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Pulleys what is going on!

  1. Aug 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Pulley system, attachment below. Would be impossible to explain without a diagram.



    2. Relevant equations

    Wa = 200N
    Va0=2m/s
    t= 2 seconds
    Wb= 40N
    μK=0.2
    g=9.81m/s^2

    3. The attempt at a solution

    Okay my attempted solution was this:

    For block A:

    40N + Fn+Tension =m1a

    T - 40N*0.2=40N/9.81

    Block 2

    200 + Fn + Tension = m2a

    200N + T = 200N/9.81

    So lost to be honest!
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2014 #2

    Nathanael

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    Recheck the forces on each of the blocks (I didn't see you include the force of friction anywhere??)

    Also, please use the correct names of the blocks ("A" and "B") to avoid confusion :tongue: (you said "Block A" and "Block 2")



    Edit:
    Hint: (in case I go to sleep before you reply) it only takes 3 equations to solve this problem:
    The "equation of forces" on block A
    The "equation of forces" on block B
    And an equation which relates the acceleration of block A to the acceleration of block B
     
    Last edited: Aug 25, 2014
  4. Aug 25, 2014 #3
    Ans:
    final velocity= v(initial) + g(Wa-2μWb)/(4Wb+Wa)t
     
  5. Aug 25, 2014 #4

    Nathanael

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    That is correct, but Stelios already knows the answer. He doesn't just want the answer, he wants to understand it.
     
  6. Aug 25, 2014 #5

    andrewkirk

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    Gold Member

    The downward force on block A is its weight minus twice the frictional force on B (twice because of the mechanical advantage of the two-fold pulley). This will be constant because (IIRC) sliding frictional force does not vary with speed.

    That is ##F_A=W_A-2\mu_k W_B##

    The mass of A is ##m_A=W_A/g##

    The mass of B is ##m_B=W_B/g##

    The acceleration of A is ##a_A=F_A/(m_A+2m_B)## The coefficient of 2 for ##m_B## is there because of the reverse mechanical advantage of the pulley: a downward force on the pulley translates to half that force on the single rope leading to B.

    The speed at time ##t## will be ##v_{A0}+a_At##
     
    Last edited: Aug 25, 2014
  7. Aug 25, 2014 #6
    Thanks Andrew, I forgot to include note the fact that their is mechanical advantage as a result of two pulleys in the system.

    And as for block a and block 2 I think it was just a typo :(
     
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