Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Why did you both put them to $100N$? Why would the force on/to the box be, say, $1N$? –  Bernhard Mar 26 '13 at 18:38
    
Because the rocket is pushing the box with 100 N? Is it not? If not, how can then the box accelerate as quickly as the rocket? –  Peatherfed Mar 26 '13 at 18:41
    
Because the box has less mass; see my response below for details. –  joshphysics Mar 26 '13 at 18:46
add comment

1 Answer

up vote 5 down vote accepted

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!

share|improve this answer
    
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
3  
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 8:25
    
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 9:09
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.