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Newton's 3rd Law is difficult to understand.

  1. Oct 2, 2006 #1
    I fully understand Newton's 1st law and 2nd law, but I'm not confident with 3rd law.
    Unlike 1st or 2nd law, 3rd law doesn't quite make sense.
    Also, I find it difficult to draw a free body diagram that constitutes 3rd law.
    Could anyone please explain Newton's 3rd law in other ways?
    Thank you.
     
  2. jcsd
  3. Oct 2, 2006 #2

    Doc Al

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    Why don't you explain in your own words what you think Newton's 3rd law says.
     
  4. Oct 2, 2006 #3

    Pythagorean

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    When you throw a ball, can you feel it pushing against your hand? When you walk, do you feel the ground pushing up on your feet? The normal force is a common introductory example of Newton's third law.

    If newton's third law weren't true, and force only worked in one direction, when you threw a ball, you wouldn't feel that force back on your hand, and as you propelled it forward, your hand would go through the ball as well, it would be paradoxal.

    An important thing to remember is that newton's law requires more than one body. A ball flying through the air is pulling on earth as much as earth is pulling on it, the ball is just less massive, and so it is more easily manipulated, but you're only looking at the force on the ball from the earth, you're not looking at the force on the earth from the ball. You're effectively ignoring Newton's third law even though it's still true, it's just useless information for a particular question about the ball's motion.
     
  5. Oct 2, 2006 #4
    Newton's 3rd law says that whenever Object A exerts force on Object B, Object B exerts a force equal in magnitude but opposite in direction on Object A, and vise versa.
    The concept itself is not that hard, but I don't know why it's really confusing and weird to represent 3rd law on free body diagram.
     
  6. Oct 2, 2006 #5

    Pythagorean

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    A free body diagram represents the forces on one body, Newton's third law represents the forces between two bodies.
     
  7. Oct 2, 2006 #6
    In free-body diagram, should I eliminate any pairs of forces that consitute Newton's third law, and only include forces constituting Newton's 1st or 2nd law?
     
    Last edited: Oct 3, 2006
  8. Oct 3, 2006 #7

    HallsofIvy

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    There is no such thing as forces "constituting" a particular law. All forces obey all three of Newton's laws.
     
  9. Oct 3, 2006 #8
    hey i think i got a easy way of showing N3L...draw a FBD of an object resting on the table...since the object is not moving,there is another force to balance it so it doesnt move rite?so this other force will be the reaction force. is this better to understand or did i just make it more confusing?:uhh:
     
  10. Oct 3, 2006 #9

    Doc Al

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    Exactly! Realize that the pairs of forces referred to in Newton's 3rd law always act on different bodies.

    When drawing a free body diagram, one choose a single body to analyze. Only show forces acting on that body. (The fact that the body, in turn, must be exerting forces on other objects is irrelevent.)

    As Halls' explained, all forces obey all of Newton's laws all of the time! The only forces that should appear in a free body diagram are the forces acting on the body you are analyzing.
     
  11. Oct 3, 2006 #10

    Doc Al

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    Unfortunately, your example is just plain wrong! (Which just shows how subtle Newton's 3rd law can be when first learning it.)

    The fact that the object on the table doesn't move has to do with the fact that the net force on it is zero. This is an example of Newton's 2nd law, not the 3rd.

    There are two forces acting on the object: (1) Gravity pulling down (2) The table pushing up. These are not 3rd law pairs!

    To analyze 3rd law pairs, always identify the two objects that are interacting. Let's analyze these two forces and find their 3rd law pairs:

    (1) Gravity is the force exerted by the earth on the object. So the reaction force is the force of gravity exerted by the object pulling up on the earth.

    (2) The table exerts a contact force up on the object. The reaction force is the contact force exerted by the object down on the table.

    A good rule to remember is that 3rd law "action-reaction" pairs never "cancel out" since they act on different bodies.
     
  12. Oct 3, 2006 #11

    Andrew Mason

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    The first and third laws are really corollaries of the second law, which says that force on a body is the time rate of change of momentum of that body: F = dp/dt.

