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Acceleration of a car pulled by a block falling (pulley)

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A car and pulley system are setup as shown. Find the acceleration of the car. The car wheels have a friction coefficient of 0.15. The car has a mass of 0.05 kg while the block has a mass of 0.02 kg.

    http://sadpanda.us/images/836771-GBD7SB9.jpg [Broken]
    *sorry if the picture is bad I tried to sketch it quickly on paint.

    I chose up and right to be positive.



    2. Relevant equations
    Fnet = ma
    Ff = μFn
    Fg = ma


    3. The attempt at a solution
    If we draw two seperate free body diagrams we would get the following forces.

    Car: Force gravity is 0.49 N
    Fg = ma = 0.05 * 9.8 = 0.49 N

    and since the car isn't accelerating vertically the normal force is the same but opposite direction.

    Ff = μFn = 0.49 * 0.15 = 0.0735 N

    We also have a tension in the rope that we don't know.

    Block: Force gravity is 0.196 N.
    Fg = ma = 0.02 * 9.8 = 0.196 N

    We also have the tension of the rope which we again don't know.

    To calculate the acceleration of the car:

    Fnet = ma
    Fgcar + Fncar + Ff + T1car + T1block + Fgblock = ma
    Since forces cancel out.
    Ff + Fgblock = ma
    -0.0735 - 0.196 = (0.02 + 0.05) * a
    -0.2695 = 0.07 * a
    a = -3.85
    a = 3.85 m/s^2 [Down]

    I have a feeling something is wrong because in the previous question it said there was a frictionless surface and I ended up getting 2.8 m/s^2 for the acceleration which is lower. I'd assume a frictionless surfaces would cause there to be a higher acceleration.

    If anyone can verify this for me and tell me what I'm doing wrong it'd be much appreciated.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 8, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The difficulty is that you are adding the friction force and the force from the block. They are in opposite directions. The force on the car from the falling block is directed to the right while the friction force is directed to the left. Assuming you take "+" to the right, the net force on the car is Fgblock- Ff.
     
  4. Feb 8, 2012 #3
    So then you mean it'd be like:

    Fgblock - Ff = ma

    0.196 - 0.0735 = 0.07a
    0.1225 = 0.07a
    a = 1.75

    Also, would I have to follow the curve of the pulley to the right (or clockwise) is positive instead of straight directions, in questions like this?
     
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