Conceptual Physics: I
You push a heavy car by hand. The car, in turn, pushes back on you. Doesn't this mean that the forces cancel each other, making acceleration impossible?
Why or why not?
A. Yes: In accord with Newton's second law, a = m F net , and since F n e t = F y o u + F c a r = 0 , no acceleration occurs.
B. Yes: In accord with Newton's first and third laws, an object continues in its state of rest unless it is compelled to change that state by forces impressed upon it. Since the net force is zero, the car doesn't move.
C. No: In accord with Newton's third law, I push the car and the car pushes with an equal and opposite force on me. Thus, with Newton's second law and a = m F , the car accelerates, but so do I in an equal and opposite direction.
D. No: An external, horizontal force is applied to the car by my push; the car accelerates (in accord with Newton's first law).
Note:
By acceleration we mean acceleration relative to the ground.
Problem credit: Conceptual Physics: Special Edition Series, Paul G. Hewitt
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7 solutions
I was confused with C and D.The weak point in C was that for both the bodies i.e. the man and the car you wrote one eqn for accln, but there should be two i.e. one for the car and one for the man. Because nothing was said on mass of the bodies so I clicked on D.
But if anyone assumes that mass of the man equals the mass of the car then we got to answer to this question C & D.C being more precise.
What do you say @John Muradeli & @Michael Mendrin
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Since the question concerns itself with the movement of the car only, our expedient is the equation for the car's acceleration only. C wrongfully expresses Newton's Second Law by stating F instead of F_net. And the mass is still irrelevant to the purpose of the problem, because unless the car is infinitely massive (or the man), some sort of acceleration must exist - which just turns out to be the answer to this problem.
I'm glad you're interested in my problems!
Cheers
Share more problems of these kind , very interesting ones
It's a very good way of presenting clear,crisp and informative....Thank you!!
If we look at the entire system, which includes the car, myself, and the planet I'm standing on while I'm pushing the car, then as the car accelerates in one direction, the entire planet Earth is accelerating in the other direction. Hence, the forces balances. To make this point a bit easier to understand, imagine that this is happening on an asteroid, and I'm pushing the car away from it. What happens to me, the car, and asteroid as I push?
Saying that an "external horizontal force is applied to the car" is a way of saying, "please disregard the other half of what's going on". But in fact, a lot of engineering is done this way anyway, for reasons of expediency.
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Well yes of course the Earth is also affected but you can go as far with mechanics analysis as bringing the entire quantum uncertainty principle or the entire universe into this. I took a simplistic view to keep the solution relevant to the problem, and also to keep it Level 2.
And OOOOH MICHAEL GOT MY PROBLEM WRONG!!! Lol jk. What did you answer, anyway?
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I picked C, "no, it wouldn't make acceleration impossible" I know this for a fact because I've pushed my own car a few times too many. I knew that D would be equally valid, but I wanted to see your explanation. What's in your favor in this argument is that it is not I, solely, that is being accelerated, It's me and the entire planet Earth, which is why I don't seem to be going anywhere when I'm pushing.
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@Michael Mendrin – Hm I see what you're getting at. But there were two flaws with that answer: First of all, acceleration equals NET force over mass, not just force. This is HUGE! Second, I do not accelerate in an equal and opposite direction - first of all that doesn't even make any sense. But we know what I meant, and we're still wrong: since I'm pushing against the ground, the vector forces would not allow an EQUAL and opposite acceleration, even though there would be a reactant acceleration
But knowing you missed out on the formula, I see you zapped through the answer. It's ok lol watever - the main thing is you know your stuff. Btw check out my updated solution - AIN'T IT COOL?? I just got a great inspiration to make more of these bad boys! All thanks to you Michael.
Cheers, friend
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@John M. – It's the forces that are in equal and opposite directions. But because of the combined mass of myself and the planet Earth, the acceleration that I (and Earth) will experience in my pushing the car is going to be extremely tiny as compared to the acceleration of the car being pushed. This is looking at the problem from the rest frame that is defined by the center of mass of the entire system, which includes the car, me, and Earth.
