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Homework Help: Sliding Block on incline

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A block with mass m = 5.00 kg slides down a surface inclined 36.9 degrees to the horizontal (the figure View Figure ). The coefficient of kinetic friction is 0.23. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 28.0 kg and moment of inertia 0.500 kg*m^2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 m from that axis.

    2. Relevant equations
    I tied solving for the sum of torques, which is T*r=I*([tex]\alpha[/tex]) where T is the tension and r is the radius of the pulley, .200 m. I think solved for T for the box. Once solving for T I got, T=mg(sin(theta)-cos(theta)*[tex]\mu[/tex]k)-m*a. Then using the rule, [tex]\alpha[/tex]=a/r, I then substituted a/r for all the [tex]\alpha[/tex]'s and then plugged my T for the torque equation into the T from the force equation.

    In the end I came up with the equation:
    mg(sin(theta)-cos(theta)*[tex]\mu[/tex]k)-m*a = (I*a)/r^2

    This yeilded me the wrong answer and I have no idea where I am wrong. I have a feeling it has to do with the mass of the pulley since it was given, but I am not sure how to relate it to the problem since they give you the moment of inertia for the pulley ???

    Any help would be great thanks !

    Attached Files:

  2. jcsd
  3. Apr 17, 2008 #2


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    Homework Helper

    Hi ASUengineer16,

    What wrong answer did you get?
  4. Apr 17, 2008 #3
    After solving for a in the final equation I got 1.1 m/s^2
  5. Apr 18, 2008 #4
    actually seems like I made a rounding error, the correct answer was 1.165 m/s^2 but it wouldn't take 1.1 when I entered it,

    and for part b, 14.6 N came out to be correct with 1.2 m/s^2 as the acceleration.
  6. Apr 18, 2008 #5


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    Homework Helper

    I was about to write back that I kept getting 1.16619 for the acceleration.

    Glad it worked out.
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