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Newton's third law problem

  1. Mar 10, 2012 #1
    When I exert a force, I will experience an equal and opposite force on two mutual bodies. However, when I break through a wall how do the free body diagrams of my hand and the wall? When the wall doesn't break, the forces on the hand is normal contact force on my hand from the wall and a force pushing it to make it stay st that position. As for the wall, its a force that I exact on it by my hand and also a force to balance the exerted force on the wall.

    But when dealing with the breaking wall case, I'm unsure what happens. Thanks for all the help!
  2. jcsd
  3. Mar 10, 2012 #2


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    Your analysis is correct in both cases: the case where the wall breaks and the case where the wall doesn't break. The third law doesn't imply or require equilibrium.
  4. Mar 10, 2012 #3
    Seems you are combining the first and the second laws. It's like when you push an object on a frictionless surface, the force you apply on the object is equal to the force the object applies on your hand. If I got you correctly, according to your analysis, your hand ( and the object) should not accelerate.

    Second's law is about net forces on the "same body". When you break through the wall, the wall exerts a force on your body, your weight and also your the floor ,through your shoes, exerts forces. The combination these forces determines your motion.
  5. Mar 10, 2012 #4
    So when the wall breaks the forces on me is the normal contact firce by the wall on me and my own pushing force on myself? Then for the wall its my exerted force with less pushing force back on my exerted force? Then what happens when I push air? I apply a force in it so Teresa another force on me so I have to apply some force on myself to counteract it. While for the air molecules they don't have that counteracting force so they simply move away? Are these assumptions right? Thanks for all the help!
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