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@Michael Mendrin – What? Who turned off the sun in the forest? I'm lost.
COMBINED MASS OF EARTH AND YOU? How does that work? Do you treat yourself and the earth as a single object? I don't think you can do that. You move independently of the Earth, and so does it. And the reason why the Earth moves so miscorscopically has to do with the fact that the ratio of force impressed by you upon it to its mass is infinitesimal:
. = M ,
(, is force , . is acceleration, expediently emphasized).
There are a total of 4 interactions (that matter): You v. Car; Car v. Earth; Earth v. You; Car v. Earth v. You. Each applies in reverse also. The last one is probably the one you have in mind: combined object interactions: Net Vector Force of Car+You on Earth and You+Earth on Car. But what Center of Mass has to do with this I have no clue.
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@John M. – Let's try to look at this in another way. Instead of my arms and legs pushing the car, we'll have a giant compressed coiled spring between 1) the car, and 2) the planet Earth. The spring forces 1) and 2) apart. The force it exerts on both 1) and 2) is exactly the same, which we'll call F. Let m = mass of the car, and M = mass of the Earth, and a = acceleration of the car, and A = acceleration of the Earth. Then F = ma = MA, and so we can see that A<<a. Thus, the car accelerates in one direction away from the car-Earth center of mass, while the Earth accelerates in the other direction away from the car-Earth center of mass. Meanwhile, because Earth is barely moving at all, the CENTER of the spring itself will appear to be moving in the same direction of the car! However, the center of mass of the spring-Earth system still moves in the direction away from the car. Work was done in sending both the car and the spring+Earth systems apart with the same resultant kinetic energies. Think of a huge cannon that recoils as it shoots a shell at terrific speed---the energy of the gunpowder is converted into kinetic energy of both. The same thing applies if the cannon used a giant coiled spring instead of gunpowder.
The whole confusion about this problem stems from the fact that I, in pushing the car, am not a simple solid body, but actually behave as an elastic solid with a changing shape. Physicists then often resort to using center of masses because it's easier to work out things that way, and look at things from work and energy point of view.
This also assumes that the spring is attached to the Earth, i.e. isn't free to bounce from it.
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@Michael Mendrin – Hmph yes indeed you aren't a simply solid body, and I think I like the center-of-mass approach. I see there's one use to this method: precision. But, there's a big hole: the center of mass of the Earth is highly unstable. The fluid motions on the surface (oceans, seas, rivers, animals, etc) and the underworld (lava, tectonic plates) make the center of mass of the Earth far more non-constant than the simple minor setbacks made by the fluid motion inside your body.
Besides, all of this can be explained with the three laws, even though superficially, sufficiently. You push the car with a horizontal vector, the car pushes back on you with a horizontal vector, you push the ground with a perpendicular vertical vector that also contains a horizontal vector, and with parallelogram rule results in a diagonal vector, and thus you accelerate a bit backwards due to this vector mishap. And one thing's for sure: a = m F .
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@John M. – Let me make one last comment on this subject. We can imagine another setup where on a frictionless track we have the following (left to right): 1) a very heavy mass 2) a light spring with a non-zero mass, and 3) an intermediate mass. Initially, the spring is compressed, and all three are in contact with each other. The spring is released. What happens next? While 1) will move imperceptibly to the left, 2) will move to to the right, the spring itself 3), if free, will also move to the right. What can be said about this setup is that momentum is conserved, and energy is conserved, so that the center of gravity of the system does not move, but it becomes messy physics to actually work out the final velocities of 1), 2), and 3).
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@Michael Mendrin – Hm Alright Michael. To be honest, the first time I read through the book I zapped past the chapter exercises all the way till Light Quanta, and realized I was really confused. So I started over, and now I'm on Chapter 7 - Energy, having encompassed Newton's Three Laws, Linear Motion, and Momentum. I did all the exercises and feel much more knowledgeable. However, now I gotta take care of other, non-optional subjects in my school, and no time. I didn't get to center of gravity yet, but when I do, I shall return to your comments.
