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Physics newtons 3rd/2nd law problem

  1. Mar 3, 2005 #1
    The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.280.

    I cant get a picture up here, but i will make the picture easy for you to imagine.

    there is a table with a pulley on 2 corners. one 2kg weight is on the middle, there is a 3kg weight hanging over the right side, there is a 1 kg weight hanging over the left side. all attached to same rope.

    thus far i have -f-T1+T2=ma or -5.488-T1+T2=2kg(a)
  2. jcsd
  3. Mar 3, 2005 #2
    3a = 3g - T(3,2)
    2a = T(3,2) - T(1,2) - 2 mu g
    1a = T(1,2) - g

    solve for a, T(1,2) and T(3,2).
  4. Mar 3, 2005 #3
    i dont understand, what are the commas for.

    I have the tensions for the string for the dangling weights, but how do i relate this to the weight on the table to find accel
    Last edited: Mar 3, 2005
  5. Mar 4, 2005 #4

    Doc Al

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    Staff: Mentor

    This is Newton's 2nd law applied to the 2kg mass on the table. Good!

    Now apply Newton's 2nd law to the other two masses. Be sure to adopt a consistent convention for acceleration: For example: the 2kg mass has an acceleration of "a" to the right, the 3kg mass has an acceleration of "a" down, the 1kg mass has an acceleration of "a" up. This is the acceleration constraint of the system, since the masses are tied together.

    You'll get two other equations, which will allow you to eliminate T1 & T2 and solve for "a".

    Note: This is exactly what stunner5000pt provided--I'm just adding a little explanation. In his notation, T(1,2) means the tension in the rope between masses 1 and 2, etc. His T(1,2) is equivalent to your T1.
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