Pully and Inclined Ramp 100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The pulley has a mass of 10 kg and a radius of 10 cm. However the pulley does not have a friction.
A). What is the Max weight the hanging mass can have so that the box remains at rest.
B). The system is at rest with the weight of the hanging mass found in Part A, but a speck of dust comes to rest on the weight on the hanging mass, making the system unstable. What is the acceleration of the box at the very moment it starts moving?
See diagram

Part A)
 Here is how I setup my equations:
 Force Gravity on Block = (100)(9.8)sin(30) = 490
 NF = 100cos(30) - Tensionsin(22) = 50√3 - Tensionsin(22)
 Mass on Ramp = Ma
 Mb = Hanging mass = Tension-(Mb)(9.8) = 0  
Net force on Box = 0, forces equation =
 Tensioncos(22) - Ma gsin(30) - ( Ma gcos(30) - Tensionsin(22) )μ = 0  
I then proceeded to plug in Ma, g = 9.8, and μ = .60
 Then using the equation above and the Hanging mass equation I solved for Mb
 I got
Mb = ~ 88 kg  
Part B)
 I am fairly confident about my answer of Part A.. but now unsure exactly how to solve part B.
 I know the radius of the pulley is 10 cm and the mass is 10 kg. 
I had to find the equations for the two tensions. However, one thing I had to form a new equation for (at least I think ) is the Normal Force. Before I knew the vertical component of the normal force would just be 0, but now since it is moving it could also change.  
My new equation is:
NF = 100gcos(30) - Tensionsin(22) = Ma A
 NF = 100gcos(30) - Tensionsin(22) - Ma A = 0
 I know I need to substitute this into my original equation, so I can use it to calculate the force of Friction
 Tension2*cos(22) - Magsin(30) - ( 100gcos(30) - Tension2*sin(22) - Ma A )μ = 0
 put this equation in terms of Tension2, but I must leave acceleration in terms of A
 Tension2 = 129A + 770
 My equation for tension1 was easier, just:
 Tension1 = Mb A - Mb A     
Now I plug them into a Torque equation taking into account the mass of the pulley and radius
 Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝  
Now here is where my struggles appear, I have two unknowns with A and Mb. I've gotta imagine that the intention of Mass B is that the speck of dust would represent just about nothing, and the point of it was to explain that the system would be in motion and I should just account for it as the same mass from before, but I don't know. ANy advice on this would be great!  
Let me know if there is anything I should elaborate on something.. ,thanks in advance!  
 A: 
Problems like these are very hard to solve without a force diagram. We consider everything to be stationary.
Now consider parts of the system in isolation.
1. The pulley:
$$\vec{F_c}+m_B\vec{g}+\vec{T}=0\tag{1}$$
We'll use it later to determine $T$.
2. The mass on the incline:
$$\vec{T}+\vec{F_N}+\vec{F_F}+m_B\vec{g}=0\tag{2}$$
3. The whole system:
$$\vec{F_c}+\vec{F_N}+\vec{F_F}+m_B\vec{g}+m_A\vec{g}=0\tag{3}$$
Now use $(3)$ to determine $F_c$:
$$\Sigma F_y=0$$
So:
$$F_c-m_Bg-m_Ag+m_Ag\cos\alpha-m_A\mu g\cos\alpha\sin\alpha=0\tag{4}$$
(Note: $\beta=22°+33°$)
As $(4)$ only contains knowns, $F_c$ can now be determined.
Now apply $\Sigma F_y=0$ to $(1)$:
$$F_c-m_Bg-T\sin\beta=0\tag{5}$$
This determines $T$.
Then apply $\Sigma F_y=0$ to $(2)$:
$$T\sin\beta-m_Ag+m_A\mu g\cos\alpha\sin\alpha=0\tag{6}$$
That determines $m_A$.
As regards your second question, if $m_B$ is too high then $(6)$ is no longer true and becomes:
$$T\sin\beta-m_Ag+m_Ag\mu \cos\alpha\sin\alpha=m_Aa\sin\alpha\tag{7}$$
Where $a$ is the acceleration.
