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Blocks connected by rope on slope

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    I just want to confirm this.

    A block m1=10 kg is hanging from a rope. the rope is attached by a frictionless pulley to a block m2=5 kg on a frictionless slope with a 40 degree angle from the ground.

    Find the tension in the rope and the acceleration of the first block.

    2. Relevant equations

    So the tension is the same for both blocks, and so is the acceleration.

    m1 * a=-m1 * g + T
    m2 * a=-m2 *g sin(40) + T

    3. The attempt at a solution

    T = m1 * a + m1 * g
    T= m2 * a + m2 *g sin(40)

    m1 * a + m1 * g = m2 * a + m2 *g sin(40)
    (m1 - m2) * a = m2 *g sin(40) - m1 * g

    a = (m2 *g sin(40) - m1 * g) / (m1 - m2)
    a = -13.3 m/s2

    I'm not sure about the negative. Because I set up the problem so - for the first block is down and + is up. and - for the second block is downslope and + is upslope. But if a is the same for both blocks, then it doesn't make sense.

    But did I do everything right?
  2. jcsd
  3. Sep 27, 2011 #2

    Doc Al

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    Staff: Mentor

    If the acceleration of m1 is down, then the acceleration of m2 must be up. Fix your equations accordingly. (With your sign convention, you can't call both accelerations 'a'.)
  4. Sep 27, 2011 #3
    So would I do a1 and a2 with a1 = - a2 ?

    Or do I have to do all the equations over again?
  5. Sep 27, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That should work.


    The trick I like to use is to guess the most likely direction and call it positive. Here I'd guess that m1 goes down, so I'd say that the acceleration of m1 was 'a' downward. And thus the acceleration of m2 is fixed to be 'a' up the incline. If you guess right, 'a' will turn out to be positive; if you guessed wrong, it will be negative. No worries.
  6. Sep 27, 2011 #5
    Okay, thanks
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