Why Did I Get the Newton's Law Problem Wrong?

[PhysicsinCalifornia]

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 $$m_b$$, rests on block A, with mass $$m_a$$, which in turn is on a horizontal tabletop. The coefficient of kinetic friction between block A and the tabletop is $$\mu_k$$, and the coefficient of static friction between block A and block B is $$\mu_s$$. 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 $$m_c$$ 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:

$$\SigmaF = m_ca$$
Block C
______
$$x:m_cg - T_c = m_ca$$
$$T_c = m_c(g-a)$$

Block B
--------
$$x: f_B = m_ba$$
$$\mu_sN_B = m_ba$$
$$y:N_b - m_bg = 0$$
$$N_b = m_bg$$

Block A
-------
$$x:T_A - f_A - f_B = m_Aa$$
$$T_A - \mu_kN_A - \mu_sN_B = m_Aa$$
$$y: N_A - N_B - m_Ag = 0$$

Max $$m_c$$
-------------------
$$T_C = m_c(g-a)$$
$$T_A = T_C$$

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

Is this correct?

 

[OlderDan]

$$a = \frac{\mu_sN_B}{m_B} = \frac{\mu_sm_Bg}{m_B} = \mu_sg$$

Is this correct?

With no slipping, the acceleration should be
$$a = \frac{m_C-(m_A+m_B)\mu_k}{m_A+m_B+m_C}g$$

 

[PhysicsinCalifornia]

With no slipping, the acceleration should be
$$a = \frac{m_C-(m_A+m_B)\mu_k}{m_A+m_B+m_C}g$$

Well, with my acceleration and your accerlation, I set those equal to each other and I got:

$$m_c = \frac{(m_A + m_B)(\mu_s + \mu_k)}{(1 - \mu_s)}$$

 

[OlderDan]

Well, with my acceleration and your accerlation, I set those equal to each other and I got:

$$m_c = \frac{(m_A + m_B)(\mu_s + \mu_k)}{(1 - \mu_s)}$$

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.

 

[PhysicsinCalifornia]

$$a = \frac{m_C-(m_A+m_B)\mu_k}{m_A+m_B+m_C}g$$

Well, I see that $$m_A + m_B$$ comes from the two blocks on top of each other because block A is at rest relative to block B.
In that case:
$$T - f_k = (m_A + m_B)a$$
$$T - \mu_kN = (m_A + m_B)a$$
$$T - \mu_k(m_A + m_B)g = (m_A + m_B)a$$
Plugging T with the equation that I had:
$$T_c = m_cg - m_ca$$ since Tension are equal
thus::

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

That matches with your a

 

[OlderDan]

That matches with your a

Then I think we are good on this one.