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Friction - 3 Blocks and a Pulley

  1. Jun 21, 2015 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/C0XYEKw

    2. Relevant equations
    Fnet = m*a
    FG = m*g
    fstatic = μ*N

    3. The attempt at a solution
    If both Block 1 and Block 2 have acceleration a in the x-direction and have an acceleration of 0 in the y-direction, then the force of static friction on Block 1 due to Block 2 doesn't seem to depend at all on what Block 2 is doing. This is because the normal force on Block 1 due to Block 2 is still the same as it is at rest and the force of static friction on Block 1 due to Block 2 is the only force acting in the x-direction on Block 1.

    I tried this by considering only Block 1 as a system.

    In the y-direction, the forces acting on Block 1 are the normal force on Block 1 due to Block 2, N12, and the gravitational force on Block 1 due to the Earth, F1E. The acceleration, ay, is equal to 0.

    Thus:

    Σ Fnet,y = 0 = N12 - F1E = N12 - m1*g

    Thus:

    N12 = m1*g

    In the x-direction, the only force acting on Block 1 is the force of static friction, which is acting in the right, positive x-direction. The acceleration, ax, is equal to a.

    Thus:

    Σ Fnet, x = m1*a = fstatic = μ*N12 = μ*m1*g

    Thus the force and direction of the force of static friction on Block 1 due to Block 2 is:

    fstatic = μ*m1*g

    This, however, is the wrong answer.

    I don't understand why though. Where am I going wrong?

    Thank you.
     
  2. jcsd
  3. Jun 21, 2015 #2

    ehild

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    Gold Member

    What is the difference between kinetic and static friction?
    Imagine that you have a block in rest on a table. The mass of the block is 2 kg, the coefficient of static friction is μs=0.5.

    You push the block with force F=1N force. Will the block move? What is the force of friction?
    You push the block with F=9 N force. Will the block move? What is the force of friction now?
     
  4. Jun 21, 2015 #3
    O.K., I realised I grossly overcomplicated things.

    Since the force of static friction on Block 1 due to Block 2 is the only force in the x-direction acting on Block 1, and Block 1 has acceleration a, then Newton's Second Law states that:

    ∑ Fx = m1*a = fs12

    Thus, the force of static friction on Block 1 due to Block 2 is:

    fs12 = m1*a

    EDIT: Responding to your original question, the force of static friction can take on any value up to a maximum value of μ*m1*g. So, when an applied force of 1 N is acting on Block 1, the force of static friction is 1 N in the opposite direction. As a result, the net force is 0 N and Block 1 does not move. If an applied force of 9 N is acting on Block 1, the force of static friction is at its maximum 9 N (if g = 10 m/s2) in the opposite direction. As a result, Block 1 travels at a constant velocity in the direction of applied force.
     
    Last edited: Jun 21, 2015
  5. Jun 21, 2015 #4

    ehild

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    That is correct now.
     
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