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Laws of Motion w/ Friction - Two Forces on Two Boxes

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A force of 38.64 N pushes and pulls two blocks as shown in the figure below. The vertical contact surfaces between the two blocks are frictionless. The contact between the blocks and the horizontal surface has a coefficient of friction of 0.18.
    The acceleration of gravity is 9.8 m/s2.

    http://img121.imageshack.us/img121/962/physicsproblem.tif [Broken]

    What is the magnitude a of the acceleration of the blocks? Answer in units of m/s2.

    2. Relevant equations

    a = (F - (µm1g + µm2g)) / (m1 + m2)

    F = m • g

    3. The attempt at a solution

    I've worked on this for a while but still cannot seem to come to the right answer. I submit answers online, so I know the following is either incorrect or not close enough.

    There are two contact forces that urge the boxes right:
    The push has a horizontal force = 38.64N ➙
    The pull has a diagonal force = 38.64 ➚
    This force will gives the horizontal force of 38.64cos(50) = 24.84N ➙
    as well as a vertical force of 38.64sin(50) = 29.60N ↑

    There is also the two opposing forces of friction for each box:
    With the vertical force on the right box, m2 must be modified
    Its original downward force is 90.16N (m2g) minus the upward force from the pull 29.60N, gives it a new downward force of 60.56N.
    Right box µ(m2)g: .18 • 60.56N = 10.9N ←
    The left box is still the same
    Left box µ(m1)g: .18 • 45.08 = 8.1N ←

    Now that I have forces calculated, I put them into the acceleration formula above. (This is where I think I'm messing up.)

    a = (38.64N + 24.84N - 10.9N - 8.1N) / (4.6kg + (60.56N/9.8m/s2))

    Simplified
    a = 44.48N / 10.78kg = 4.13m/s2

    And I was told that was incorrect.

    That is as right as I think I've been able to get it.
    Thanks for any help in advance!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 23, 2009 #2
    The mass of m2 doesn't alter due to the vertical force. Just simply the normal force, which combines with this new force to offset gravity and keep it from accelerating through the table. Friction, as you predicted, decreases due to this change, though.

    Otherwise, very nicely done.
     
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