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Need help with newton's law problem!

  1. Oct 12, 2006 #1
    Hi, I needed to calculate this for my quiz couple days ago, but I didn't get the right answer. Professor told me that I set it up correctly, so I don't understand why I got it wrong. Can anyone help? I resolved this problem, so I just need to verify the answer. Thanks

    Q: Block B, with mass [tex]m_b[/tex], rests on block A, with mass [tex]m_a[/tex], which in turn is on a horizontal tabletop. The coefficient of kinetic friction between block A and the tabletop is [tex]\mu_k[/tex], and the coefficient of static friction between block A and block B is [tex]\mu_s[/tex]. A light string attached to block A passes over a frictionless, massless pulley and block C is suspended from the other end of the string. What is the largest mass [tex]m_c[/tex] that block C can have so that blocks A and B still slide together when the system is released from rest?

    Here is what I have so far:

    [tex]\SigmaF = m_ca[/tex]
    Block C
    [tex]x:m_cg - T_c = m_ca[/tex]
    [tex]T_c = m_c(g-a)[/tex]

    Block B
    [tex]x: f_B = m_ba[/tex]
    [tex]\mu_sN_B = m_ba[/tex]
    [tex]y:N_b - m_bg = 0[/tex]
    [tex]N_b = m_bg[/tex]

    Block A
    [tex]x:T_A - f_A - f_B = m_Aa[/tex]
    [tex]T_A - \mu_kN_A - \mu_sN_B = m_Aa[/tex]
    [tex]y: N_A - N_B - m_Ag = 0[/tex]

    Max [tex]m_c[/tex]
    [tex]T_C = m_c(g-a)[/tex]
    [tex]T_A = T_C[/tex]

    Now I need to solve for a:
    [tex]m_Aa + \mu_kN_B + \mu_sN_B = m_c(g-a)[/tex]
    [tex]m_Aa + \mu_k[g(m_A + m_B)] + \mu_s(m_bg) = m_cg - m_ca[/tex]
    [tex]a = \frac{\mu_sN_B}{m_B} = \frac{\mu_sm_Bg}{m_B} = \mu_sg[/tex]
    [tex]m_A(\mu_sg) + \mu_k[g(m_A + m_B)] + \mu_s(m_Bg) = m_cg - m_c(\mu_sg)[/tex]
    [tex]m_A\mu_sg + \mu_k[g(m_A + m_B)] + \mu_sm_bg = m_c(g - \mu_sg)[/tex]
    [tex]m_c = \frac{m_A\mu_sg + \mu_k[g(m_A + m_B)] + \mu_sm_Bg}{g - \mu_sg}[/tex]

    Is this correct?
  2. jcsd
  3. Oct 12, 2006 #2


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    With no slipping, the acceleration should be
    [tex]a = \frac{m_C-(m_A+m_B)\mu_k}{m_A+m_B+m_C}g[/tex]
  4. Oct 12, 2006 #3
    Well, with my acceleration and your accerlation, I set those equal to each other and I got:

    [tex]m_c = \frac{(m_A + m_B)(\mu_s + \mu_k)}{(1 - \mu_s)}[/tex]
  5. Oct 12, 2006 #4


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    So, I take it you accept my equation for (a). Can you see where it comes from? You should be able to reproduce it from your free body equations.
  6. Oct 12, 2006 #5
    Well, I see that [tex]m_A + m_B[/tex] comes from the two blocks on top of each other because block A is at rest relative to block B.
    In that case:
    [tex]T - f_k = (m_A + m_B)a[/tex]
    [tex]T - \mu_kN = (m_A + m_B)a[/tex]
    [tex]T - \mu_k(m_A + m_B)g = (m_A + m_B)a[/tex]
    Plugging T with the equation that I had:
    [tex]T_c = m_cg - m_ca[/tex] since Tension are equal

    [tex](m_cg - m_ca) - \mu_k(m_A + m_B)g = (m_A + m_B)a[/tex]
    Solving for a gives:
    [tex]a = g\frac{m_c - \mu_k(m_A + m_B)}{m_A + m_B + m_c}[/tex]

    That matches with your a
  7. Oct 12, 2006 #6


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    Then I think we are good on this one.
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