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Another tension problem

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A rope hangs from a tree branch, and a bunch of bananas of mass M are tied halfway up the rope. A monkey of mass m hangs on to the end of the rope. Assume the monkey, bananas, and rope are not moving. Neglect the mass of the rope.
    1)Draw a FBD of all the forces acting in the monkey. use this to find the tension of the rope just about the monkey.
    2) Now think of the monkey and the bananas as a system. Draw a FBD of all the forces acting on this combined system of bananas and monkey, and find the tension in the rope above the bananas.
    3)Draw a FBD of the forces acting on the bunch of bananas, and find the tension in the rope above the bananas. How does this tension relate to the answer in part (b)


    2. Relevant equations
    f=ma



    3. The attempt at a solution
    I am getting very confused on this problem because all of my FBD's are coming out the same!
    for the monkey: I have mg pointing down in the negative j direction, and "t1" pointing up in the positve j direction. I then calculate t1=mg

    for the monkey and the banana as one system: i have "t2" pointing up in the positive j direction and m(total)g pointing down in the negative j direction.
    I the calculate t2=m(total)g

    ....and I have the same thing for the fbd of the bananas.
    If anyone can clarify this problem for me if would be great! I don't understand how these tensions differ. I know the masses are different, but those are being multiplied by zero anyway since there is no acceleration.
     
  2. jcsd
  3. Oct 23, 2008 #2

    alphysicist

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    Homework Helper

    Hi ScullyX51,

    What equation did you get for the free body diagram of the bananas? It should be different than the other two.

    I'm not sure what you mean here. The acceleration is zero, but you have already set that equal to zero when you calculated these expression. So for the first two:

    t1= m1 g
    t2= (m1+m2) g

    with (m1=monkey mass, m2=banana mass); there is nothing to set to zero here, so t1 and t2 will be different.

    (In the full Newton's equation these would be:

    t2 - m1 g = m1 a
    t2 - (m1+m2) g = (m1+m2) a

    and then you would set a=0, but that has already been assumed in your results).
     
  4. Oct 23, 2008 #3
    ok. so I did what you said and I have m1= monkey mass and m2=banana mass. Then from equation 1 of the tensions above the monkey:
    t1=m1g
    I solved for m1...and got m1=t1/g
    I then used this value to find the tension in the second part which is the tension of the rope above the bananas:
    the equation is: t2-(m1+m2)g=0
    then when I substitute in m1, I have: t2-(t1/g+m2)=0....and the g's cancel out.
    so now we have: t2-t1+g=0
    and for the final answer of the tension I have: t2=t1+g. is this right? or does it not make sense to substitute the m1 in?
    As for the last part...the tension just above the bananas, I still have no idea. I don't see how this is any different then the tension I just solved for because there's still a tension for in the positive j direction, and there is still the mass of the monkey and bananas and g in the negative j direction. Am I missing something. Thanks again for all your help. :)
     
  5. Oct 23, 2008 #4

    alphysicist

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    Homework Helper

    These last two lines do not look right; what happened to the m2?

    However, since the problem statement only mentioned the masses, I would think they want the tensions in terms of the masses, so your result:

    t2=(m1+m2)g

    from the third line up is the answer they are looking for.

    For the free body diagram of just bananas, there is more than one tension acting. After you write down the equation, you can solve for t2 and compare the result to what you calculated for part 2.
     
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