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Newtons Third Law

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Two blocks are lined up in a row. Block A a mass of 1kg, and Block B with a mass of 2kg. If block a is A is pushed forward by 12N, how much force does Block B exert on Block A. Friction can be nelected.




    2. Relevant equations
    Newtons Third law
    F(A on B)=-F(B on A)
    For the system of A:
    [tex]F_{Force on A}-F_{B on A}= m_{a}a_{a}[/tex]
    For the system of B:
    [tex]F_{A on B}=m_{B}a_{b}[/tex]




    3. The attempt at a solution
    I know something is wrong because when I plug in values I get [tex]F_{B on A}=0[/tex]

    I think it may be just me but I've had some trouble with Newtons Third law, I can't stand the book we are using: Physics for engineers and scientists by knight, any suggestions?
     
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2

    Doc Al

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    Staff: Mentor

    Good.
    I'd write that as:
    [tex]F_{Force on A}+F_{B on A}= m_{a}a_{a}[/tex]
    Good.

    Now what can you say about their accelerations?
     
  4. Feb 16, 2007 #3
    Well I know that the Acceleration of B is going to be less than A. Since A acts on B's mass of 2kg, B will act on A less than 12N.
    Right?
    But isn't [tex]F_{B on A}[/tex] acting in the other direction to make it [tex]-F_{B on A}[/tex]
     
    Last edited: Feb 16, 2007
  5. Feb 16, 2007 #4

    Doc Al

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    Oh really? So you're saying that A will pass through B? :wink:
    12 N is the applied force on A, not the force that A and B exert on each other.

    By definition, the force on A due to B is [tex]F_{B on A}[/tex]. So the net force on A will be the applied force (12N) plus the force of B on A.

    Let's get specific. Assume A is on the left and B on the right. And let's take to the right as the positive direction. Let's call the magnitude of the force of B on A (and A on B) to be Fn. So [tex]F_{B on A}[/tex] = -Fn, since that force points to the left. And [tex]F_{A on B}[/tex] = +Fn, since that force points to the right.

    It's certainly true that [tex]F_{B on A} = -F_{A on B}[/tex]. Make sense?
     
  6. Feb 16, 2007 #5
    Got yah on that.

    Ok, so by definition. of [tex]F_{B on A} = -F_{A on B}[/tex]
    The agent acts on A with 12N, then A must act on the agent with 12N, right(neglecting friction)? The force that A imposes on B:
    [tex]F_{AonB}=m_{a}a_{acceleration of a}[/tex]
    I get it conceptually, but I guess I have trouble representing it mathematically.
     
    Last edited: Feb 16, 2007
  7. Feb 16, 2007 #6

    Doc Al

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    That's true not by definition, but by Newton's 3rd law.

    That's certainly true, but we don't really care about what happens to the agent of that force in this problem.
    Careful: you mixed up your A's and B's a bit.

    Since the only force acting on block B is the force from A, then [tex]F_{A on B}[/tex] is the net force on B. Then [tex]F_{A on B} = m_b a_b[/tex] follows from Newton's 2nd law.

    But return to the accelerations for a minute. How does the acceleration of A relate to the acceleration of B? Hint: Use common sense!
     
    Last edited: Feb 16, 2007
  8. Feb 16, 2007 #7
    The accelerations are going to be the same?
     
  9. Feb 16, 2007 #8

    Doc Al

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    You got it.
     
  10. Feb 16, 2007 #9
    So.
    Acceleration is going to be [tex]a=F/m=12N\_div1Kg=12m/s^2[/tex]
    [tex]F_{A on B} = m_b a_b= (2kg)(12m\s^2)=12N[/tex]
    then this equals [tex]-F_{B on A}[/tex] ?
     
  11. Feb 16, 2007 #10

    Doc Al

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    That acceleration is incorrect. When you apply Newton's 2nd law to block A, you must use the net force on block A. You just used the applied force of 12 N.

    (Note that you can also treat block A and B together as a single system. What's the net force on that system? What's its mass?)
     
  12. Feb 16, 2007 #11
    ok..
    [tex]F_{AB}=(m_{a}+m_{a})a_{AB}[/tex]
    The force on A in contact with B is 12N so the net force is 12N.
    [tex]a_{AB}=F_{net}divided(m_{a}+m_{b})=4m/s^2[/tex]
     
    Last edited: Feb 16, 2007
  13. Feb 16, 2007 #12

    Doc Al

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    Yes, that's the correct value for the acceleration of the blocks. Now you can solve for the force they exert on each other.
     
  14. Feb 16, 2007 #13
    Thanks Doc...did I really make this problem harder than it really was?
     
  15. Feb 16, 2007 #14

    Doc Al

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    Problems often appear much easier after you've figured them out! :smile:
     
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