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Frictionless incline, 2 masses.

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass 3.7 kg on a frictionless plane inclined at an angle of 30 degrees is connected by a cord over a massless, frictionless pulley to a second block of mass 2.3 kg. What are the acceleration of the blocks? The direction of acceleration of the hanging block and the tension of the cord.


    2. Relevant equations
    Mg-T = Ma
    T-mg sinx = ma

    3. The attempt at a solution
    I drew my free body diagram and identified the force that was going to pull the block down the inclined plane. Because the mass of the block on the plane is greater than that suspended over the plane I figured that the acceleration of the hanging block is upward.

    Using the two relevant equations I came up with g[(msinx-M)/(M+M)] = a
    I feel like this should be right and for an answer I keep coming up with 0.801. The correct answer is 0.735 Help please :)
     
  2. jcsd
  3. Oct 1, 2009 #2
    You are confusing me with you use of M and m. I just worked the problem and got .735 m/s^2 as the acceleration. By the way, the direction you chose ends up being the wrong way, so your acceleration will just be negative, but it is still correct. Back to the problem, the equation you have for a looks correct except for the denominator. The denominator should be mass 1 plus mass 2, not two times one mass. However, since I am confused about M and m, I think you may have mixed them up. Start the problem over again, labeling one mass m1 and the other mass m2. Use this terminology throughout your work and see what happens because your equation looks correct.
     
  4. Oct 1, 2009 #3
    I apologize for the confusing notation.

    It turns out my entire problem was a copying error. When I was dividing by the two masses I used 2.3+3.2 instead of 2.3+3.7. Thank you for you help though in reassuring me that I was on the right path.
     
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