which vehicle will stop first?

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Which vehicle will stop first

The heavier one
4
9%
The lighter one
35
76%
They'll stop at the same time
7
15%
 
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askvictor
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which vehicle will stop first?

Postby askvictor » Wed Jun 13, 2007 7:06 am UTC

Two otherwise identical vehicles, one carrying more weight than the other. Both hit the brakes at the same time. Which one will stop first?

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Postby Hawknc » Wed Jun 13, 2007 7:14 am UTC

The lighter one. F = ma, so assuming they both have the same braking capacity F, the one with greater mass will have correspondingly lower deceleration.

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Postby MalaysianShrew » Wed Jun 13, 2007 7:21 am UTC

See, you pulled out F=ma. I was gonna pull out ke=1/2*m*v^2. Since the heavier one has more kinetic energy it will take that much more energy to stop it. And since the brakes have the same stopping ability, it will take longer for it to impart that extra energy.

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Postby ArmonSore » Wed Jun 13, 2007 7:27 am UTC

Actually...

Forces in y direction:
N - mg = m*a_y = m*0
N = mg

Forces in x direction:
Friction = m*a_x
Friction = constant*N = constant*m*g
constant*m*g = m*a_x
constant*g = a_x

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Postby adlaiff6 » Wed Jun 13, 2007 7:59 am UTC

If they're skidding, they will stop at the same time. If they're rolling (but braking with the same power), the lighter one will stop first.

ArmonSore, your calculations are great as long as the normal force matters (which is when the friction is between the tyres and the road, when it's skidding). When the wheels are still turning, the friction is between the disc and the pads, so the normal force depends on how much pressure your foot is applying (which we were assuming was equal). The heavy one will start skidding first, but at that point everything goes to hell because coefficients of friction change too.

EDIT: What ArmonSore said is actually a pretty neat effect, and is the reason why roads can be angled one static amount around a curve and work for all masses of vehicle (within large enough limits, and assuming all other properties---velocity, tyres, etc.---remain equal).
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Postby skeptical scientist » Wed Jun 13, 2007 8:11 am UTC

But adlaiff, the amount of force one can put on the brake pads before the wheels go from turning to skidding certainly depends on the (static) friction between the tires and the ground, which is again proportional to the normal force, right? So if you are braking as instructed in order to stop as fast as possible, which is to brake as much as possible without skidding, the force between the brake pads and the wheels should be the same either way, no? That would tend to support the hypothesis that they will stop at the same rate.

However, this seems to be the type of situation where the nonlinearity of friction might come in to play, so I'd feel more comfortable with some experimental verification to back this up. Does anyone have a reference to actual experimental testing?
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Postby askvictor » Wed Jun 13, 2007 8:57 am UTC

in an ideal world I'da thought that they would stop at the same time provided that the wheels didn't lock (i.e. we are dealing with the static friction of the wheels rolling on the road rather than the dynamic friction of the wheels skidding on the road). I'm not certain what would happen if the wheels did lock.

Anyone have a car with anti-lock brakes who wants to test this out? Actually, a car without anti-lock brakes would be nice too for the other trial.

Now, should it be easier, harder or more difficult to lock the wheels while braking with the heavier load? After a certain point you would have to start thinking about how much heat the brakes can dissipate...

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Postby ArmonSore » Wed Jun 13, 2007 9:05 am UTC

Yeah, wheel turning is quite a bit more complicated, since static friction is variable. I wonder what one would do in that situation, as far as mathematics goes.
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Postby askvictor » Wed Jun 13, 2007 9:12 am UTC

ArmonSore wrote:Yeah, wheel turning is quite a bit more complicated, since static friction is variable. I wonder what one would do in that situation, as far as mathematics goes.


I'da thunk the opposite - as long as the the brakes are being applied at the maximum point before locking the wheel, the friction (x) force will be equal to the maximum possible static friction, which is proportional to N. So as long as you are holding the brakes just at that point (and the brakes themselves are capable of locking the wheel, but are not) the stopping distances should be the same. If the brakes are being applied at a lesser rate, then things change completely.

