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(adsbygoogle = window.adsbygoogle || []).push({});

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?

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