Thank you and sorry for giving you a hard time. Peace.
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@John M. – Oh you can give me all the hard time you want. Peace.
@Michael Mendrin – Either way it seems a lot of people are getting this wrong: only about 1/3 are right. Seems I got myself a good trick question :)
But really, the true problem with C is the lack of n e t on F . The second part of the argument is alright, but D does the best job.
But thanks for the asteroid example - it just inspired a problem for tomorrow. I'll be sure to quote you.
Cheers
But the forces can never cancel each other as they are acting on two different objects.
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Hehe this is about as short of an answer as you can give. But to someone just finished with Newton's Third Law, this is as much of a cancellation as -5 and 5. It's a common misCONCEPTION that people have. And that's not just me - this problem has a 142:55 unsolved to solved ratio.
Cheers
I don't like question even though I got it right because it's not clear what it's getting at. If I have a firm foothold on the ground, then I can accelerate the car as in D, and by the way, this is Newton's second Law. If I don't have a firm foothold, then I will accelerate in the opposite direction and my acceleration will be far greater than the car's because I have a much lower mass, as anyone who has tried to push a car that's stuck on ice for example will know.
D is the solution Explanation: external f is applied to car. Car also pushes back. But with help of friction we add the mass of earth to us. Thus their is conservation of momentum as well. We also move backwards but because we have added earth's mass to ourselves using friction, our acceleration is less while Car's is more and we see car moving ahead
By Newton's 3rd law, if object 1 exerts a force on object 2, there's an equal and opposite force exerted by object 2 on object 1. In this case, the net force upon the car is the EXTERNAL force applied by the person. Acceleration would be impossible if the external forces applied upon the whole SYSTEM (person and car) cancel out.
That is a nice way to summarize it ;)
The force pushing back on you is inertia, the reaction to acceleration of a mass. So of course the car is accelerating - if it wasn't, there'd be no inertial reaction.
Tricky part of the question is this "Thus, with Newton's Second Law, and , the car accelerates, but so do I, in an equal and opposite direction". I'm not accelerate with an equal magnitude because is necessary count friction force. Also, if we don't count friction force, my acceleration will be m m y m a s s F = a m i n e and car aceleration will be m c a r m a s s F = a c a r so both acceleration are diferent.
Even though it contradicts logic and everyday experience, the answer that the net force is zero and thus no acceleration occurs (A) seems quite reasonable. But that's because we're missing an important tool in our analysis: systems.
A system is a defined set of objects; it's of your own making. For example, we can have a system of a train and a passenger. In this system, no motion occurs (assuming the passenger is seated). However, bring the rail into the system, and both the train and the passenger are moving with respect to the rail. A system may be as tiny as an atom or as large as the universe.
So, back to our problem. Here, the acceleration that occurs is with respect to the ground. When you push the car, you apply a horizontal force on it. Now, in a system of the car and you, no motion relative to the ground occurs. This is because the net force in the system is zero. Notice also, when the car moves, if you keep pushing, you will just be stuck to the car and keep up with it, thus not moving at all with respect to it.
Now if we redefine our system and include the ground in it, we have a different picture: You exert a horizontal force on the car, the car exerts a vertical force on the ground, and the ground exerts an equal and opposite force on the car. The NET force in the system, however, is horizontal, in the direction of your push: the vertical forces cancel, and the only EXTERNAL force on the car left is your push. And we know, from Newton's First Law, that if there is an external force acting on an object, it has to move.
Notice also that if our system was simply the car and the ground, the car would still move with respect to the ground: we'd see a horizontal external vector, and no equal and opposite external vector to cancel it.
Either way, Newton was right, and so was Einstein, the car accelerates:
It's all relative.
The answer is D .