If there are two people on ice skates named $A$ and $B$, and $A$ pushes on $B$ with a force $F_A$, then both of them will experience a force of magnitude $F_A$ in opposite directions. I'm confused as to what will happen if $B$ is also pushing at the same time with a force $F_B$. If $A$ pushes on $B$, he experiences an opposite force because of his 'crime' (the act of pushing on $B$), but now $B$ is also pushing at the same time with a force $F_B$, so maybe the forces add up on each of them? I'm not sure. I would really appreciate any help!
[Physics] Two people pushing off each other
forcesfree-body-diagramnewtonian-mechanics
Related Solutions
It is always best to draw a diagram to convince yourself of things in a case like this.
This is intended to represent a steady state situation: nobody is moving / winning. As you can see, there are two horizontal forces on A: the floor (pushing with 100N) and the rope (pulling with 100N). There will be two vertical forces (gravity pulling down on center of mass, and ground pulling up) to balance the torques - I did not show them because they are not relevant to the answer.
Now I drew a dotted line between A and B. Consider this a curtain. A cannot see whether the rope is attached to B (an opponent) or a wall. A can measure the tension in the rope by looking (for instance) at the speed at which a wave travels along the rope - or by including a spring gage.
Now ask yourself this question: if A feels a tension of 100N in the rope (this is the definition of the force on A), and can confirm (by looking at the gage) that the tension is 100 N, but he cannot see whether the rope is attached to a ring or to an opponent, then how can the tension be 200N? If I pull on a gage with a force of 100N, it will read 100N - it cannot read anything else (in a static situation, and where the gage is massless, ... )
I think I understand the source of your confusion based on the earlier q/a that you referenced - so let me draw another diagram:
In this diagram, I have move the point of attachment of the rope with which A pulls B away from B's hands, to his waist. Similarly, the rope with which B pulls on A is moved to A's waist.
What happens? Now there are two distinct points where A experiences a force of 100 N: one, his hands (where he is pulling on the rope attached to B's waits); and another where the rope that B is pulling on is tied around his waist.
The results is that there are two ropes with a tension of 100N each, that together result in a force of 200N on A (two ropes) offset by a force of 200N from the floor, etc.
This is NOT the same thing as the first diagram, where the point on which B's rope is attached is the hands of A - there is only a single line connecting A and B with a tension of 100 N in that case.
As was pointed out in comments, you can put a spring gauge in series with your rope to measure the tension in it; and now the difference between "a single person pulling on a rope attached to a ring at the wall (taken to be the dotted line) and two people pulling across a curtain (so they cannot see what they are doing) is that in one case, a single spring (with spring constant $k$) expands by a length $l$, while in the second case you find a spring that's twice as long, with constant $k/2$), expanding by $2l$.
These are all different ways to look at the same thing.
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!
Related Question
- [Physics] Person Pushing a Block vs. People Pushing off Each Other – Newton’s Third Law
- [Physics] Free body diagram in sliding blocks situations
- [Physics] Newton’s Third Law
- [Physics] Two People Pushing Off of Eachother, Newton’s Third Law, and Unbalanced Force
- [Physics] Is this the right way to calculate the force on a screw
Best Answer
General Nonsense
A force is a current, like an electric current. A current is the rate of flow of a conserved quantity from place to place. The force is the current of momentum from one object to another. Like a regular electric current tells you how much charge is going from point to point, a force tells you how momentum goes from point to point.
When you have 20 Coulombs of charge at one point, and 1 second later, you find 20 coulombs at another point attached by a wire, you can say that 20 Amperes of current moved through the wire. It's not 40 coulombs of current, even though one point lost 20 Coulombs, and the other point gained 20 Coulombs. This is what makes currents currents--- you count the flow of the conserved quantity from one place to the other, you don't double count, by saying "oh, 20 Coulombs disappeared here, and 20 appeared there, so it's 40 Coulombs overall". The flow is defined by how much of the conserved quantity crossed an imaginary surface separating the two points, counting once, with one sign for one direction, and the other sign for the other direction. This is why Newton's 3rd law is true--- the force from A to B is opposite to the force from B to A just because it is a current, and the flow in opposite directions is opposite.
The same is true for a hose carrying water from the main to your yard, at a rate of 8 kg/s. In one second, the main loses 8 kg of water, your yard gains 8kg. But the total rate of flow is not 16 kg/s, just 8. The rate of loss of water in the main is equal to the rate of water mass gain in the yard. For water, you don't get confused with double counting.
