Elementary physics question

[jasonpatterson]

Did you mean "does not affect the force", given that that's what your example shows?
Dave

Good grief. I need to stop posting in this thread, it is sapping my physics mojo... Thanks for the correction.

 

[X]

Hmmm... Well, I guess I was wrong about that aspect. Though given that we're talking about airplanes, where the normal force against the ground is a function of aerodynamics and not simply mass, we should probably be careful about assuming too much about how braking power scales with mass. But it still leaves the question of other forces (reverse thrust and drag from those flaps they stick up).

"those flaps they stick up" are called spoilers. They disrupt the airflow over the wing, which would normally provide lift. They are the reason the plane stays down after landing. By disrupting (or "spoiling") the airflow, they remove the wings ability to produce lift.
That's their main function. The spoilers will provide some drag, but it's not their primary purpose. If you want, I could probably dig out the equations to determine how much drag they produce, at different speeds.

Because of this, you probably could get away with assuming the weight of the plane as the normal force against the ground, and ignore the lift.
If nothing else, at least you'll be estimating a longer run than needed, putting you on the safe side.

I think we need to construct an experiment. Who wants to burn some rubber?

I'd love to. But my Lincoln doesn't do burnouts. The rear tires have so much traction that they push the front ones along (locked up). however, a little rain makes a difference. It's raining now...

 

[CORed]

The energy dissipation should not be the limiting factor for a decent set of brakes. Brakes on a consumer vehicle are engineered so they can easily apply as much force as we like, enough to lock the wheels and keep them locked at highway speed. I can lock the wheels in a Chevy minivan (~3000 kg) at 100 km/hr, so brakes should not be the limiting factor unless they are broken. The issue is the road-tire friction force. Tires do not have a constant coefficient of friction, it decreases as the load increases by a small amount. Breaking distances for heavier vehicles is maybe 20% more as a result.

Of course, I don't know what sort of breaks a 4000 kg car would have, let alone an 8000 kg one.

Regarding the 10 kg bowling ball, it should be pretty close to terminal velocity after 100m, so seeing air resistance shouldn't be a challenge. 100m is roughly 30 stories, an impressive height to be working with. Of course, a 100 kg bowling ball is pretty impressive too.

All of those weights are a bit out of whack. 4000kg and 8000kg is more in the range of a medium duty truck than a car. I'm not much of a bowler, but a quick look at some online sporting goods stores confirmed my recollection that bowling balls top out at 16 pounds (~7.3 kg), so even the 10 kg is on the heavy side. I don't even want to think about how sore my shoulder would be after a couple of games with the 100 kg ball, if I could even lift it, which I couldn't. Not more than once, anyway.

 

[Kuko 4000]

You've all been very helpful, I'm back tomorrow to make sure I've understood correctly, thanks!

 

[Kuko 4000]

Ok, let's see if I've understood the answers correctly. Then I can move on to trying to understand the reasons behind. I guess the first thing I need to do is to really understand the terms used here, such as dynamic friction force, kinetic energy and retarding force, I would appreciate some other pointers on where to start from.

Guybrush Threepwood has correctly pointed out that the air resistance is the same for both balls, but will decelerate the more massive ball less, therefore that ball will strike the ground first. However, over a 100m drop I suspect that the difference in impact times would be very small, and careful experiment design would be needed to measure it at all.

So, the heavier ball will land first. But the difference would probably be impossible to tell without the use of some tools. I had always thought that Galileo proved that objects fall at the same speed regardless of their weight. I would appreciate a correction for this as well.

My follow-up question is:

How high would the drop have to be for the difference to be apparent (I don't know the suitable word, but so that a normal person could see the difference reliably)? Or alternatively, how much heavier the other ball would have to be for the difference to be apparent when dropped from the original height of 100 m?

