Dynamics of two pulleys I have been trying to do this exercise for 2 days but I'm sure that I'm negleting something... 
Here is the sketch of the situation: 


$$m_A= 4M \\
m_B= M \\
m_C=2M$$
All the strings cannot be extended and have no mass. The pulleys $c_1$ and $c_2$ have no mass. 
The table is rough.
The absolute value of acceleration  is the same for A and B. 
Find the dinamic friction coefficient between A and the table, and the ratio between the tensions. 

My attempt:
$T-2Mg+U=2Ma$ (equation for the body C)
$T-Mg+U=Ma$ (equation for the body B)
$U-F_a=4Ma$ (equation for the body A) ($F_a$= friction force)
From eq 1) and eq 2) I obtain $a= g/3$
But then I can't use this result for the third equation because I don't know the value of U!
I'm sure that I'm neglecting something.. but what??
 A: If you take the marked portion of the image then there is a property for pulleys in which forces get magnified.

In your case, the tension on the strings that hold B and C is T
By the property of magnification, The tension U becomes equal to $$U=T+T=2T$$
So the third equation becomes, $$2T-F_a=4Ma$$ 

Here is another example of magnification of forces by pulleys:

A: You messed up on  the kinematics.  The key to solving a pulley problem is to get the kinematics correct.  If a is the acceleration of body A to the right, what makes you think that the upward acceleration of body C is a?  What makes you think that the upward acceleration of body B is a?  If the downward acceleration of mass C relative to pulley c2 is a*, what is the total downward acceleration of mass C in terms of a and a*?  What is the acceleration of mass B in terms of a and a*?
Also, the force U is not acting on either B or C.  Have you drawn free body diagrams for these two objects.  Have you drawn a free body diagram for pulley c2?
ADDITION: 
The absolute accelerations of A and B cannot be the same unless the lower pulley c2 is not rotating.  If that were the case, then the tensions in the two sections of the string going over c2 would have to be different.  So, I am going to assume that masses A and B do not have the same acceleration, and that the downward acceleration of mass C relative to pulley c2 is a*.  In this case, the downward acceleration of mass C is (a + a*) and the downward acceleration of mass B is (a - a*).  
FORCE BALANCES:
$$U-F_a=4Ma\tag{FB on Mass A}$$
$$2T-U=0\tag{FB on pulley c2}$$
$$2Mg-T=2M(a+a*)\tag{FB on Mass C}$$
$$Mg-T=M(a-a*)\tag{FB on Mass B}$$
My first two force balance equations are the same as yours, but the 3rd and 4th are not because, if the pulley c2 rotates frictionlessly, the accelerations of masses B and C cannot be a.  
We have 4 equations in the 5 unknowns, $F_a$, T, U, a, and a*.  So the best we can do is to eliminate T, U, and a*, and express #F_a# in terms of a (the value of which would have to be measured experimentally).
NEW ADDITION
Yikes, I just realized that it actually is possible (in a very special case) for the absolute values of the accelerations of A and B to be equal and for the pulley c2 to still rotate with no friction (with the tensions in the two strings on either side of the pulley being equal to T).  This would be the case if we somehow observed experimentally that the accelerations of masses B and C relative to pulley c2 are exactly equal to 2a.  
In this case, we would have that $a^*=2a$.  This would mean that the overall resultant downward acceleration of mass B would be -a, which is equal in absolute magnitude to the acceleration of mass A.  If we make this substitution into the force balances for masses B and C, we then obtain:
$$2Mg-T=2M(3a)\tag{FB on Mass C}$$
$$Mg-T=-Ma\tag{FB on Mass B}$$
This reduces the force balance equations to 4 equations in the four unknowns $F_a$, T, U, and a.  So, the problem is now determinate.  However, for all this to work, we would somehow have to know (observationally, for this particular set of masses) that the upward acceleration of mass B is exactly equal to the acceleration of mass A to the right.  We could not know this by applying first principles.  So, in a way, this is kind of a contrived and unrealistic problem.  
Solve the equations for the four unknowns, and see if the match they answer in your answer key.  I'm confident they will.
