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Simple mechanics problem.

  1. May 28, 2008 #1
    1. The problem statement, all variables and given/known dataA uniform rope of mass m and length L is attached to a block of mass M. The rope is pulled with force F. Find the tension at a distance x from the end of the rope.



    2. Relevant equations T(total)=Ma, F-T(total)=ma & T(total)/L=T(x)/x



    3. The attempt at a solution

    T(total)=Ma, F-T(total)=ma => T(total)=(MF)/(1+mM)

    This is the part I'm confused about. I assumed the following proportion. T(total)/L=T(x)/x
    This implies, T(x)=(T(total)/L)x= (M/(L(mM+1)))x

    Am I right in my reasoning?
     
  2. jcsd
  3. May 28, 2008 #2

    G01

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    Hmm. I think my answer is different from yours, and I am not sure about your reasoning. I don't think you can assume that proportion.


    Start the problem by considering the force at a point x. This force, T(x), is the total force, F, minus the force needed to accelerate the length of rope in front of it. What is this force needed to accelerate the segment of rope before point x?

    HINT: You are going to need to use the mass per unit length of the rope, m/L.

    Also, you will end up with a factor of "a', the acceleration? How can you express this in terms of F, m, and M?

    HINT: Consider the motion of the box.
     
  4. May 29, 2008 #3

    tiny-tim

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    Hi torquerotates! :smile:

    There's no such thing as total tension.

    You can't add tension as if you were adding mass or forces.

    The tension varies along the rope. It has an average, but not a total.

    Hint: divide the problem into two sections:
    i) the block plus part of the rope
    ii) the rest of the rope.

    Apply good ol' Newton's second law to i) and ii) combined, and to i) and ii) separately. :smile:
     
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