg)

 

[Bugz]

is that some variation of the suvat equation?

sadly it doesnt take into consideration the friction coefficient

thanks anyway. and noblok, i think there is some discrepancy with your math there. in that im not sure how you managed to square the mass

I can't find anything that uses mass and the things you want.

All I know is that the mass is considered to be evened out by the other things in the formula (I always believed it was the friction but i'm reading here that it's caused by the inertia & the deceleration force).

Whatever the case, it's a fkin confusing topic...

 

[noblok]

Yes, but in transposing it you have to multiply 1/(2mv^2) with 1/umg. Since both m's are in the lower part of the fraction they have to be multiplied and they don't cancel eachother out.

2X = 1/2
X = 1/2*1/2 = 1/4 and not X=(1/2)/2 with the 2's cancelling eachother out.

Or did I completely misread the original formula and was it 0.5mv^2 instead of 1/(2mv^2)?

 

[tris-]

Or did I completely misread the original formula and was it 0.5mv^2 instead of 1/(2mv^2)?

sorry ye its 0.5mv^2.

 

[crispy]

Well i can't really help much if i dont know what dgu is but both sides of the equation is proportional to the mass of the object. So in the special case you got there where udg = 1/2v^2 then the mass doesnt matter.

But this isnt the case for your example with your car. You car might look like this udg < 1/2v^2 so there the stopping distance yould increase. But if you slided your box across the floor of sandpaper then udg = 1/2v^2, or if you glue something to the floor then udg > 1/2v^2....

So in conclusion, its only true for certain obects at certain speeds that the mass is not in the equation :>

 

[Ezteq]

dunno about that but ez @ 11.5 stone took longer to stop wobbling after doing the butt dance than Ez @ 10 stone, the total time to wobblage ratio has approximately cut down by about 21.43 minutes.

 

[tris-]

tHERE IS NO TIME FOR COMIC RELIEF

 

[noblok]

Are you absolutely sure mass should make a difference? This seems to imply that your formula is correct and mass does not make a difference.

 

[tris-]

this is the problem you see. on paper it shouldnt make a difference but im pretty sure more mass = longer to stop

 

[old.Tohtori]

tHERE IS NO TIME FOR COMIC RELIEF

Then i shouldn't drop my pants and show my willy?

 

[UrganNagru]

Looked at this and at first thought there must of been a mistake in the maths, but no you've got it correct. The only explanation I can give is that its because the equations is to simple to take into account energy lost due to friction in the car itself and possible sources of potential energy stored in rotating masses such as in the wheels and the flywheels cranks etc not really sure if that would make a susbstantial difference but its the best I can think of at the momment.

How ever i think if you use the more simple equations of motion you can prove mass would affect stopping distance, as F=ma acceleration can be replace by f/m. Plug that into V^2=U^2+2AS to get V^2=U^2+2(f/m)S. Doing that the mass can't be canclled out so mass must affect stopping distance. However other than the crappy reason I gave earlier can't really explain why the energy method wasn't working.

 

[tris-]

using suvat it would make a lot more sense. but apparently all the required formula are what ive listed about :\

 

[UrganNagru]

ah reet so you've been told to calculate the braking distance from just the cars mass and velocity?
If so I'll try messing about with the numbers a bit more n see if I can come up with owt.


Edit: Just had an idea do you know if the car is traveling at a constant velocity, do you know the force being exerted by the car at this point (yes I remember you post a week or so ago something to do with cars and force hoping its relevant to this) and can we ignore air resistance? If so it might work .

 

[tris-]

i have created a formula to check stopping distances and take into account mass -

stopping distance = ((0.5*mass*velocity)/(u*mass*9.81)+(1.5*velocity))

that end bit is to 'add a thinking' distance.

we pretty much have to work out the d under different friction co-efficients and velocity and state if added mass changes the distance. this is done in excel which i have placed a field where you can alter mass on the fly.

the only formulas we were told to use were umgd = 0.5mv^2

re arranged can give - d= 1.5V+ (V^2/2ug).

also as background info it gives formula for friction force (umg) and normal force (mg). cant see much use for those though (except umg, which i arrived at on my own before i noticed it was stated :\)

 

[tamtap]

afaik the stopping distance of a vehicle should increase with mass of the car.

i.e. a car of 1500kg should take longer to stop than a car of 1000kg.

problem is, on paper the mass seems to make no difference.

using the formula

umgd = 1/2mv^2

(u = friction coefficient)

d = 1/2mv^2 / umg
d = 1/2v^2 / ug

which shows mass will cancel out. except using only those formulas im supposed to show increasing mass will make a difference . is it even possible?

mass allways makes a difference.

Your question bears no relation to this though. e.g. take a 1 ton car and put 3 foot wide breakk pads on it and it will stop_much_ sooner than a car with standard discs/pad.

Go back to school and tell your teacher instead of diddling the art teacher to frame a real question.

Tam

 

[UrganNagru]

Bugger in that case I'm stumped mate, was hoping you could of use the max engine force from the question you had the other week. That way you could just put the frictional force equal to that so 1/2mv^2=fd. But thats not the case and it makes quite alot off asumptions to simplify things too.

 

[UrganNagru]

mass allways makes a difference.

Your question bears no relation to this though. e.g. take a 1 ton car and put 3 foot wide breakk pads on it and it will stop_much_ sooner than a car with standard discs/pad.

Go back to school and tell your teacher instead of diddling the art teacher to frame a real question.

Tam

Your right, unless the 3 foot break wide break pads were made out of jelly. Though the car would be travelling pretty slowly considering it would running on its brakes disks which would have to be bigger than its wheels to makes the most of those pads.

 

[Everz]

ignore.

 

[tris-]

Your question bears no relation to this though. e.g. take a 1 ton car and put 3 foot wide breakk pads on it and it will stop_much_ sooner than a car with standard discs/pad.

Go back to school and tell your teacher instead of diddling the art teacher to frame a real question.

Tam

my question bears no relation to putting bigger break pads onto cars.

im also lost on this school and art teacher shit.

i wonder why the fuck they ask if mass makes a difference when its obvious that from the formula given, it will cancel out.
its either a trick question, or somehow they can all merge into a super mega formula which makes mass count :\

 

[Everz]

my question bears no relation to putting bigger break pads onto cars.

im also lost on this school and art teacher shit.

i wonder why the fuck they ask if mass makes a difference when its obvious that from the formula given, it will cancel out.
its either a trick question, or somehow they can all merge into a super mega formula which makes mass count :\

trick question as far as i see, literally tried everyway to rearrange the formulae without the m canceling and thus so far seem impossible.