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Angular acceleration and tension

  1. Mar 9, 2005 #1
    A block of mass 2.00 kg and a block of mass 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having a radius of .25 m and mass 10 kg. These blocks are allowed to move on a fixed block-wedge of angle 30 degrees. Friction = .360 for both blocks.

    Ok I don't have a scanner but do you get the picture? The 2 kg block is sitting on a flat surface and the other is sitting on a 30 degree surface attached by a string that goes over a pulley at the change in slope point.

    I basicly know the set up but having problems with making a equation. The first part is to find the acceleration of the blocks. This would be m_2*g*cos(30)-friction of M_1- friction of M_2. Is this right?

    The second part is the tension of the strings on both sides of the pulley. All I can think of is T=m(g-a)

    Please help need to understand finals next week.
     
  2. jcsd
  3. Mar 9, 2005 #2
    Ok tension on M_1 has something to do with the pulley and T on m_2 should be greater but what is the equation? M_2 is the equation above but what is the other?
     
  4. Mar 9, 2005 #3
    You are forgetting the angular acceleration of the pulley. The pulley is not massless, so it has a moment of inertia which you can determine from its mass, size and shape.

    The linear accelerations of the two blocks are equal, but the tensions in the string on each side of the pulley are not equal, so you can make two equations for the accelerations of the two blocks, and one for the angular acceleration of the pulley.
     
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