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Forces and Tension

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The 1.0 kg block in the figure is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is mu_k = 0.380.

    knight_Figure_08_27.jpg

    What is the acceleration of the 2.0 kg block?


    2. Relevant equations
    a = F/m
    F_friction = mg * mu_k

    3. The attempt at a solution
    I found in a previous question that the tension of the rope holding the 1kg block, and thus the force of friction on the top of the 2kg block, was 3.72N (mg * mu_k = 9.8 * .38 = 3.72).

    So the frictional force on the bottom of the 2kg block:

    mg * mu_k = 2 * 9.8 * .38 = 7.448

    Those two forces hold back the 20N force pulling to the right, so the total force acting on it to the right is:

    20 - 3.72 - 7.448 = 8.832

    So using a = F/m:

    a = 8.832 / 2 = 4.416 m/s^2

    However, 4.416 is incorrect. Any help you could give would be appreciated. Thanks!
     
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 26, 2010 #2
    The upper friction done by the 1-kg block is okay (3.72N), but I see a problem in the friction on the bottom, you should check what's the real mass affecting to that second friction.
     
  4. Mar 26, 2010 #3

    rl.bhat

    User Avatar
    Homework Helper

    Hi Dormin, welcome to PF.
    Net friction force on the lower surface is due to the normal reaction of 3 kg mass. It is in the opposite direction of 20 N. Frictional force on the upper surface of 2 kg block is due to 1 kg mass. It is also in the opposite direction of applied force.
    Now find the net friction force on the lower block which opposes the applied force.
     
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