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Isaac Newton Contest Question.

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Two blocks are in contact on a frictionless table. A horizontol force is applied to one block as shown in Fig 7. If m1 = 2.0 kg, m2 = 1.0 kg, and |F|applied = 3N, find the force of contact between two blocks.

    ----> --------
    | A | B |

    note: the height of B is a little smaller.
    A= m1
    B = m2.
    The arrow is the force applied.



    2. Relevant equations



    3. The attempt at a solution
    Fnet(1) = Fapp - Fm2
    -> Fapp = m1a + Fm2 (eqn 1)

    Fnet(2) = Fapp + Fm1
    Fapp = m2a - Fm1. (eqn 2)

    Fm2 + Fm1 = a(m2-m1)
     
    Last edited: Jan 28, 2008
  2. jcsd
  3. Jan 28, 2008 #2

    Tom Mattson

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    Good.

    No. Fapp doesn't act on block 2. Only Fm1 acts on block 2.

    Also, you don't need 3 variables. Newton's 3rd law says that the action of block 1 on block 2 is equal and opposite to the reaction of block 2 on block 1. So Fm1 and Fm2 have the same magnitude, but opposite directions. So just call their common magnitude F. Then you have 2 equations in 2 unknowns. (F and a).
     
  4. Jan 28, 2008 #3
    From post 1 EQN1: Fapp = m1(a) + Fm2
    EQN2: m2(a) = Fm1
    but Fm1 = Fm2 (action rxn) = F
    therefore

    Fapp = m1(a) + m2(b)
    3 = a(2 + 1)
    a = 1 m/s^2

    therefore back to EQN 1
    Fapp = m1(a) + F
    3 = (2)(1) + F
    F = 1N?

    Can anyone confirm this answer?
    thanks (its not in the back of the textbook)
     
  5. Jan 28, 2008 #4

    Tom Mattson

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    That's what I got too.
     
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