This one's more interesting. It depends on whether the brakes are applied hard enough to initiate a skid. If not, then GT is correct in that the force on the two cars is always the same, so the lighter car stops first, in half the distance of the heavier car. However, if the brakes lock the wheels of both cars, then their deceleration is caused by the dynamic friction force between the tyre and the road surface, which is directly proportional to the weight carried by the tyre; so in a skid, the force on the lighter car is only half the force on the heavier, and both will therefore stop in the same distance.

It's also possible for the heavier car to stop first. If the force applied by the brakes is greater than the limiting friction force between the tyres of the lighhter car and the road, but not greater than that between the tyres of the heavier car and the road (which is twice as large), then the lighter car will skid, but the heavier car won't. The lighter car will be decelerated by at most the dynamic friction force, whereas the heavier will be decelerated by up to the limiting friction force, which is greater than the dynamic friction force. Therefore, if the brakes are applied by the same amount on both cars, such that the braking force isn't quite high enough to cause the heavier car to skid, then the heavier car stops first.

Dave

So, the answer seems to be that if the cars skid, they will stop at the same time. If the cars don't skid, the lighter car will stop first, in half the distance of the heavier car. However, the heavier can stop first if the lighter car skids and heavier doesn't.

Phew. Thanks everyone. I hope I got the answers right, now to really understand the reasons behind.....

 

[sol invictus]

So, the heavier ball will land first. But the difference would probably be impossible to tell without the use of some tools. I had always thought that Galileo proved that objects fall at the same speed regardless of their weight. I would appreciate a correction for this as well.

They fall at the same rate if gravity is the only force acting on them. Gravity has a very special property - that its force is directly proportional to the mass of the object. Since F=ma, that means the acceleration due to gravity is independent of the mass of the object.

But the acceleration due to air resistance is not independent of the mass. If the balls are the same shape and size the force of air resistance doesn't depend on the mass at all, and that means it affects the heavy ball less (since a/F/m). Hemce the heavier ball lands first.

How high would the drop have to be for the difference to be apparent (I don't know the suitable word, but so that a normal person could see the difference reliably)? Or alternatively, how much heavier the other ball would have to be for the difference to be apparent when dropped from the original height of 100 m?

For a ball about 10cm in diameter, terminal velocity in meters/second (the speed at which air resistance equals the force of gravity) is very roughly 30 times the square root of the mass in kg. So for a 1kg ball it's about 30 m/s. The ball reaches 30 m/s in a bit over 3 seconds, during which time it would fall about 50 m. After that, the difference should be pretty obvious (because a lighter or heavier ball would rapidly fall behind or get ahead of the first). So for example it should be pretty easy to see the difference between a 1kg and 2kg 10 cm ball dropped from 100m (they'll hit the ground a half second apart, very roughly)..

 

[Ziggurat]

So, the heavier ball will land first.

Because of air resistance, which is frequently ignored for simplicity. In the absence of air resistance, they fall at the same rate. In fact, showing a feather falling in a vacuum is a popular intro physics demonstration, and was even replicated by astronauts on the moon.

I had always thought that Galileo proved that objects fall at the same speed regardless of their weight.

One of Galileo's early biographers claims he demonstrated this by dropping balls from the Tower of Pisa, but it is likely that Galileo never actually performed the measurement, but merely discussed it as a thought experiment. He did state that gravitational acceleration should be constant regardless of the mass of an object, as long as you ignore air resistance, and that's true, but like many of his ideas, he never rigorously proved it. He was often right, but not always (for example, he though the period of a pendulum was independent of its amplitude, but that's only true in the small-oscillation limit).

How high would the drop have to be for the difference to be apparent (I don't know the suitable word, but so that a normal person could see the difference reliably)? Or alternatively, how much heavier the other ball would have to be for the difference to be apparent when dropped from the original height of 100 m?

It's not simply a matter of mass, but also of density, since air resistance depends upon area. One way to look at the effect of air resistance is in terms of terminal velocity: if the two objects have very different terminal velocities, and the lower terminal velocity object has enough time to get to a large fraction of its terminal velocity, you'll notice a difference easily. If neither object gets to a significant fraction of its terminal velocity, or if the terminal velocities are very close together, it may be hard to detect any difference.