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Tension in a pulley system

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data

    In the figure below, M1=1.00 kg, M2=2.00 kg and M3=4.00 kg. Theta is 30.0°. The pulley and all surfaces are frictionless. Find the tension in the two strings and the direction (entire system to the left/counterclockwise or entire system to the right/clockwise) and magnitude of the acceleration.

    Diagram attached.

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    m1=A, m2=B, m3=C
    I drew a free-body diagram for each of the blocks and I came up with these equations.
    A=T1-W=ma
    B=-T1-W+Fn+T2=ma
    C=-T2-W+Fn=ma (I rotated it to have the block on a straight surface)

    After doing these steps I am completely lost. I think I did my force equations correctly, but if someone can help me with this, step by step, it would be GREATLY appreciated.
     

    Attached Files:

  2. jcsd
  3. Dec 14, 2011 #2

    ehild

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    Remember, force is vector quantity. The components add up separately.
    You can take this as a one-dimensional problem, considering the motion along the string. The force components parallel to the strings add up to ma, where a is the acceleration in the direction of the string: If A accelerates upward, B will accelerate to the right and C accelerate down the slope.
    The normal force components cancel with gravity. You do not need to include them into the equations.

    C feels a force component along he slope due to gravity. That is missing from equation C.


    ehild
     
  4. Dec 14, 2011 #3
    so you would separate them into Fx and Fy vectors?

    Fx = -T1+T2 =ma
    Fy = T1 - W = ma

    does that look correct?
     
  5. Dec 14, 2011 #4

    ehild

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    I said to relate the acclerations and forces parallel with the string for each blocks. I do not understand what your your equations mean. Do they refer to the same mass???

    ehild
     
  6. Dec 14, 2011 #5
    they were the x components from all 3 blocks and the y components from the blocks. i'm not sure how to do what you're saying to do, any help?
     
  7. Dec 14, 2011 #6

    ehild

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    These are your equations from post 1.

    A:T1-W1=m1a
    B:-T1[STRIKE]-W+Fn[/STRIKE]+T2=m2a
    C=-T2[STRIKE]-W+Fn[/STRIKE]+Fslope=m3a

    Omit the Fn-W terms.

    Do you know what force acts downward along the slope, because of gravity?

    ehild
     
  8. Dec 14, 2011 #7
    kinetic friction? i thought it would have been mgsin∅
     
  9. Dec 14, 2011 #8

    ehild

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    It is not kinetic friction, but it is m3gsin(theta). Substitute the values given for the masses and for theta and write up the equations again with the numbers.


    ehild
     
  10. Dec 14, 2011 #9
    A:T1-(9.8)(1.00)=(1.00)a
    B:-T1+T2=(2.00)a
    C=-T2+(4.00)(9.8)(sin(30))=(4.00)a

    do these look correct?
     
  11. Dec 14, 2011 #10
    When you have calculated the acceleration a useful double check is to look at the whole system and recognise that T1 and T2 are INTERNAL forces. There are only 2 external forces (W1 and Fslope) and it is the resultant of these 2 forces that cause acceleration of the TOTAL mass. You should get the same acceleration !!!but this will not give you T1 and T2.
    You need the full analysis as outlined by ehild.
     

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  12. Dec 14, 2011 #11

    ehild

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    Good!Now the trick comes: Add all three equation. What do you get?

    ehild
     
  13. Dec 14, 2011 #12
    T1-(9.8)(1.00)=(1.00)a
    T1+T2=(2.00)a
    T2+(4.00)(9.8)(sin(30))=(4.00)a

    T1-(9.8)=(1.00)a --- T1 = (1.00)a/(9.8)
    T2+19.6=(4.00)a ---- T2 = (4.00)a/(19.6)

    Insert T1 and T2 into 2nd equation

    (1.00)a/(9.8) + (4.00)a/(19.6) = (2.00)a
     
  14. Dec 14, 2011 #13

    ehild

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    You missed some minuses. And you did not isolate T1 and T2 correctly.


    ehild
     
  15. Dec 14, 2011 #14
    T1 = (1.00)a + 9.80
    -T2 = (4.00)a - 19.6

    that better?
     
  16. Dec 14, 2011 #15

    ehild

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    OK. Substitute T1 and T2 into the second equation.

    ehild
     
  17. Dec 14, 2011 #16
    i come up with

    -(1.00a + 9.80)-(4.00a-19.6)=2.00a
    -29.4 - 5a = 2.00a
    -29.4 = 7a
    -4.2 = a

    does that look correct? would that be the magnitude of the acceleration?
     
  18. Dec 14, 2011 #17

    ehild

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    No... How did you get -29.4?

    Resolving the parentheses: -1.00a - 9.80-4.00a+19.6=2.00a

    ehild
     
  19. Dec 14, 2011 #18
    sorry, math error. i end up with a = 1.4

    (-9.80+19.6) = 9.8 --- (-1.00a-4.00a) = -5a
    9.8 - 5a = 2.00a
    9.8 = 7a
    1.4=a
     
  20. Dec 14, 2011 #19

    ehild

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    Correct. And the direction you assumed proved to be correct.
    Now find the tensions.


    ehild
     
  21. Dec 14, 2011 #20
    T1 = 1.00*1.4+9.80 = 11.2
    -T2 = 4.00*1.4-19.6 = -14

    (thanks for all your help and patience so far btw.)

    i may be jumping ahead, but in order to find the direction don't you need x and y?
     
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