In the case of 1 dimensional motion, there is a conserved quantity--- the momentum. The momentum associated to a velocity in the positive x direction is positive, and the momentum associated to a velocity in the negative x direction is negative, so unlike Energy (or nonrelativistically, mass), the conserved quantity is signed. It's like charge, not like mass. The rate of momentum flow is in momentums/second, or Newtons. It's annoying that the unit of momentum is not named after anybody, I'll call it "the Galileo". One Galileo is 1 kg times 1 m/s. A Newton is a Galileo per second, it's Galileos of momentum going from one body to another.
When you have two people on two ends of a rope in a tug of war, pulling with equal strength at 10 N, 10 Galileos of momentum are flowing from one to the other every second, through the rope. The tension in the rope is 10N, and the force on opposite sides is opposite sign, and equal to 10N. Just like the hose, if you slice at any point, 8kg/s of water is moving through the hose, when you slice the rope, anyplace you slice it, there is 10N of tension of each end on the other, which is the current of momentum through the rope.
Your thing
Your confusion comes from the issue of reconciling momentum flows with exertion-feeling, that feeling that one is exerting oneself when one is pushing. The exertion-feeling is not directly related to force, rather to energy consumption and production, and without a good model, it is easy to get confused about the forces.
Consider two astronauts in space, pushing on each other. I will model the astronauts as two identical blocks of mass M, and they have a compressed spring attached to a plate. The plate can be attached to the other astronaut, to the other block directly, or you can attach the plate to the other plate. Further, each block can "choose" whether to push--- meaning that there is a removable rigid rod attached to the plate, which prevents it from moving, which you can remove at any time to allow either astronaut's spring to do work.
I will consider four situations, and then the situation should be clear.
In all situations, each spring starts out compressed an amount x, and has spring constant k. The total spring constant in the first case comes from adding springs in parallel, the total spring constant is 2k, and the total energy is
$$ kx^2 $$
The two forces from the two springs always add up, the total force at the initial time is $2kx$, and you get double the force, as your intuition suggested for two astronauts pushing on each other's chests.
The work done by the force converts all the energy to kinetic energy, so that the final velocity (which is equal by symmetry) is:
$$ Mv^2 = kx^2 $$
So that
$$ v = \sqrt{k\over M} x$$
That's the velocity the two astronauts have when they shove off each other pushing on each other's chest, with their arms modelled as springs. That's situation 1.
In situation 2, only one of the astronauts pushing, the force is halved, the initial energy is reduced by a factor of 2, and the final velocity is reduced by a factor of $\sqrt{2}$, which comes from the energy law:
$$ Mv^2 = {kx^2\over 2} $$
Next, consider situation 3. Here the two plates are attached to each other. The springs are in series, so that the effective spring constant is halved. But when both springs are compressed, the amount of displacement is doubled, so that the total energy in the springs is
$$ {k\over 2} {(2x)^2\over 2} = kx^2 $$
just as for the first case. The final velocity is the same as in case 1. But the force is completely different! The momentum flowing through one spring is the same as the momentum flowing through the second spring, and the force is kx, not 2kx.
The final velocity, though, is unchanged--- when both are pushing on each other, they end up moving equally fast even though the force is exactly the same as if only one were pushing. The reason is that the displacements are larger by a compensating factor, so that the work done by the forces is greater.
The situation 4 gives the same velocity as when only one was pushing before, but the force is still exactly the same--- the force is kx, whether both springs are pushing, or only one is pushing. This is the counterintuitive thing that confused you--- it depends on whether you are pushing at the same spot or at a different spot. When the momentum flow is through both force sources, the force doesn't add up over the two sources, but the work still adds up.
In case 4, the final velocity is still reduced by $\sqrt{2}$. So while the end velocity doesn't depend on where you push, only whether both people are pushing, the force you generate to do the pushing does depend on where exactly you are pushing.
This is what is leading to the intuition paradox.
When the two people are pushing on each other's chest, the force is doubled. When they are pushing on each other's hands, the force isn't doubled, but the displacement is doubled, so that the work is still doubled. The work is what is felt as the exertion, and to produce the same final velocity (although not the same initial force) you need to exert yourself twice as hard when pushing alone.
But if you have a rope you want to break, and you're not strong enoug, and you have a friend, tie it to a wall and ask him to pull on the same end as you. Don't put him on one end and you on the other.
This is the correct answer, I wrote nonsense previously. The gist of the other answers to the other problem was not wrong, but they didn't consider the issue of pushing through the same link, or pushing through parallel links. It makes a difference in the force.