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Postby ArmonSore » Wed Jun 13, 2007 9:18 am UTC

I'm not understanding you. If the brakes lock the wheel, then won't we have skidding? This would take us back to the case where we have kinetic friction, which is constant, and takes us back to them both stopping at the same time.

Edit:
Sorry, I misunderstood you. I'd like to think that if they both have the brakes on maximum that they'd both have maximum static friction. But I'm not sure that's the case. I'd like to see some proof.
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Postby askvictor » Wed Jun 13, 2007 9:35 am UTC

ArmonSore wrote:I'm not understanding you. If the brakes lock the wheel, then won't we have skidding? This would take us back to the case where we have kinetic friction, which is constant, and takes us back to them both stopping at the same time.

Edit:
Sorry, I misunderstood you. I'd like to think that if they both have the brakes on maximum that they'd both have maximum static friction. But I'm not sure that's the case. I'd like to see some proof.


My proof is in wikipedia:
The static friction force must be overcome by an applied force before an object will move. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: Fmax = μN. It is important to realize that when there is no sliding occurring, the friction force can have any value from zero up to Fmax. Any force smaller than Fmax attempting to slide one surface over the other will be opposed by a frictional force of equal magnitude and opposite direction. Any force larger than Fmax, will overcome the force of static friction and cause sliding to occur. The instant that sliding occurs, kinetic friction is applicable, and static friction is no longer relevant.


So if any larger force were applied to the brakes, they would exceed Fmax. But a larger force is not being applied; exactly Fmax is being applied. Which is proportional to N.

Note that I'm not talking about the brakes being on maximum - that would result in skidding. I'm talking about they brakes being at the point before they lock the wheel.

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Postby ArmonSore » Wed Jun 13, 2007 9:50 am UTC

Ok, I understand your argument now. If the brakes are capable of causing skidding, they're capable of causing just less than that, yielding maximum static friction.

But what if the brakes aren't capable of causing skidding, and the tires are turning when we're not sure they're at maximum static friction. What happens then?
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Postby askvictor » Wed Jun 13, 2007 12:29 pm UTC

ArmonSore wrote:Ok, I understand your argument now. If the brakes are capable of causing skidding, they're capable of causing just less than that, yielding maximum static friction.

But what if the brakes aren't capable of causing skidding, and the tires are turning when we're not sure they're at maximum static friction. What happens then?


Then the lighter one would stop first. Actually, it would depend on whether the brake in the heavier vehicle was applied with a correspondingly increased amount of force; in that case they should still stop in the same time. If the same amount of brake force was applied to both, then the lighter vehicle would stop first.

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Postby ArmonSore » Wed Jun 13, 2007 5:20 pm UTC

Could you back that up with some mathematics?
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Postby Macbi » Wed Jun 13, 2007 5:37 pm UTC

Does the position of the weight in the car matter?
If it was above one of the wheels the friction would be greater there.

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Postby ArmonSore » Wed Jun 13, 2007 6:08 pm UTC

I've been thinking a bit about the problem. Rotation and movement at the same time always confuse me. Let's assume that the brakes are applied, but don't stop the tires from rotating. The tires must still be rotating in the same direction. Wouldn't this mean that they are still pushing back on the road, causing the road to push the car forward through static friction? So then I don't think that static friction can slow the car down. Could someone correct me on my logic here?
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Postby DonChubby » Wed Jun 13, 2007 6:12 pm UTC

The breaks apply friction to the wheels, making them rotate slower, and, eventually, stop.
Or did I miss something?
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Postby ArmonSore » Wed Jun 13, 2007 6:18 pm UTC

DonChubby wrote:The breaks apply friction to the wheels, making them rotate slower, and, eventually, stop.
Or did I miss something?


That's the rotational aspect of the problem, which I grasp pretty well. But when the wheels slow down, do they cause the road to apply a static frictional force opposite of the translational motion of the car? That's what I'm confused about.

