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Movable pulley and two connected particles on horizontal tables

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Two particles of masses 6 kg and 8 kg rest on two horizontal tables.The coefficient of friction between both particles and their respective tables is (1/2).The particles are connected by a smooth inextensible string which passes over smooth pulleys and under a smooth movable pulley of mass M kg

    Show that none of the particles will move if M is less than or equal to 6


    2. Relevant equations
    F=MA

    3. The attempt at a solution
    t = Tension
    a = acceleration of 6 kg particle
    b = acceleration of 8 kg particle

    for the six kg particle 3g - t = 6a
    a = (3g - t)/6

    for the eight kg particle 4g - t = 8b
    b = (4g-t)/8

    for pulley M

    2t -Mg = M(a+b)/2


    4t -2Mg = M(3g-t)/6 + M(4g-t)/8

    96t -48Mg = 4M(3g-t) + 3M(4g -t) multiplied my 24................

    96t = 48Mg + 12Mg -4Mt + 12Mg - 3Mt

    96t + 7Mt = 72Mg

    t(96 + 7M)= 72Mg

    t = 72Mg / (96 + 7M)

    Since its not accelerating

    2t - Mg = (3g - t) + (4g -t) Sum of the forces are equal

    4t= Mg + 7g

    288Mg = (Mg + 7g)(96 + 7M) Substituting value for t

    288Mg = 96Mg + 672g +7Mg^2 + 49Mg

    7Mg^2 - 143Mg +672g = 0

    68.6M^2 - 1401.4M + 6585.6

    When I use the minus b formula i get M = 13.1 and 7.32
    I'm convinced I've done this completely wrong.I was hoping I would get a solution of M = 6.
    Any help would be appreciated.
     
  2. jcsd
  3. Jun 1, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Keep it simple.

    What's the maximum amount of tension that the 6 kg particle can support before it begins to move? Assuming everything is in equilibrium, what would the mass of the hanging pulley have to be to create that tension? Can the other mass support that tension without moving?

    Do a similar analysis for the 8 kg particle.
     
  4. Jun 1, 2013 #3
    Since the limiting friction of the 6 kg particle is 3g would that not mean that the tension would have to be 3g? Also would that mean the tension in the 8 kg particle would be 4g?.I thought since its the same wire it would have to have the same tension every where?
     
  5. Jun 1, 2013 #4
    I think I get it now.Since the tension is split between the two sides of the wire of the pulley would it have to be to be 6 since 6g/2 = 3g
    which would mean the 6kg particle wouldn't move because the tension is just enough to cancel out the limiting friction............... 3g -3g = 0 N
    and the 8kg particle also wouldn't move since the tension is less than the limiting friction 4g - 3g = g N
     
  6. Jun 1, 2013 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly. And if the mass of the pulley were greater than 6 kg, there could not be equilibrium.
     
  7. Jun 1, 2013 #6
    Alright, thanks for the help.
     
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