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Acceleration assumed in the wrong direction again

  1. Feb 20, 2007 #1
    Acceleration assumed in the wrong direction again!!

    1. The problem statement, all variables and given/known data
    See page 1, this is the problem statement, figure and the solution as supplied in the books manual.

    2. Relevant equations
    3. The attempt at a solution
    I solved this problem as seen on page 2, clearly, I assumed the acceleration in the wrong direction. But why is this a problem? Friction is not a factor here and I think the directions for the accelerations that I assumed are reasonable.

    How can I avoid making this kind of mistake on an exam?? I always seem to pick the wrong direction for the acceleration.

    Attached Files:

  2. jcsd
  3. Feb 20, 2007 #2

    Doc Al

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    Staff: Mentor

    Once you set up your constraint equation, everything that follows must be consistent with it. Realize that the signs of Aa, Ab, and Ac are positive when Xa, Xb, and Xc are increasing. Your force equations for A & C contradict that assumption.
  4. Feb 20, 2007 #3
    So when you are setting up the constraint equation, what determines the sign of your X's are not the axes you establish but wether or not the distance X is increasing or decreasing. For example, in this problem, I established my x-direction as positive to the left, but this has nothing to do with Xc being positive or negative. When I assumed the distance Xc positive, I also assumed that the the Xc distance was increasing and that block c was moving to the right (which contradicts my sign convention, left being positive). Am I making sense? This is why I think I keep getting confused.
  5. Feb 20, 2007 #4

    Doc Al

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    Staff: Mentor

    Yes. Remember that the constraint is just a way of saying that the cord length doesn't change.
    Xc is defined (see the diagram) as the horizontal distance between C and the first pulley. Ac will be positive if Xc increases, which means that it accelerates to the right. Your sign convention doesn't matter, as long as you are consistent.

    Want to take "to the left" as positive? No problem. The force will then be +T and the acceleration will be -Ac.

    Want "to the right" to be positive? Again, no problem. The force is now -T and the acceleration is +Ac.

    Once you solve for Ac, you will find the actual acceleration.
    I think you're getting the idea.
  6. Feb 21, 2007 #5
    Thanks you,.
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