3
$\begingroup$

This is the absolute basic of a physics and yet 2 hours of googling fails to find an answer ! So ignoring all vertical movement and just concentrating on the horizontal movement:-

  1. A man who weighs 75 kg jumps off a pier (steps off horizontally) with a force of 150N. Ignoring gravity he accelerates by 2 meters per sec per sec.

  2. A man who weighs 75kg jumps off (steps off horizontally) a boat that weighs 75 kg with a force of 150N. Ignoring gravity he accelerates by 2 meters per sec per sec. So does the boat (it accelerates by same amount), but in the opposite direction.

  3. What is wrong here? Im sure the boat and the man would end up at half the speed in each direction compared to example 1. Example 2 sounds like free energy compared to example one. What am I missing, apart from a working version of google, and an easy way to format this text?

  4. p.s. I think one of my key confusions is in example 2, is the acceleration relative to a point in space or is the acceleration relative to the other body. I.e. do they accelerate away from each other at 2m per sec^2 or 4m per sec^2? I guess my other confusion is if they are accelerating away at twice the rate then I appear to be doing twice the amount of work for the same amount of force applied which seems odd, and feels like free energy.

$\endgroup$
9
  • $\begingroup$ What is the question? $\endgroup$ – my2cts Jun 12 '20 at 10:13
  • 1
    $\begingroup$ acceleration (force) is not a conserved quantity. Conservation balances should be done with energy and/or momentum $\endgroup$ – anna v Jun 12 '20 at 10:55
  • $\begingroup$ Force and acceleration are instantaneous quantities. Knowing only the applied force and the resulting acceleration at some particular instant in time will not tell you the final speed of the man or the boat. Here's a clue: The man's legs can only extend so far. What are the implications of that limited extension when the man jumps off the boat, which moves backward because of the applied force, vs. the case where he jumps off of the unyielding pier? $\endgroup$ – Solomon Slow Jun 12 '20 at 11:15
  • $\begingroup$ So in both my examples I am pretty sure the mans legs would be in contact and accelerating for the same length of time, I am pretty sure the force applied would be the same, the impulse the same, the energy used the same and the total momentum the same and the work done the same. So basic questions are is he actually still accelerating at the same rate in example 2? Is that relative to the same point in space, or is it that his acceleration in example 2 is actually an acceleration relative to the boat rather than a general accel in a direction, which is what is shown in all the diagrams ? $\endgroup$ – MajorTom Jun 12 '20 at 12:39
  • $\begingroup$ Actually Im not sure if the momentum would be the same, so there is another question would it be the same between the 2 examples ? How would you calculate the end speeds if in both cases if say the force lasted for 1 second ? Ill start you off with the first one it would be pier 0 m/s and man 2 m/s. But what about the second example of man and boat going in opposite directions ? Is it that the length of time available for acceleration is less for some reason ? $\endgroup$ – MajorTom Jun 12 '20 at 12:59
1
$\begingroup$

Good question, though you have stated it a little imprecisely I think. So first I will state the question more precisely.

Scenario 1. Man leaps from pier, such that the force between himself and the ground, in the horizontal direction, reaches 150 N (for some length of time to be discussed).

Scenario 2. Man leaps from a boat having the same mass as the man, such that the force between himself and the boat, in the horizontal direction, reaches 150 N (for some length of time to be discussed).

You are correct to suppose that in both cases the acceleration of the man, relative to some convenient frame of reference such as planet Earth, reaches the same value in the two scenarios at the moment when the force reaches 150 N. But in the second scenario the man will find it harder to get the force to reach that value, because the boat is being pushed away as his legs get straighter. He could do it, but he would find that he had to expend more energy.

To find the final velocity of the man in the two cases, you can use either energy or momentum. In terms of momentum, what you need to know is the length of time for which any give force is applied. Leaping from a pier, a man can get to some given force pretty quickly, before his body has moved very much, and then maintain that force for, let's say, half a second. So he gets the momentum approximately 75 kg m/s. Leaping from a boat, the force will rise from zero more slowly, so will take longer to get to 150 N. So he will only manage to apply 150 N for some shorter time. His final momentum will therefore be lower than in first case.

Next let's think about energy. Now we have to consider the distance over which the centre of mass of the man moved while the force applied to the centre of mass was 150 N. Leaping from the pier, this distance is approximately the distance by which his legs extended from the crouch to the leap. Leaping from the boat, since the boat moves away, his centre of mass moves by about the half the distance of the first case. Therefore the energy delivered to the man in the second case is about half that of the first case. On this estimate the man does the same work in the two cases, but when leaping from the boat half of the work goes to kinetic energy of the boat, and half to the man, whereas in the first case the man gets all the kinetic energy (the change in motion of the pier being negligible).

$\endgroup$
1
  • $\begingroup$ I was kind of thinking in terms of the horizontal acceleration and ignoring all vertical movement ! So like stepping off a pier or a boat. I think the force applied would be pretty instantaneous. I think from what others have said the relative acceleration between the 2 bodies in the second case (stepping off a boat) would be twice that of the first case, so it shines twice as bright for half as long. I.e. in my second example I get twice the accel but for only half the length of time. So the legs would be in contact applying the same force but accelerating twice as quickly for half as long. $\endgroup$ – MajorTom Jun 13 '20 at 12:10
0
$\begingroup$

To avoid getting sidetracked into questions of how muscles work, let's replace the man with a sack of potatoes with a mass of $75$ kg and let's suppose the force is provided by a compressed spring which produces an initial force of $150$ N when it is released.

