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Easy question confusing me

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Two boxes are sitting side on a frictionless surface. The box on the left has a mass of 11kg and the box on the right has a mass of 17kg. If a 90-N force pushes on the 11-kg box from the left, what is the force exerted on the 17-kg by the 11-kg box?

    A) 55 N
    B) 43 N
    C) 68 N
    D) 35 N

    2. Relevant equations
    F=ma

    3. The attempt at a solution

    F=MA
    90=(11+17)a
    a= 3.2 m/s^2
    Since you have to find out the force exerted by the 11-kg box
    F=11(3.2)
    F=35N

    This was a question on a quiz. All my friends got 55N.
     
  2. jcsd
  3. Feb 1, 2009 #2
    1...3.2m^s^2 is the correct acceleration for both boxes together
    2....If you now draw afree body force diagram for both boxes you will see that the 11Kg box has two forces on it the resultant being 90-F where F is the force that the 17 Kg box puts on the 11Kg box
    3....There is just one force on the 17Kg box ie F (Newtons 3rd law)
    4...In your final step you chose to consider the 11Kg box but you should have wrote
    90-F=11a.If you had chosen the 17Kg box you could write F=17a
     
  4. Feb 1, 2009 #3

    Delphi51

    User Avatar
    Homework Helper

    Oops, I missed that last bit. Yes, 17x3.2 rather than 11x3.2.
     
  5. Feb 1, 2009 #4
    Thanks for the reply
    I don't understand what the force by the 17kg box on the 11kg box is.
     
  6. Feb 1, 2009 #5
    The force wil be the same ie 55N. The left hand box pushes to the right with 55N and the right hand box pushes to the left with 55 N(3rd law )The result is that the left hand box has two forces :90 N pushing to the right and 55N pushing to the left making a resultant of 90N-55N.The right hand box has the single force of 55N pushing right.It would probably help if you sketched it out and marked in the forces.
     
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