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Finding the tension in rope of pulley with 2 blocks.(lots of my work shown)

  1. Aug 13, 2011 #1
    I am new here so if any other info is needed lmk. :bugeye:
    Ok I have been working on this problem for a very long time and there is a possibility that my book's answer is wrong. If that's the case I am going to be very mad ahaha. I put up the diagram and my work is after the passage. I don't think my logic is wrong but I might have missed something.

    http://img196.imageshack.us/img196/5859/mmspicture8o.jpg [Broken]

    The block on the left ledge is 10.0kg and the block on the right is 3.00kg.

    Uploaded with ImageShack.us

    1. The problem statement, all variables and given/known data
    In the drawing, the rope and the pulleys are massless, and there is no friction. Find a) the tension in the rope and b) the acceleration of the 10kg block. (Hint: theh larger mass moves twice as far as the smaller mass) The books answer is a) 13.7N and b) 1.37m/s^2.
    I am not worried about answer b, because I cant get the first part!

    (m1) mass 1=10kg (m2)mass2=3kg
    g=gravity 9.8
    a=acceleration
    Heres my work.
    Ok so since there is no friction the 10kg block's only forces are the weight pointing down 10kg*9.8m/s^2=98N and its going to have positive acceleration to the right.

    equation 1: for 10kg block is just T(tension)=m1*a

    Now the fun begins with the 3kg block. Now originally I didn't catch that there must be 2 tensions because 2 ropes are attached to the 3kg box. All the 3 tensions should be the same force. The acceleration is negative since its going down.
    This is probably where there could be a mistake. Should I add mass 1 and 2 in 3kg equation or just have m2
    For 3kg there is only movement in y axis so equation 2 is: 2T-m2*g=-a(m1+m2)

    Then i add in equation 1 from 10kg into equation 2.
    2(m1*a)-m2*g=-a(m1+m2)
    2(m1*a)+a(m1+m2)=m2*g
    a(2*m1+m1+m2)=m2*g
    a=m2*g/(3*m1+m2)

    a=3kg*9.8m/s^2 / 3*10kg+3kg
    a=.891m/s^2

    plug into equation 1 to solve T. T=m1*a T=8.91N

    Now I tried various methods which usually works but I have tried everything and can't get 13.7N I also tried not including mass1 in the second equation and i still get a different answer.

    2T-m2*g=-a(m2)
    a=m2*g/(2*m1+m2)
    a=3kg*9.8m/s^2/2*10kg+3kg
    a=1.28m/s^2
    plug in for t and get T=12.8N
    Now I thought 12.8 could possibly be it due to imprecision but I used 4 sig figs then rounded to 3sig figs so it is not imprecise.(I hate sig figs)

    I assume all the tensions will be same because if they are different I cant solve because there aren't enough variables. Thanks for any help or guidance. Im going to sleep soon but I will be on here tomorrow morning. Now if the accelerations are different, even when i plug them in how can i solve to get an acceleration?
    1. The problem statement, all variables and given/known data

    After I solve this my life will be complete...
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 14, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    You took that block 1 accelerates to the right, very correctly. So m1a1 =T.

    As for block 2, collect all forces and equal the resultant to m2a2. What forces act on block 2?

    The sign of a2 will be the same as the sign of g, as block 2 can not move upward. You can chose the sign of g either positive or negative, but think: if the block on the table moves to the right, block 2 moves downward. If block 2 moves downward by distance y, it pulls both piece of string by y so block 1 moves by 2y to the right. Moving to the right means positive displacement for block 1, that corresponds to a downward displacement of block 2. So the appropriate choice for block 2 is to take the downward direction positive.
    In case of such problems with bodies connected to a string, the motion of all bodies are determined by the motion of the string, and you can decide the sign of forces with respect to the motion of the string.

    ehild
     
  4. Aug 14, 2011 #3
    With all due respect ehild, I would like to add my own input.


    Whatever you do, take each mass as an individual!

    Use F = ma on both masses as you did, but don't combine mass.

    You understand that the tension is the same and is twice the magnitude for the 3kg mass. Which is good.


    "2T-m2*g=-a(m1+m2)"
    I would like to mention 2 things if I may.

    It's not 2T - m2*g. It's m2*g - 2T
    And as I mentioned before. don't combine masses.


    Lastly, the acceleration.

    As ehild said, you don't particularly need to apply signs to the acceleration. Just keep it simple, and keep everything positive.

    You should also note that the 3kg mass has only half the acceleration of the 10 kg mass given that tension is twice as strong.


    Well I hope this helps you!
     
  5. Aug 14, 2011 #4
    Thanks ehild and darth frodo. It didn't occur to me that the acceleration of 3kg would be 1/2 which makes sense since it has less mass than the 10kg block. The reason i picked negative acceleration is because for some reason my book( cutnell an johnson) had it in the examples. Would it really matter if I always keep the bottom negative and top positive? Thanks again for the help, I am getting a head start for my class that starts in a week.
     
  6. Aug 14, 2011 #5
    No the reason it has 1/2 the acceleration is because Tension is twice as strong. Mass is irrelevant.
     
  7. Aug 14, 2011 #6
    ok so generally in a pulley system the more tension ropes an object has, the slower its acceleration?
     
  8. Aug 14, 2011 #7
    Yes typically, but 2 is probably the most you will come across.

    Remember: Same rope same tension.
    Different rope different tension
     
  9. Aug 14, 2011 #8

    ehild

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    Gold Member

    The acceleration of the sliding mass is twice of the hanging one as the latter one pulls two pieces of string. The vertical pieces of the string become longer by y each, the length of the string does not change, so the horizontal part has to get shorter by 2y. Do the following experiment. Fold a sting and lay it on a table and fix one end. Pull out as it is shown by 2 cm. How much does the other end move?

    ehild
     

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  10. Aug 14, 2011 #9
    :tongue:thanks for the clarification
     
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