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Block Motion and Normal Force

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    You are pushing four blocks of masses 4, 7, 11, and 17kg positioned next to each other on a frictionless surface.
    Find the force of contact between the 7 and 11kg blocks when you push the blocks with a 190N force to the left.
    They are arranged as such: |17||11||7||4|


    2. Relevant equations
    F = 190N = (17+11+7+4)a
    a = 4.87

    3. The attempt at a solution
    m4a = F - F7on4
    F7on4 = 190 - (4x4.87) = 170.52N
    F4on7 = -170.52N
    m7a = F4on7 - F11on7
    F11on7 = -170.52 - (7x4.87) = -204.12N
     
  2. jcsd
  3. Oct 8, 2013 #2

    Doc Al

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    Staff: Mentor

    You're making a sign error. Note that acceleration is to the left.
     
  4. Oct 8, 2013 #3
    So F7on4 should be -170.52 given that I have chosen the left direction as positive?
     
  5. Oct 8, 2013 #4

    Doc Al

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    Staff: Mentor

    Yes. F7on4 points to the right, which is negative in your convention.
     
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