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I want to start off by saying that I've looked around for other explanations, but I've not really found any satisfying ones. My question is basically the whole "Why can anything move at all?" question, with a little twist. Refer to this picture:

Now I understand that opposite forces act on different objects, so in the top one, the rocket would have a net force of a 100 N to the right, and accelerate, as per Newton's Second Law. However, in the bottom example, wouldn't the box push back on the rocket, equal and opposite, as the rocket pushes on the box, therefore cancelling the forward push from the exhaust, making the rocket not move at all, whilst the box gets a net force of a 100 N?

This isn't true obviously, but why not? Also, in this earlier Phys.SE post there is a great answer depicting a finger and a matchbox. Now this post is very related to mine. I'm wondering why in that picture, the force of the finger pushing on the matches doesn't equal the force of the muscles pushing forwards in the finger? Surely when I'm having a force pushing the finger forwards, that same force applies to the matches?

Community
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Peatherfed
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4 Answers4

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I'm not sure where that picture is coming from, but it's misleading at best and here's why.

Let's say that the rocket expels some stuff (like the flaming gases in the picture), then the force of the rocket on that stuff will be $-F$, say. By Newton's third law, the force of that stuff on the rocket will be $F$. Now let's consider the system consisting of the rocket plus the box. The net external force on this combined system is $F$ because there is nothing external to the system exerting a force on either the rocket or the box besides the gases. Assuming the rocket and the box are in rigid contact, the acceleration of each object equals the acceleration of the whole system which is given by Newton's second law as $$ a = \frac{F}{m_\mathrm{rocket} + m_\mathrm{box}} $$ Now consider the system consisting of only the box. The only external force on this system is the force $f$ of the rocket on the box, so that acceleration of the box must also satisfy $$ a = \frac{f}{m_\mathrm{box}} $$ Combining these results gives $$ f = \frac{m_\mathrm{box}}{m_\mathrm{rocket}+m_\mathrm{box}}F $$ and therefore $$ f < F $$ In other words, the contact force between the rocket and the box is less than the contact force between the gaseous exhaust and the rocket!

joshphysics
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  • Ok, that seems nice. I understand that the contact force should be less, e.g. if the rocket had a mass of 10kg (small rocket) and the box a mass of 1kg, then the force between the rocket and the box is 10 N. However, doesn't Newtons third law say that forces are equal and opposite? I.e. that the force that the rocket applies to the box (100N) is opposite and equal to the force that the box applies to the rocket (100N). Note that I understand that the forces affect different objects. – Peatherfed Mar 26 '13 at 18:55
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    The force that the rocket applies to the box must equal the force that the box applies to the rocket, yes, but the force that the gas applies to the rocket has nothing directly to do with the force that the box applies to the rocket! Newton's third law only applies in pairs that are two halves of the same physical interaction. – joshphysics Mar 26 '13 at 18:58
  • Why doesn't the force the gas applies to the rocket have anything to do with what force is being applied to the box? If the rocket is being pushed with a force of a 100N, isn't the box also being pushed with this force? How is force otherwise "lost" along the way? – Peatherfed Mar 26 '13 at 19:21
  • Try thinking about it this way. The box has less mass than the rocket. If the box were to experience the same force as the rocket, then its acceleration would be greater, and it would detach from the rocket. As long as the box and the rocket stay in contact, they will have the same acceleration, and therefore the force on the box must be smaller. I'm not sure what you mean by "how is the force lost along the way?" Newton's laws show that the force on the box must be smaller. Perhaps you are referring to some material model that explains what's happening microscopically? – joshphysics Mar 27 '13 at 08:25
  • So when calculating this I have to do it as if it were one object? I can't do it separately?

    Also, why then, when for example throwing something, do we use the force instead of the acceleration during the throw?

     – Peatherfed Mar 27 '13 at 09:09
  • As long as the rocket and box are in contact, they will have the same acceleration. Otherwise, they won't be interacting at all, and the box experiences no force from the rocket (ignoring gravitational interactions of course). – joshphysics Mar 27 '13 at 10:12
  • Sorry but I don't understand the last question about throwing. – joshphysics Mar 27 '13 at 10:53
  • Nevermind that question, I figured that one out my self. However, is there no way to find the force on the box by the rocket without finding the mass of both objects? – Peatherfed Mar 27 '13 at 15:35
  • Not if you're only given the force of the exhaust on the rocket. – joshphysics Mar 27 '13 at 19:15
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The problem is you're assuming the box has the ABILITY to push back with equal and opposite force. Newton's 3rd Law only applies when both objects are CAPABLE of pushing back with the same force. In your box example, Newton's Second law actually comes into effect: The force the rocket exerts cannot be balanced by the box (friction, gravity, etc.), so you're left with an unbalanced force to the right. This means both objects will move to the right.

Daniel
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In my answer, I will stick to the diagram: the box is capable of applying 100N of force leftward. From the start:

(1) Ignition, Blast off: 100N of thrust accelerates rocket to the right.

(2) Rocket collides with box of unknown mass--which is ok, because its mass does not matter.

(3) Box now applies 100N of "box-force" to the left, balancing the 100N of thrust to the right:

(4) All acceleration stops, but velocity does not.

The question is "does the rocket not move at all". It moves, but no longer accelerates, as the net external force is 0N--so it cruises along at the velocity of 1st-box-contact.

Now if this were a physically sensible problem, one would just place the box AT THE BACK OF THE ROCKET, for an incredible specific impulse ($I_{sp}$).

JEB
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  • I should remark: force is not impulse. You cannot talk about what happens to velocity or acceleration with two different forces without specifying the duration that they are applied. It's much easier to understand everything in terms of momentum. If both forces are instantaneous with same Δt, rocket has momentum p = 100N Δt after 1 and receives impulse -100N Δt from the box, so the rocket would stop, the box flies away (think Newton's cradle) assuming there is no deformation. In general Δt will be different, so the rocket and box will cruise at different velocity after collision. – Lukas Berns Feb 01 '18 at 06:13
  • The reason acceleration stops is not because the 100N cancel each other. In fact they can be very different. Acceleration simply stops because the force is applied for a finite time Δt only (essentially while the two objects have contact). – Lukas Berns Feb 01 '18 at 06:14
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Look at the extremes (boundries). Anchor the box to the earth. Now the rocket and box go nowhere, because the box applies an equal and opposite force to the exhaust. Now make the box very tiny. the acceleration is a=F/M, where M=the rocket mass. Now make the box 1/10th the rocket. then a=F/(M+m). the acceleration slows down as the mass (m) of the box increases. The Force applied by the rocket exhaust has not changed. Only the "collective" mass of the rocket and box has changed.

Arthur Watkins
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