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Newton's third law when moving our body

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    When I push my belly forward, a force is exerted on my belly by my muscles. So there will be a reaction force acting on the muscles right? But eventually it has a net forward force. So if we treat the belly as one object (muscle fat everything in the belly) then where is the reaction force from my belly swinging forward.

    2. Relevant equations

    None
    3. The attempt at a solution
    I'm thinking it should be the air behind it. Since my force is forwards, so the object directly behind my forward force is air? But I'm not very sure about this though. Thanks for the help physicsforums! :smile:
     
  2. jcsd
  3. Jun 14, 2012 #2
    Hi sgstudent!! :smile:


    Ah, but that would mean you cannot push you belly forwards in vacuum(Well, not you, as blood pressure would make you burst, but something similar, say a robot copy). I'm pretty sure the robot can 'belly dance' in vacuum.

    When you try flinging your belly forwards, the forces on the belly come from other parts of the body, near it. Try it out(I did :tongue2:), and then decide what would experience the reaction force. :wink:
     
  4. Jun 14, 2012 #3
    Oh, the hips received the backwards force and it was also lodged backwards! Is that right? But in that case, won't my body not move. I understand that the hips and the belly experiences are two bodies, but combining them would make a 0 net force. If we test the hips and belly as one object... Thanks for the help Infinitum!
     
  5. Jun 15, 2012 #4
    Take for example two spherical masses in vacuum(no friction, blah blah). Now there is equal and opposite gravitational force on each of them. Does the net force here equal zero? :wink:

    Yes, combining them does make the net external force zero. But, does this mean that the masses don't move?
     
  6. Jun 15, 2012 #5

    HallsofIvy

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    In fact, this is a serious problem when working in space. You cannot move in space by simply "walking" or making swimming motions- there is nothing to press back against (as your feet against the ground) or move backwards (as water from your hands and feet) as you move forward.
     
    Last edited: Jun 15, 2012
  7. Jun 15, 2012 #6
    oh it makes sense, thanks Infinitum! Also, using the principles of newton's third law how do we explain the instance a table collapse due to weight? In my example, the table's threshold is 10N of force and the table is massless. So at 10N, the four legs of the table exerts a total of 10N of force to counteract the 10N force acting on it right? But after adding an additional 1N of force, at the first instance, are these forces acting on the table correct?
    11N of force acting downwards on the table by object
    10N of force acting upwards by the earth onto the legs of the object
    So the table has a net force of 1N and so does the object?

    Thanks for the help!
     
  8. Jun 15, 2012 #7
    How?? If the force applied is increased by 1N then the reaction force also increases by 1N...It doesn't take the table 'time' to react, its instantaneous.

    Also, you must note that the two forces you have mentioned there are not action-reaction pairs. The reaction due to the force by the object on the table, is the force the table exerts on the object.
     
  9. Jun 15, 2012 #8
    Oh, but if the maximum amount of force the table can take is 10N then won't it break when 11N is used instead? So at the first instance where t=0s will that be correct?

    Thanks Infinitum!
     
  10. Jun 16, 2012 #9
    At the first instant, yes. Immediately then, the table collapses and the reaction is no longer 11N, as the force isn't 11N now.
     
  11. Jun 16, 2012 #10
    You may wonder how the car "knows" to push back on you
    with the same magnitude of force that you exert on it. It may help
    to remember that the forces you and the car exert on each other
    are really interactions between the atoms at the surface of your
    hand and the atoms at the surface of the car. These interactions
    are analogous to miniature springs between adjacent atoms, and a
    compressed spring exerts equally strong forces on both of its
    ends.
     
  12. Jun 17, 2012 #11
    Oh wait, so when 11N of weight is used, will the normal force be 10N or 11N at first? Thanks!
     
  13. Jun 17, 2012 #12
    11N for the first instant. Why do you think should it change??
     
  14. Jun 17, 2012 #13
    Isn't it because the table can only withstand a maximum of 10N as the maximum force that the legs of the table can produce is 10N upwards (to oppose the 10N of force acting on the table by the object)? But then would not really make sense because if not the table would not have a net force...

    So at the first instance the normal force is 11N, but as it breaks the normal force becomes 0 they both fall through? Could you explain the mechanics in detail? Thanks for the help!
     
  15. Jun 17, 2012 #14
    This is it! :wink:

    When you place the block on the table, the table doesn't give away the same instant that you place it(so it can exert 11N reaction). Just after the object is kept, the table breaks down, and both the tabletop and block go free-falling their way to the floor, during which the normal reaction is zero.
     
  16. Jun 17, 2012 #15
    Oh okay :smile: I get it thanks Infinitum!
     
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