If not then I'll assume that only air resistance and rolling friction are to blame for slowing down the vehicle.
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Postby askvictor » Wed Jun 13, 2007 11:08 pm UTC

ArmonSore wrote:
DonChubby wrote:The breaks apply friction to the wheels, making them rotate slower, and, eventually, stop.
Or did I miss something?


That's the rotational aspect of the problem, which I grasp pretty well. But when the wheels slow down, do they cause the road to apply a static frictional force opposite of the translational motion of the car? That's what I'm confused about.

If not then I'll assume that only air resistance and rolling friction are to blame for slowing down the vehicle.


Then how does any car slow down?

When a wheel is rolling, the surface of the wheel that is touching the road is stationary with respect to the road. That takes a little while to get your get around. But at any instant, the part of the wheel touching the road is not moving. And the top of the wheel is moving twice as fast as the vehicle is. Think about it this way: the car is moving at velocity v. The wheels are rotating; with respect to the car, the bottom of the wheel is going at velocity -v. But with respect to the road, the two cancel out, so the bottom of the wheel is stationary. Do you understand up to here?

Now, consider any instant in time. the thing touching the road (the bottom of the wheel) is stationary. So the only horizontal force is static friction, equal to (at most) kN.

i think.

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Postby ArmonSore » Wed Jun 13, 2007 11:39 pm UTC

@askvictor
I completely and totally understand that we're dealing with static friction. The wheels are rolling, pushing back on the road, and so the road pushes back on the wheels thanks to newton's third law. No big deal.

What I'm driving at is that when the brakes are turned on they will cause the wheel to slow down, but not to change direction. If the wheels are turning in the same direction they must still be pushing back on the road, causing the road to push forward on the car with static friction. Hence it seems to me that static friction isn't causing the car to slow down. The slowing must therefore be due to the effects of air resistance and rolling friction alone.

In other words I'm proposing that the car slows down only because static friction has been decreased, and dissipative forces are now greater than the static friction (which is still pushing the car forward).

But I don't trust that result.
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Postby gmalivuk » Thu Jun 14, 2007 2:11 am UTC

ArmonSore wrote:If the wheels are turning in the same direction they must still be pushing back on the road, causing the road to push forward on the car with static friction.


No, because the wheels are "trying" to decelerate, so the force they put on the road is forward, so the road pushes backward, slowing the car. This is why the wheels continue sliding forward on the road if they do lock up.

This question really depends on a lot of things not stated in the problem. So let's start out by idealizing the situation a bit.

If they're skidding, and the tires have the same coefficient of friction for both pressures, etc, then yes, they stop at the same time. Friction force is proportional to the mass, so deceleration is the same.

For rolling, let's suppose that the wheels and the road are simply incapable of slipping. Perhaps your wheels are actually gears and the road has corresponding teeth that fit into them, or something. Then the slowing force is the discs on the wheels. Ignore the actual mass of the wheels, and suppose equal force is being applied to the discs. This translates to equal friction force slowing both cars, so the lighter one stops sooner. (This is closer to what would really happen, since we don't like the wheels to lock as we lose control. If you're in a small car, you can stop if there's a cow in the road. If you're driving a road train, you have to hope the cow moves or at least doesn't come up through your windshield, since you're definitely not going to stop in time.)
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Postby adlaiff6 » Thu Jun 14, 2007 6:58 am UTC

skeptical scientist wrote:But adlaiff, the amount of force one can put on the brake pads before the wheels go from turning to skidding certainly depends on the (static) friction between the tires and the ground, which is again proportional to the normal force, right? So if you are braking as instructed in order to stop as fast as possible, which is to brake as much as possible without skidding, the force between the brake pads and the wheels should be the same either way, no? That would tend to support the hypothesis that they will stop at the same rate.

However, this seems to be the type of situation where the nonlinearity of friction might come in to play, so I'd feel more comfortable with some experimental verification to back this up. Does anyone have a reference to actual experimental testing?

Yes. The heavier vehicle can brake harder without starting to skid. I was assuming they were braking the same amount as well as being otherwise identical.

Hmm, I like this problem better. I think this would be modeled with the force of static friction between their tyres and the road, and therefore they would stop at the same time. Can someone confirm it for me?
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