A bike going down a hill would usually reach terminal velocity quickly, and therefore it’s a useful illustration of why a heavier bike rider descends faster. But this approximation is not strictly necessary.
We’re not talking about riders in vacuum.
No. Just because two effects are numerically identical does not mean you can ignore both. The identical force has a more significant effect on the smaller object.
Because one is pulled downward more strongly than the other.
I think you are confusing this greater force with greater acceleration. Acceleration here is due to gravity and is constant. Therefore velocity can be calculated given distance and time, and is likewise unaffected by mass. Mass is playing a part, but you are miss-applying Newton’s second law.
Let’s look at it: F = m * a, where (a) is going to be the gravitational constant (g).
F = M * g
Pick any mass you like, as large as you like, and you will see F increase and balance the equation, but (g) remains constant. A bigger (F) only equals a bigger (a) if 1) you hold mass constant, and 2) we aren’t talking about gravity, where (g) is constant by definition. 9.8m/s/s
Mass factors in as more of a first law consideration. A [heavy] body in motion tends to stay in motion [more so than a lighter body.]
So yea. I do dissent. I’m am correct, and you are wrong.
Is not that same sphere pulled downward more strongly in a vaccuum?
Tuck, for the case of two unequal weight spheres falling through air: the air resistance will be the same, but the force of gravity will be higher for the heavier sphere because it’s, well, heavier. Gravity is also higher in the case of the vaccuum, but the difference with air is that the force due to air resistance will be smaller in proportion, for the heavier ball vs. the smaller ball.
For the question of heavier bikes, it’s the same thing, and I’ve seen it many times. On organized distance bike rides, the riders tend to naturally sort themselves out by average speed over the first few miles (the people riding around you are the ones who go about the same speed as you, since you all started about the same time). When I’m near a tandem bike, they go much faster on the downhill slopes because of their weight. Of course, I go faster on the uphills (since our average is the same and they’re faster downhill, it follows that I’m faster uphill). I’ve also seen it simply with heavier riders, but the effect is less noticeable.
Is it not also pulled downward more strongly than the other in a vaccuum?
CurtC – if you put two of you on a tandem, you should go up hill faster.
The people weigh twice as much.
The bike weighs less than twice as much.
Your power output should be doubled. Or at least close to doubled, withing the limitations of the tandem drive train.
Going UP hills is TOTALLY about power to weight ratio. And the amount of power that an “average” heavy rider generates doesn’t make up for the extra amount of power he generates over the “average” light rider.
Exceptions abound.
Totally anecdotal info to follow:
I ride a road bike and a tandem. Combined weight on the road bike is around 220 pounds; on the tandem, 360.
When on the tandem, we engage in “tandem leap frog,” where we are passed by single bikes going up, and we pass them going down.
While on my single bike, I have to spin out to reach the low 40 mph range while descending a smooth, straight 7% hill. However, on the tandem we coast into the high 40s, when the “automatic speed govenor” (my wife) kicks in and we start to brake–and are passed by other tandems that get into the mid 50s.
On long gradual descents, where we can tick along easily at 38mph, we seem to pick up a long train of single bikes who tuck in to draft us after we scoot by them.
As I said, anecdotal. I’ve always assumed that the greater mass was better able to overcome the air resistance, but I was a psych major, y’know.
Actually, no. scr4 is correct. He’s talking about forces due to air resistance.
To simplify the problem, think about two blobs that are falling through the air. The blobs are the same size and shape, but different masses – call the masses m and M. Each blob has two forces on it: the drag force (F[sub]D[/sub]) and the body force due to gravity. If the blobs have the same velocity, the drag force is the same for each. The body force, however, is different. In one case, the body force is (mg), and in the other it’s (Mg).
So the total force on each blob is different also: In one case it’s (mg-F[sub]D[/sub]), and in the other it’s (Mg-F[sub]D[/sub]).
Since acceleration is equal to force divided by mass (F=ma, right?), the acceleration of the two blobs is different. a[sub]m[/sub] = (g-F[sub]D[/sub]/m) and a[sub]M[/sub] = (g-F[sub]D[/sub]/M). In the absence of air, the drap force is zero, and the acceleration of each simplifies to just g, as you would expect. But with air resistance, the acceleration is not constant, and it’s not the same for the two blobs.
