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Friction problem with pulleys

  1. Nov 15, 2005 #1
    The block with mass m2=0.50kg is found to have a speed of 0.30m/s after it has dropped 0.80m. how large a (kinetic) friction force retards the motion of the block with mass m1=2.0kg?

    (side note from teacher)
    To solve this problem, you can use a string constraint. the total length of the string (L) is constant. However, as block m1 moves to the right, the portion of the string that is horizontal shortens by delta y and this length is distributed over the portion of the string taht is vertical according to the relationship; |delta x|=1/2|delta y|. By taking a derivative with respect to time twice, we have a1=2a2.


    i think i did this completely wrong

    i drew a free body diagram for each,for 1st free body diagram (mass1) and found m1 had Fn of 19.6

    on the 2nd free body diagram (mass 2), i had Fg-Ft = ma
    then, 4.9 N - 2Ft =0 , Ft = 2.45, then i said the Ft in the first free body diagram was also equal to 2.45, and therefore the kinetic friction force equals 2.45

    its probably wrong, since i didnt use any info my teacher gave in the side note
  2. jcsd
  3. Nov 16, 2005 #2


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    You put the accln of m2 as zero - it should be a2.
  4. Nov 16, 2005 #3
    okay, i did it like u said, but now im stuck

    for the mass 1 fbd, i got
    Ft - Ff = a1

    for the mass 2 fbd, i got
    Fg - 2Ft = ma2 (is it 2Ft???)
    then isolate for a2
    4.9 - 2Ft = 0.5a2
    9.8 - 4Ft = a2

    and the equaion from fbd #1:
    Ft - Ff = ma1
    since a1 = 2a2
    Ft - Ff = 4a2 (1)
    then i substituted a2 into equation (1)
    Ft - Ff = 4(9.8 - 4Ft)
    Ft - Ff = 39.2 - 16Ft
    17Ft = 39.2 + Ff

    okay im stuck, how do i find Ft???
  5. Nov 16, 2005 #4


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    You have,

    9.8 - 4Ft = a2 (from fbd #2)
    Ft - Ff = 4a2 (from fbd #1)

    eliminate Ft from these two eqns, giving you,

    9.8 - 4Ff = 17a2

    Now use the info about the movement of m2 given in the question to work out a2.
    Substitute for a2 and solve for Ff.
  6. Nov 16, 2005 #5
    so, a2 = 0.30m/s / 0.80m = 0.375 m/s^2
    and then sub it in to get Ff = 0.856 N
    oops, thats not the acceleraton
    hmm, u need kinematics??
    is the veloctiy given v1 or v2
    a = 9.8 m/s^2
    d= 0.80
    if velocity given is v2, so v1 = 0?
    Last edited: Nov 16, 2005
  7. Nov 16, 2005 #6
    actually i have a question, i used 2Ft in one of the equations, and im wondering if this is right.

    for this step i used 2Ft :
    for the mass 2 fbd, i got
    Fg - 2Ft = ma2 (is it 2Ft???)
  8. Nov 16, 2005 #7


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    Yes, that's right.

    Nope, that's wrong :smile:
    use the kinematic eqn,
    v² = u² + 2as
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