Momentum vs Friction Problem

[Cor80]

I'm trying to get the following to make sense - I have 2 vehicles traveling at the same speed. both have 4 wheels. both don't have any special brake assistance. only diff is the weight. Vehicle A = 4x heavier than vehicle B. Now say they both do a full brake lock on all 4 wheels starting at the same line. Will they travel the same distance due to friction or will A travel further?

 

Does momentum play a sizable roll in the dragging distances?

[timo]

Since your problem is talking about forces (which you get from weight and friction coefficient) and distance, your best guess is to go via energy (force * distance) rather than momentum (force*time). It is possible to solve your question using momentum and momentum does -from the standpoint of modern physics- play the fundamental role in describing dynamics of a system. But in your case, you don´t really need to consider momentum (though you can if you want to).

[Cor80]

Since your problem is talking about forces (which you get from weight and friction coefficient) and distance, your best guess is to go via energy (force * distance) rather than momentum (force*time). It is possible to solve your question using momentum and momentum does -from the standpoint of modern physics- play the fundamental role in describing dynamics of a system. But in your case, you don´t really need to consider momentum (though you can if you want to).

 

OK...let me rephrase. Does the extra weight on A have an effect on the stopping

distance of the 2 vehicles if both have all wheels locked (max friction) ?

[timo]

Newtons Law: a = F/m

Frictional force: F = m*g*c, where c is the friction coefficient.

 

Shouldn´t be too hard to find it out from these two equations.

[Cor80]

thanx. if u substitute F=m*g*c into a=F/m then the m gets eliminated.

am i wrong to say that the mass doesn't effect the dist?

 

It still doesn't sound logic. a 10ton car stopping in the same dist as a 1ton car???

[Cor80]

maybe these things is not suppose to be logic?

[timo]

You get the distance d from E=F*d and the initial conditions E=mv² for both cars. Again, m will cancel resulting in the same distance. It does indeed not sound logical (when you use the term "logical" in the sense of "intuitive" - in reality, logic is something different) but it´s the correct result of the calculation.

 

I do not know to what extent a truck really stops as fast as a car, but in reality there are a few technical parameters (e.g. the tires) that need to be taken into account. They might make a difference, but I´m no car manufacturer, so I cannot judge that.

[Cor80]

Yes, again m is canceled.

 

ok, here's where I come from - Someone made the statement that 2 identical vehicles, identical velocities, with identical tires on identical surfaces with only the weight that's different will still get to a stand still at the same point after dragging their tires from the same identical start.

U see, only the weight can vary. no need for any specifications.

[Cor80]

I've tested this now.

 

I took 2 diff weights on 2 identical pieces of paper and slide them on the same surface with the use of only one hand (to cancel out that either one gets diff in force). Now this may not be super accurate but it does give an good example. In all the cases I got roughly 25% diff. That a continuous result.

 

The theory doesn't come to the practice here.

[Bignose]

But, this is exactly the same problem as dropping two objects weighing different weights in a vacuum. And that experiment was done in one of the few places it could be done on the moon. A feather and a hammer dropped from the same height on the moon fell at identical speed. And a feather and a hammer do not weight the same!

 

The experiment is the same as yours with only a few modifications. The initial velocities are the same, the initial energies are potential instead of kinetic, and the force is the same, g. The frictional force is modified by a frictional coefficient, but otherwise, they are exactly the same.

 

Your experiment with a paper and two weights is definitely flawed. 1) Using your hand it is impossible to give the objects the exact same initial conditions 2) Were both the objects the same in every way except mass? Different sized objects would behave differently. Heck, if they were different objects they would behave differently.

 

You should be able to try this out at home: Open your window and drop a basketball and a baseball. They should impact the ground as very similar times. You can use most any ball, golf ball, tennis ball, etc. While you are doing it, think of the story of Galileo doing pretty much that exact same experiment from the top of the leaning tower of Pisa. Then, try other objects. Think about why the astronauts could use a feather on the moon, but why that would not work here on Earth.

[Cor80]

ok - I give up (for now). My explanation is that although things is supposed to be calc theoretically it doesn't always seem logic in practise.

 

I'll try again later

[NeonBlack]

Your intuition is correct.

Normally, if two cars are travelling at the same speed, it will take longer for the heavier one to stop if they have the same breaking force.

Is this problem, the cars are not stopping as a car would normally stop. They are locking their wheels and stopping by sliding friction between the tire and road, not by friction in the brakes.

In this case, the heavier car will have 4 times more energy, but they stopping force will be 4 times greater and the stopping distance will be the same.

I think they reason you are having trouble with this problem is tat it doesn't exactly describe a real world example. If cars stopped like this, the tires would not last very long.

[Bignose]

Please try the dropping balls experiment on your own. You will be able to achieve much more consistent results. Gravity will accelerate the objects much more consistently than your hand will be able to. All you have to do is time it so that you drop both objects at the same time. While not easy, it is far easier than pushing two objects at the same time with exactly the same force. In fact, you can probably very quickly improvise a device that would allow you to drop two objects at exactly the same time.

 

You are right in that the theoretic calculations are not perfect. Even the dropping balls example is not perfect, since two different balls will have two different drags. However, the difference is drags is very, very small, you should see both balls hit the ground very closely to the same time.

 

To really show this using the experiment you were trying to perform, you need to build some device that will push your two objects at exactly the same speed. You must also have some method of verifiying that the two initial speeds were exactly the same, like a high speed camera or something. It can be done, but under more exacting and rigorous standards, your hand is not going to cut it.

 

Athiest brought up a good point, that just because your intuition doesn't like the result, certianly does not mean it is wrong. Your intuition has not been tuned towards solving physics problems, and, at least regarding solving physics problems, your intuition is unreliable.

[Jacques]

From wiki:

This approximation mathematically follows from the assumptions that surfaces are in atomically close contact only over a small fraction of their overall area, that this contact area is proportional to the normal force (until saturation, which takes place when all area is in atomic contact), and that frictional force is porportional to contact area. Such reasoning aside, however, the approximation is fundamentally an empiricle construction. Rather than a physical law, it is a rule of thumb describing the approximate outcome of an extremely complicated physical interaction. The strength of the approximation is its simplicity and versitility--though in general the relationship between normal force and frictional force is not exactly linear

Maybe the contact area is already saturated in the case of the ligther car...

The area of contact of the tire is small compared to the weight of a car , and ruber maybe is easily saturated...

[mysolipsis]

could possibly be that the difference in contact area is so small in comparison. when figuring out the contact area, things like the elasticity of the rubber which are steel belted radials (usually) and also how the air pressure get displaced. now it should make more contact area with the vehicle that weighs more, however, if the difference is so small it becomes almost negligible by comparison.

[spikerz66]

simpky put they will stop in the same distance bc the more massive cars' mass will be met with an equal amount of friction.

small car= less friction