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Block and disk on double inclined plane

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data

    https://courses.edx.org/asset-v1:MITx+8.MechCx_2+2T2015+type@asset+block/rolling_quiz_3.svg

    A block of mass mb and a disk of mass md and radius r are placed on a symmetric triangular slope connected with a massless string over a massless pulley as shown above. The string is connected to a center axle of the disk so that the disk is free to rotate. The moment of inertia of the disk about its axle is I=1/2*m_d*r^2. The coefficient of static friction between the slope and the block/disk is 0.05 and the coefficient of kinetic friction between the slope and the block/disk is 0.15. The angle θ is 30∘.

    Reminder: sin(30∘)=cos(60∘)=12 and cos(30∘)=sin(60∘)=3√2

    Find the maximum ratio m_b/m_d such that the disk still rolls without slipping up the hill.

    2. Relevant equations
    Just to be clear I'm using:
    t = torque and
    T = tension

    t_net = I*alpha
    F_net = m*a

    3. The attempt at a solution
    Since its the maximum ratio m_b/m_d, the static friction must be at its max: mu_s*m*g*cos(theta) so

    t_net = r*mu_s*m_d*g*cos(theta) = I*alpha = 1/2*m_d*r^2*a/r = 1/2*m_d*r*a

    This simplifies to:

    2*mu_s*g*cos(theta) = a

    Then for the disk the addition of N+F_g = m_d*g*sin(theta) and:

    F_net = T - m_d*g*cos(theta)*mu_s - m_d*g*sin(theta) = m_d*a
    T = m_d*g*cos(theta)*mu_s + m_d*g*sin(theta) + m_d*a

    and for the block (I'm not sure if I've messed up the signs here, I guessed that the positive axis should point towards the acceleration):

    F_net = m_b*g*sin(theta) - T - m_b*g*cos(theta)*mu_k = m_b*a
    T = m_b*g*sin(theta) - m_b*g*cos(theta)*mu_k - m_b*a

    m_d*g*cos(theta)*mu_s + m_d*g*sin(theta) + m_d*a = m_b*g*sin(theta) - m_b*g*cos(theta)*mu_k - m_b*a

    then I plugged in 2*mu_s*g*cos(theta) = a and simplified:

    m_d(mu_s + tan(theta) + 2*mu_s) = m_b(tan(theta) - mu_k - 2*mu_s)

    and finally:

    m_b/m_d = (mu_s + tan(theta) + 2*mu_s)/(tan(theta) - mu_k - 2*mu_s) = 2.22193

    But I just realized I forgot a mu_s in my original solution so is this correct?
     
  2. jcsd
  3. Aug 23, 2015 #2

    TSny

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    Overall, your work looks correct and I think your final answer is correct. I noticed a couple of trivial typos:
    Also, I don't follow your statement:
    But I don't think you used this.
     
  4. Aug 23, 2015 #3
    Yes it was correct, thank you.
     
  5. Aug 23, 2015 #4

    haruspex

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    It is physically impossible for the kinetic friction coefficient to exceed the static friction coefficient. Did you quote them correctly?
     
  6. Aug 23, 2015 #5
    Yes, I copied and pasted it... thats strange
     
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