Perhaps I did misundertand the post, but even as I read it again, I see someone correctly identifying that force increases with mass, but extending this to mean that this same increase in force has caused the higher velocity. Have a look:
Air resistance is mentioned, but only but only in terms of it not increasing as quickly as what seems to me the erroneus attribution of accelaration due to mass.
I agree with the remainder of your post, since it is clear your F=ma is in the opposite direction to your F=mg. My apologies to scr4 if I have continue to misundersand his/her meaning.
I’m 99% certain that scr4 is saying exactly the same thing that I am. The key is this little bit, from what you quoted: “the heavier object has a larger mass-to-surface-area ratio.” His post is cast as a description of the scale effect. So he’s not saying that the higher body force causes a higher velocity per se, but the rather that the body force is proportionately larger than the drag force, and thus results in a higher velocity.
Or, to cast it ina different light, remember that I was claim above that the total acceleration on a body is a[sub]M[/sub] = (g-F[sub]D[/sub]/M). Well, F[sub]D[/sub] is proportional to area, so when scr4 talks about the “mass-to-surface-area ratio,” he’s talking about F[sub]D[/sub]/M. In other words, as you scale a blob up, the mass-to-surface-area ratio changes, which means that F[sub]D[/sub]/M changes, which means that a[sub]M[/sub] changes, which means that the maximum velocity changes.
scr4 is indeed correct. I think at this point it is useful to get some forumlas on the table.
I am assuming that the wheels roll without slipping, the bearings in the bike are frictionless and the biker and bikes are identical except for mass.
For you playing along at home draw your best representation of a biker riding down a hill (don’t draw the hill though) and a similair one next to it.
Got that?
Ok good now lets start adding forces to the one on the left. First lets add the force due to gravity which is massacceleration due to gravity pointed straight down. Next add a an air resistance force parallel to the hill pointing up the hill. There is also a friction force but assuming we have the same bike it is inconsequential so we can ignore it. Same goes for the Normal force from the hill onto the wheels.
Now draw an = in between the two pictures and draw a force parallel with the hill pointing down the hill.
(waits patiently)
Ok good we have now drawn our free-body diagram with all forces. Let us write the equation for the force parallel to the hill with down positive. From our left picture we have massacceleration due to gravitysin(theta) - Force due to air resistance and from our right we have massacceleration of the rider. Putting these together we have mgsin(theta)-Far=ma where m=mass, g=gravitational acceleration Far=force due to air resistance, theta=angle of the hill and a=acceleration of the body
Now Far is governed by the equation Far= 1/2CpAv^2 where C=a coefficient dependedant on the material of the object i.e. rough objects fall slower than smooth objects, p=density of air, A=crossectional area and v=velocity. Now in our situation here we can ignore p,C and A becuase they are the same.
Our equation now looks like mgsin(theta)-Zv^2=ma where Z is some constant that equals 1/2CpA. solving for a we get gsin(theta)-Zv^2/m=a. The greater the mass of the object the smaller the term Zv^2/m is and consequentially the higher the acceleration of the biker. In this case the acceleration for both bikes is equal (g*sin(theta)) but the acceleration due to air resistance depends on the mass.
Lets go ahead and draw another free body diagram for this situation. Draw two balls and the one on the left draw an arrow straight down and label it mg and on the right ball draw an arrow straight down and label it ma. Writing the equation down we get mg=ma. The masses cancel out so the acceleration of the object is equal to g. However the value of the force is equal to mg so it depents on mass. If you apply the same force to say a couch and a empty cardboard box you of course know that the cardboard box will accelerate much faster than the couch.
Now in a case with air resistance you need to add a Far to the left ball. Using the same reasoning as we did with the bike this force is equal to Zv^2. Writing Newtons 2nd law we get mg-Zv^2=ma. Solving for a we get a=g-Zv^2/m. Again the Zv^2/m term depends on mass so the two objects accelerate at a different rate. As before the acceleration due to gravity is the same for both balls but the acceleration due to air resistance is different.
