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How a net force acts on 'Earth'?

  1. Jul 9, 2016 #1
    Hi, Newton's 2nd Law states that when a net force acts on a body, it causes acceleration in the body...This is well for forces acting on bodies in space...But what about on Earth??...On Earth when you apply force to an object, the air around it pushes back with same force!!!, So where is the concept of net force or 'unbalanced' force???...Kindly answer this question...
     
  2. jcsd
  3. Jul 9, 2016 #2
    The 3rd (EDITED) law implies that the force you apply to the object is equal to the force the object applies to you, it's not equal to the force the air applies to the object.

    To determine the motion of the object, we calculate the net force on the object. In this case, it's the force you apply to the object minus the force of the air on the object. If there is a net force, the object will accelerate.
     
    Last edited: Jul 9, 2016
  4. Jul 9, 2016 #3
    Thanks for the quick reply...Suppose an object applies a force of 20N on the air particles...As a reaction, the air particles apply the same 20N force on the object...So where is the net force???...Should'nt the object be balanced by same force from both sides???
     
  5. Jul 9, 2016 #4

    CWatters

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    What do you mean by "the air pushes back with the same force"? That's only true in certain situations. Please give an example.
     
  6. Jul 9, 2016 #5
    Again, the motion of an object is determine by the net force on the object.
     
  7. Jul 9, 2016 #6

    NascentOxygen

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    When I push on a basket-ball or a boomerang, air molecules move apart to let the object accelerate.
     
  8. Jul 9, 2016 #7
    I mean that when an object moves through the air, it comes in contact with air particles or the particles present in air...In other words it applies force to those particles...According to third law, should'nt the particles apply the same force back on the object...So the forces are 'balanced' in this situtation...For example, If a bullet is fired with 50N force, it applies the same force to air particles as it comes in contact with them...As a reaction, the air particles should apply the same force to the bullet.i.e.50 as action and reaction are equal...So there is a 'balance' of forces in this situation...Is'nt there???!!!
     
  9. Jul 9, 2016 #8
    But they would still come in contact with the basketball of the boomerang...So should'nt there be an action-reaction between the objects and air molecules???
     
  10. Jul 9, 2016 #9
    Lets take an example of a ball...You throw the ball in the air with 10N force...The ball applies same 10N force to air molecules...Equal reaction causes a 10N force on the ball in OPPOSITE direction...(Taking gravity out of account)....SO there is a positive 10N force on ball and a negative 10N force on ball...So where is the net force???
     
  11. Jul 9, 2016 #10

    NascentOxygen

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    It's almost too small for me to notice where I cite for these specific examples.
     
  12. Jul 9, 2016 #11

    NascentOxygen

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    If the ball transferred all 10N to the air then there would be no net force on the ball, therefore it would have zero acceleration and would be moving at unchanging speed. When we observe this condition in a falling body we call it terminal velocity because there is [obviously] zero net force acting on the body.
     
  13. Jul 9, 2016 #12

    Janus

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    The force between air and ball will be drag force and that is a function of how fast the ball is moving through the air. When the ball just starts to move the air will only push back with a force much less than the 10N. It won't be until the ball is moving at a considerable speed that the drag force will equal 10N.
    You apply 10N to the ball. The air pushes back via drag (which has no relation to the force you are applying), and it is the difference in these forces that results in the net force.
     
    Last edited: Jul 9, 2016
  14. Jul 9, 2016 #13
    Third law is the most difficult for newcomer to understand. You are best to learn to draw vector diagrams of the forces.

    On face value acceleration would be impossible due to the conundrum of the OP. The problem lies in what bodies you are assigning the forces to.

    Vector diagrams will clarify your thinking.
     
  15. Jul 9, 2016 #14
    You have to be careful here because when you "throw" a ball you are not continuously applying a force. Let's say instead that you are pushing an object with a continuous force. In your specific example there is no net force, so no acceleration. But if you increase the force on the object, at some point the air molecules cannot provide an equal force, so you could get acceleration. Note again that the 3rd law says:
    • your force on the object = the object's force on you
    • the object's force on the air = the air's force on the object
    It does not say that your force on the object = the air's force on the object.
     
  16. Jul 10, 2016 #15
    Thanks for the answer....You said 'at some point the air molecules cannot provide an equal force'....Can you kindly state the reason as to why the air molecules would not be able to react with equal or same force???
     
  17. Jul 10, 2016 #16
    You said ''When the ball just starts to move the air will only push back with a force much less than the 10N''...Why is that the air will push back with lesser force...Should'nt action-reaction be equal???As the ball moves through air, the ball applies 10 N force to air molecules it comes in contact with...So why is that the air molecules would not push back with same force???
     
  18. Jul 10, 2016 #17
    Why cant we observe the same phoenomenon to an object being pushed FORWARD (not upward)???
     
  19. Jul 10, 2016 #18
    You said''When the ball just starts to move the air will only push back with a force much less than the 10N''...Can't you state it like this..
    Your force on object=The force the object applies to the air molecules=The force the air molecules apply BACK on the object...???
     
  20. Jul 10, 2016 #19

    CWatters

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    I think you are confusing three forces..

    1) The force the air applies to the object (example = drag)
    2) The force the object applies to the air
    3) The force applied to the object (example = engine or rocket thrust).

    1 and 2 are equal due to Newtons third law.

    The net force acting on the object is the vector sum of all external forces (eg 1 and 3). Remember that Force 1 acts in the opposite direction to Force 3 so the sign of Force 1 is negative with respect to Force 3.

    If the net force is >0 the object will accelerate
    If the net force is <0 the object will decelerate
    if the net force = 0 the object will maintain the current velocity.
     
  21. Jul 10, 2016 #20

    CWatters

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    No.

    The force you apply to an object is not always the same as the force the air molecules apply back on the object.

    The force you apply to an object is not always the same as the force object applies to the air molecules.
     
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