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Pulleys and blocks - find acceleration

  1. Jan 18, 2008 #1
    1. A pulley A on the end of a spring balance. On 1 end is a 16kg mass. On the other is pulley B. On one end of pulley B is a mass of 4kg. On the other end is pulley C. On one end of pulley C is a 1kg mass. On the other end is a 7/3kg mass. All pulleys are massless and frictionless. Find accn. of each block



    2. Relevant equations
    1.Too many pulleys= big confusion! (A very relevant equation here!) Just kidding.

    Fnet = ma




    3. I got the constraint equation as 4a(16) + 2a(4) + a(1) + a(7/3) = 0(no.s in bracket indicate accn. of block of that mass.

    Now what?
     
  2. jcsd
  3. Jan 18, 2008 #2

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    Start with the pulley C. Find the net force on it and the accns of the blocks hanging from it by the usual method. I do hope you know that. Then work your way up, one by one.

    In your eqns, you have assumed a single accn 'a'? Why?
     
  4. Jan 18, 2008 #3

    Single accn. a? Whats that supposed to mean? I've assigned each of them a seperate accn but just named them a(16) for convenience. For example I know a(16) refers to the accn of the block of 16kg.
     
  5. Jan 18, 2008 #4
    Ok. Now I've modified my accn.s. Here goes what I've done. I know its wrong.

    Starting from the inside,
    T3-10=a1-a4+a16
    70/3-T3=7/3(-a1-a4-a16)

    ==> 20=5a16-5a4-2a1

    Next,
    40-T2=4(a16-a4)
    T2-100/3=10/3(a16+a4)

    ==> 10=11a16-a4

    And finally,
    160-T1=16*a16
    T1-220/3=22/3*a16

    ==> a16=26/7 m/s/s which is wrong!!
     

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  6. Jan 18, 2008 #5

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    It looked like a*16, that's why I asked the Q. It's all right.
     
  7. Jan 19, 2008 #6
    check my answer tho. Why am I getting the wrong one?
     
  8. Jan 19, 2008 #7

    rl.bhat

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    T3-10=a1-a4+a16
    70/3-T3=7/3(-a1-a4-a16)

    How did you wright the above equations? Why the signs of a1 and a16 are changed in the second equation?
    If T1, T2 and T3 are the tensions in the strings tied to the masses 1kg, 4 kg and 16 kg, then a(1) = (T1 - 10)/1 = (70/3 - T1)/7/3. Find T1
    a(4) = (40 - T2)/4 = (T2 - 2T1)/(10/3). Find T2
    and a(16) = (160 - T3)/16 = (T3 - 2T2)/22/3. Find T3.
     
  9. Jan 20, 2008 #8
    Thank you very much. I'm done.
     
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