If the sack of potatoes is on a pier (or on the ground) then the spring exerts a force of $150$ N on the sack and (by Newton's third law) on the pier. The man has a mass of $75$ kg and so initially accelerates at $2$ metres per second per second. The pier has (effectively) the mass of the whole Earth, so its acceleration is negligible. If the sack travels a small distance $d$ (small enough so that the force exerted by the spring does not change) the work done by the spring is $150d$ Joules. This energy was originally potentially energy in the compressed spring, and becomes the kinetic energy of the sack. Note that the pier does not move (or, to be precise, its movement is negligible) so the spring does no work on the pier, and the kinetic energy of the pier is negligible.

If the sack of potatoes is on a boat (or on a trolley) which has a mass of $75$ kg then the spring exerts a force of $150$ N on the sack and on the boat. Again, the man has a mass of $75$ kg and so initially accelerates at $2$ metres per second per second. But this time the boat has a mass of $75$ kg so its acceleration is $2$ metres per second per second in the opposite direction. This time, when the sack travels a small distance $d$ (small enough so that the force exerted by the spring does not change) then the boat also travels the same distance $d$ in the opposite direction, so the total work done by the spring is now $300d$ Joules. So both the sack and the boat accelerate and gain kinetic energy, but the work done by the spring is twice what it was in the first case.

$\endgroup$
9
  • $\begingroup$ is acceleration a conserved quantity? $\endgroup$ – anna v Jun 12 '20 at 10:56
  • $\begingroup$ Isnt this just bolstering the free energy theory ? I would imagine a spring would have a set amount of energy in it ? I know in the case of the spring or a person leaping the acceleration will only be being applied as long as things are making contact. Im still kind of struggling getting it all to add up though ! $\endgroup$ – MajorTom Jun 12 '20 at 12:24
  • $\begingroup$ no because in case 1 the other 150d went to earth. but dissipated $\endgroup$ – José Andrade Jun 12 '20 at 13:03
  • $\begingroup$ @anna v No, the accelerations of the sack of potatoes and the boat only have the same magnitude in the second example because they have the same mass. If the mass of boat were $150$ kg then its acceleration would only be $1$ metre per second per second. $\endgroup$ – gandalf61 Jun 12 '20 at 13:54
  • $\begingroup$ @gandalf61 thenks, I did not read it carefully thinking it copied the question. Your example is confusing I would think, as ot is not stating this last clearly. $\endgroup$ – anna v Jun 12 '20 at 13:57
0
$\begingroup$

As @anna_v has addressed several times in the comments, your confusion is caused by acceleration not being a conserved quantity. That it, the sum of accelerations is not necessary the same at different times. The total momentum and energy is, however, conserved at all times. As a small exercise, it might be fruitful for you to compute the total momentum of the system and check that it vanishes at all times.

$\endgroup$
4
  • $\begingroup$ I still dont know which bit is wrong though ! Is it that in my second example the 2m /s^2 acceleration of the person is relative to the boat and is not an acceleration in a direction relative to a point ? Is it actually the case that the person and boat accelerate away from each other at 4m /s^2 or 2m/s^2 (Assuming the same force is applied by the person in both examples)?You say total momentum is conserved. I would have thought he would have more momentum when he jumped off the pier than when he was stationary on it ? $\endgroup$ – MajorTom Jun 12 '20 at 15:54
  • 1
    $\begingroup$ @MajorTom In the second example the boat and the person are accelerating at $2$ m/s^2 in opposite directions, so, yes, the acceleration of the person relative to the boat is $4$ m/s^2. The total momentum of the boat and the person is always $0$ - remember that momentum, like velocity, is a vector quantity, so the boat's momentum plus the person's momentum nets to $0$. This is also true in the first example - the earth's momentum plus the person's momentum nets to $0$ - the earth has a very large mass, and so ends up with a very small velocity. $\endgroup$ – gandalf61 Jun 12 '20 at 17:02
  • $\begingroup$ You are exactly correct @MajorTom in point 1 and 2 of your original question. However, there is no contradiction between these two statements. Regardless whether you jump off a pier or a boat you gain the same final velocity. The object you jump from does not matter. The only thing that matters is the force acting on you. Newtons equation $F=ma$ only involves quantities associated with you. $F$ is the force acting on your, $m$ is your mass and $a$ is your acceleration. There is no pier/boat in this equation. I hope this clarifies your question :) $\endgroup$ – Michael Iversen Jun 12 '20 at 18:32
  • $\begingroup$ Something is still wrong there because if do have the same acceleration in both, then the relative acceleration between the 2 items will be twice as big in the second example, so the length of time I am accelerating for would be reduced (probably halved in this case?). Assuming your legs only stretch the same amount in both examples ! So the end velocities would probably be different. This would probably make the total momentum between the 2 different too as I think that proportional to v^2. $\endgroup$ – MajorTom Jun 12 '20 at 22:47
0
$\begingroup$

In the first case the man does a force in the pier and the pier does a force in the man. Both of 150 N. Of course only during the leg's impulse.

In the second case, the man does a force in the boat and the boat does a force in the man. Both of 150 N.

No difference concerning the third law.

And third law is conservation of momentum: $$F_{12} = -F_{21} => \frac{\partial \mathbf p_2}{\partial t} = -\frac{\partial \mathbf p_1}{\partial t} $$

About conservation of momentum, the second case is easy because the objects have the same mass. In the first one, the pier is attached to the Earth, so all the Earth is accelerated backwards by the force. But as its mass is so big, the acceleration is negligible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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