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Drawing a free body diagram problem

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.80 kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 86.0 N.

    Draw the free-body force diagram for the bucket.
    Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

    Okay, I calculated the weight of the bucket of water which is w = -47.04N . If I were to make a free body diagram, the vector of w would be smaller than that of the Tension correct? Am I supposed to do something with the 86 N they gave me?

    [tex]\Sigma[/tex]Fy = 86N + -47.04N = 38.96N

    In this case, the vector of w would be longer than that of tension?
  2. jcsd
  3. Oct 13, 2009 #2
    Yes, the weight vector should be shorter than the tension vector. I don't know why they give you 86N breaking point, unless there is more to the problem? It is not required to analyze the situation qualitatively.

    Is there another part that asks you to find the maximum upward acceleration before the rope breaks?
    Last edited: Oct 13, 2009
  4. Oct 13, 2009 #3
    Oh, wow, yeah. It says

    Apply Newton's second law to the bucket and find the maximum upward acceleration that can be given to the bucket without breaking the cord.

    I didn't see it until I scrolled down.

    I know I have to use F=ma, but would I do

    38.96N=18kg*a ?
  5. Oct 13, 2009 #4
    Haha nice.

    You know that the force of gravity is 47.04N downwards and the maximum tension force is 86N. Because the initial 47.04N is required to counter gravity, the rest of that 86N causes your upward acceleration. You calculated 38.96N correctly, just remember that the mass of the bucket is 4.8kg not 18kg.
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