    It follows from F=dp/dt that if F=0 on a single body then dp/dt = 0. So there can be no change in velocity of the body on which no force acts (assuming its mass remains constant). [first law]

    It also follows from F=dp/dt that if F=0 on a system of bodies (ie a system on which no external force acts), then [itex]dp/dt = \sum dp_i/dt = 0[/itex] [third law].

    This necessarily means that in a system of two interacting bodies on which there is no external force acting, [itex]dp_1/dt + dp_2/dt = 0[/itex] so [itex]dp_1/dt = -dp_2/dt[/itex] or [itex]F_{1on2} = -F_{2on1}[/itex] [the usual statement of the third law]

    Since the interaction times are necessarily the same for each body, at all times the loss of momentum of one body is equal to the gain of momentum of the other. This is the important result of the third law.

    I hope that helps.

    AM
     
    Last edited: Oct 3, 2006
  13. Oct 3, 2006 #12
    Doc i still dont quite get what u mean...the force acting on the object is the normal reaction since the object is acting a force on the table???whats wrong with that?
     
  14. Oct 3, 2006 #13

    Doc Al

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    Every force is a "reaction" force to some other force. You started out by saying:
    Which implies to me that you think the two forces acting on the object, which balance each other, are action-reaction pairs. Which is not so. Or did you mean something else?
     
  15. Oct 3, 2006 #14
    oh...i get what u mean now i shld hav said the pair is the table and the object instead of the object alone rite? hmm....sry i thought it was easier to understand it that way......:cry:
     
  16. Oct 3, 2006 #15

    Andrew Mason

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    Right. The normal force of the table on the object (mg) is the 'third law reaction force' of the weight of the object on the table (-mg) (up is +). The magnitude of each is equal to the gravitational attraction force between the earth and the object but the source is not gravity but the mechanical forces of the table and the object.

    The gravitational attraction of the earth on the object is one force and the attraction of the earth to the object is the equal and opposite force. These would be the only forces if the object was in free fall.

    When the object is resting on the table, there is this gravitational force (third law) pair plus the (third law) pair of mechanical forces between the object and the table: ie. the normal force of the table on the object and the equal and opposite weight of the object on the table.

    AM
     
  17. Oct 3, 2006 #16

    Doc Al

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    Generally the term "weight" refers to the gravitational force exerted by the earth on the object. I wouldn't use that term to also refer to the force with which the object presses down upon the table, else you are inviting confusion. (I'm certainly not suggesting that you are confused!) Weight is mg, but the force that the object exerts on the table only equals mg in the special case of vertical equilibrium.

    Perhaps that contact force between table and object could be called the apparent weight.
     
  18. Oct 3, 2006 #17

    Andrew Mason

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    What about 'weightless' astronauts orbiting the earth? The gravitational force is still mg but we say that the weight is 0. You could also say that the astronaut's weight is mg but its apparent weight is 0. It depends on how you define weight.

    The fact is that there are some subtle differences in how the term 'weight' is used. Some would say it is simply the gravitational force on an object. Others would say that that 'weight' of an object is the mechanical force that an object exerts on another by virtue of its position in a gravitational field. Or, similarly, one could say that weight is the reaction to the mechanical force that is required to keep an object from accelerating in a gravitational field.

    So it ultimately depends on how you precisely define 'weight'. I would say that weight best defined not as the gravitational force itself but the reaction to the mechanical force(s) that prevents gravitational acceleration.

    AM
     
  19. Oct 3, 2006 #18

    Doc Al

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    You have a good point. Personally, I think the term "weightless" is the source of much confusion--I would never use it without the quotes to emphasize its nonstandard usage. You're right, it all depends on how you define weight. Most of the standard textbooks I am used to (Halliday & Resnick, for example) use weight to refer to the gravitational force exerted on an object, clearly distinguishing it from apparent weight, which they define as the magnitude of the contact force that supports an object. Using these definitions, as you note, "weightlessness" is the absence of apparent weight (the supporting force), not gravitational force (real weight :wink: ).

    It's a matter of semantics, I suppose. Perhaps I'll start using "gravitational force" instead of "weight".
     
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