Hope that helps.
Waverly, you are ignoring air resistance. This is not a valid approximation for this problem, and therefore your answer is incorrect.
Yes, I undersatnd F=ma and F=mg. Acceleration is a=f/m=mg/m=g, which is independent of mass. Which is why everything falls at the same rate - in vacuum!
Now add air resistance, so that net force is F=mg-r(v) where r(v) is the air resistance of the object at speed v. Then acceleration is a=f/m=(mg-r(v))/m=g-r(v)/m. Do you see that if two objects had the same r(v) but different m, they would accelerate at different rates?
Or to put it in English: the same force acting on a more massive object has a less significant effect.
No, no I am not.
I can see now we are saying the same thing. Your earlier post just wasn’t clear to me.
I can also see that my spellcheck isn’t working, so maybe my posts aren’t much clearer.
Trunk, You want authoritay? The authoritay of this Ivy-league physics degree (and not from some afraid-of-numbers-school like Brown, either) says scr4 is right. You want, I’ll get my father’s PhD in physics to weigh in. So respect!
And the tandem slower-uphill faster-downhill phenomenon is most likely explained by the fact that a tandem is not that much lighter than two single bicycles, but has only a tiny bit more air resistance than one single bicycle.
Therefore going uphill, where it’s slow enough that air resistance doesn’t matter much, and it’s all about pushing mass against gravity, the tandem has little advantage. But going down, where it’s mostly about air resistance, the tandem has a huge advantage (for the same reason that the cannonball falls faster than the nerf ball).
That’s exactly what I wanted.
Someone dropping PHAT IVY on my ass and jamming the physics degree up in my grill!
Argument by authority sucks!
If you read CurtC’s post a bit more closely, you’ll see that he’s comparing tandems and singles with the same average speed, so no explanation is necessary. Probably the single riders are just stronger climbers.
I’m confused as to what your position is. Are you still claiming that even if you take air resistance into account, two objects with identical size and different mass will fall at the same rate? Or are you retracting your earlier accusation that I am incorrect?
Holy shit. I’m beginning to wonder, typos aside, if I’m conversing in English. I said, and I believe a quick perusal of my posts will bear this out, that:
Sorry. I’m not capable of stating it more clearly.
That’s clear enough. And your point 4 is a good one. On the other hand, your earlier assertion that “traction is higher for the heavier rider” is questionable.
This topic, or at least some variation on the “do heavier things fall faster” argument comes up about once a fortnight around here. The suggestion that in air things of the same size fall at the same speed regardless of differing weights is one that an amazing number of people will fight to the death to defend. Die they always do eventually, but it’s a remarkably persistent meme.
What I find interesting is that it tends to be people who have at least 8/10ths of a clue who are dead wrong about it (see below). They have enough physics to have heard the “all things fall at the same speed in a vacuum” rule and they apply that willy nilly. Another interesting thing is that while it is common for uneducated people to apply common sense and ignore counter-intuitive science, it is so much rarer to see educated people insisting on a wrong answer as a result of misapplication of counter-intuitive science in defiance of common sense.
You can’t be too sure. I once had dinner the night before a trial with an expert witness (a Phd in fluid dynamics, fer chrissakes) and one of my hobbies (water rockets) came up and we got onto this whole topic and he insisted that all else being equal, heavier things don’t fall faster in air. Eventually I managed to convince him but not before it got down to writing formulae on napkins.
He was not at all happy about having his ass handed to him in something so close to his own field by a lawyer.
Physics degree? Pishaw. I am going to throw my PhD in Aerospace Engineering in here to say that you are… umm… absolutely right Zut. Your equations are also the best mathematical explanation in this thread.
I am also going to expand this to say that exactly the same thing applies to two uniform cylinders of different density (and therefore mass). The fact we are creating both rotational and translation motion comes out in a wash if the cylinders have the same dimensions.
One last consdieration: The heavier rider, though still at an advantage downhill, will be at less of an advantage than this initial glance would suggest. You see, most people become larger with increasing mass, and have correspondingly larger drags. Drag is extraordinarily difficult to determine mathematically, but suffice to say that the big guy’s edge will be slightly reduced.