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3 Pulleys System, find the acceleration of both mass.

  1. Jul 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is a picture that I putted together(see attached picture).
    m1= 0.235 kg
    N = m1g = 2.303 N
    m2= 0.350 kg
    m2g = 3.430 N
    fk= 0.590 N

    By measurement, I confirmed (d1)=4(d2).
    Therefore, (v1)=4(v2), a1= 4(a2).

    2. Relevant equations
    ƩFx=(T1)-fk=(m1)(a1)
    ƩFy=(m2)g-(T2)/2=(m2)(a2)
    (T2)-4(T1)=0

    3. The attempt at a solution
    Try to find a1 and a2.
    [(m2)g-(m2)a2)]/2-fk=(m1)4(a2)
    1.715-0.175(a2)-0.590=(.235)(4)(a2)
    1.125=1.115(a2)
    (a2)=1.01 m/s^2
    (a1)=4.036m/s^2

    please correct me if I am wrong. Thank you.
     

    Attached Files:

  2. jcsd
  3. Jul 1, 2012 #2

    ehild

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    Hi Bonten!

    The expressions in red are not right. Think them over.


    ehild
     
  4. Jul 1, 2012 #3
    I got hang up on this all weekend.
    1)would it be ƩFy=(m2)g-(T2)=(m2)(a2) ?
    2)would (T2)-2(T1)=0 ? since T2 exerted the force from (m2)(g2) to the pulley (P1) which divided the tension by 2.
    3)but the T2 got divided by 2 by the pulley P2 which directly attached with m2. here is the diagram, it applies with p1 pulley as well.
     
  5. Jul 1, 2012 #4
  6. Jul 1, 2012 #5

    ehild

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    No. Two pieces of string act upward, each with tension T2. m2g-2T2=m2a.

    Yes, the pulley is massless so the sum of forces acting to it must be zero. 2T1 acts to the left and T2 acts to the right: T2-2T1=0.

    Your diagram applies for equilibrium, when W-2T=0, so T=W/2.



    ehild
     
  7. Jul 1, 2012 #6
    It can be seen clearly in your diagram, 2×T2 is opposing the motion of M2.
     
  8. Jul 1, 2012 #7
    this is system not quite in equilibrium, since the downward motion of m2g will create an acceleration for the whole system, but a2 will be 4 times smaller than a1.
     
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