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B Action-Reaction Pairs....

  1. Jul 21, 2016 #1
    I am having a little bit trouble with understanding action-reaction...I am gonna discuss three scenarios...Please tell me about the action-reaction in them
    Okay...let's assume about a body A that is positively accelerated upto a certain velocity...This body A is in an isolated system...
    Ist: If this body A enters in a field of air, it will actually PASS THROUGH (ofcourse with some decreased velocity)
    2nd: If the body A enters a field of water, again it will pass through(although with an even more decreased velcity than that in Ist)
    3rd:If the body A tries to go through solid( let's say wall or table), it will not pass through...
    My question is that, in all of these scenarios,the body A will come in contact with some molecules or particles and in all cases the force it applies on these molecules is same...(So reaction will also be same)...So why is that the body A can pass through less molecule area(air) and will not pass through more molecule area(solid)???
     
  2. jcsd
  3. Jul 21, 2016 #2

    jbriggs444

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    The force that the body applies will not be the same in the three cases.
     
  4. Jul 21, 2016 #3
    Can you please elaborate as to why the force will not be same??? Is it because less number of molecules???
     
  5. Jul 21, 2016 #4

    jbriggs444

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    It is not the number of molecules exactly. It is the way they are structured. A solid is not simply a very dense gas. It is rigid. The molecules are interconnected. If you push on one side, all the other molecules move as well. That means that as you push on one face of a solid, that face cannot simply move out of your way. It will stay right there. Think of running head first into a sheet of tissue paper or into a brick wall. Which one involves more force on your nose?
     
  6. Jul 22, 2016 #5
    The force is equal to a change in momentum over time. Which has more momentum: moving air, moving water, moving a table?
     
  7. Jul 22, 2016 #6
    Okay...Let's take another example...Consider a person standing on a thin wooden plank...The amount of force by the person on the plank is equal to that person's weight...But that plank pulls down until it can support the person...So at first, when the person just mounted on the plank, when the plank is just going to pull down, is the amount of force applied by the person on the plank at that specific time actually lesser than the actual force? (the total weight of that person)...So that when the plank is finally pulled down, that's the moment the person's force on the plank is 'maximum'(the actual weight of person)???...I hope you understand the confusion...
     
  8. Jul 22, 2016 #7

    A.T.

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    When the person accelerates down, the force is less than the weight. When it accelerates up, it's more than the weight.
     
  9. Jul 22, 2016 #8

    sophiecentaur

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    I think this is getting away from what "Action - Reaction Pairs" (or Newtons Third Law pairs) is all about. Whatever force is acting on an object, there is another force (equal and opposite) acting on whatever it is that is causing that force. The actual value of the force will depend on the circumstances (as in the three cases that were discussed originally but the same principle applies. The multiple forces which are involved in equilibrium situations are an entirely different matter and all of the forces acting in that case will consist of third law pairs - tension on the tether / gravitational pull / rocket engine thrust etc will all be acting on the fixing point, the Earth and the particles that the engine is ejecting as well as on the rocket. Whether or not you have equilibrium forces on the rocket is a different issue.
     
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