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Homework Help: Pulley Block Mass Problem

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Two blocks are connected over a pulley. The mass of block A is 10 kg and the coefficient of kinetic friction between A and the incline 0.20. Angle [tex]theta[/tex] of the incline is 30 degrees. Block A slides down the incline at constant speed. What is the mass of block B?

    mech041fig01.jpg

    2. Relevant equations
    Fs=mgsin[tex]theta[/tex]

    Sin[tex]theta[/tex]= Fs/mg

    3. The attempt at a solution

    I tried using substition in order to find the second mass, but that route bore no fruit.

    9.8 x 10 x sin 30= 49
    Fs=49
    T- mgsin[tex]theta[/tex]=ma
    T-49=ma
    T=ma+49

    -T+m(subscript 2)g= M(subscript2)a (we substitute t for ma+49)
    -(m(subscript 1)a+49)+m(subscript 2)g= M(subscript 2)a
    -msub1-49+msub2g=msub2a

    all that boils down to is

    (9.8m(sub2)-49)/(10+msub2)= a

    How exactly am I supposed to find the second mass without knowing the acceleration?
     
  2. jcsd
  3. Oct 6, 2007 #2
    when you found the Fs that was static friction? because the block is under a constant velocity so it is actually kinetic friction.

    and also you did 9.8x10xsin30 but you didn't include the frictional constant in there.
     
  4. Oct 6, 2007 #3
    Alrighty. But even if I do include the kinetic friction, how would I find a?
     
  5. Oct 6, 2007 #4
    well the problem states that Block A (m2) is sliding down at a constant velocity. constant velocity means what type of acceleration?

    and if block A has acceleration (a) and we're assuming that the string doesn't stretch what does that say about block B's acceleration?
     
  6. Oct 6, 2007 #5
    Constant velocity means no acceleration
    And ah, ok :)
     
  7. Oct 6, 2007 #6
    yes also your calculation for friction was wrong it should be:

    [tex]\mu_kN[/tex]

    and you know N as one of the components of mg and you know the downward is mgsintheta
     
  8. Oct 6, 2007 #7
    Right right. Thanks so much :)
     
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