Three boxes and a pulley. Why is the tension calculated differently? Three boxes A, B and C are connected by a wire. Boxes B and C are on a table. Box A is suspended across a pulley. Assume no friction and that the mass of the wire is negligible.
Diagram:
[C]--[B]--(pulley)
___________  |
  |    |    [A]
  |    |    


MA = 3 kg
MB = 4 kg
MC = 2 kg

How big are the forces operating horizontally on B?
My textbook gives the following solution:

SC = 6,5N.
SA = 20 N

I believe this was calculated as follows:

a = (9.81 m/s2 * 3 kg) / 9 kg = 3,27 m/s2
SA = (MC + MB) * a = 6 kg * 3,27 m/s2 = 19,62 N = 20 N
SC = MC * a = 2kg * 3,27 m/s2 = 6,54 N = 6,5 N

So the the tension caused by C is proportional simply to the mass of C. But, the tension caused by A is proportional not to the mass of A, but to the summed masses of C and B. Why is this?
To clarify my question, why is SA not simply:

SA = MA * a

The answer is probably that A pulls on both B and C together, but then why not:

SC = (MB + MA) * a

 A: To clarify my question, why is $S_A$ not simply $M_{A}a$
It is.
First of all, are you sure the book says the net force on Box A is 20 N and not that the tension in the string is 20N?
You said you believe
$$S_{A} =(M_{C}+M_{B})a=19.62$$
But that is not the force on box A. It is the tension in the string. The tension in the string is causing the acceleration of the combination of the two boxes B and C. 
Do a free body diagram on box A. You have the force of gravity acting downward of $M_{A}g$ and the tension in the string acting upward of 19.62 N. So the net force on box A is
$$M_{A}g-T=(9.81)(3)-19.62= 9.81 N$$
That is equal to, as you said you believe, $M_{A}a=(3)(3.27)=9.81N$.
So you were right about the force on A. If the book said the net force on box A is 20 N, it's wrong. 20 N is approximately the tension in the string.
Hope this helps.
A: By string constraints we have  acceleration of all the blocks equal to  say some value A. 
  Your clam that S= (Ma)A 
Is wrong as Newton's  second law states that the net force on an object is proportional to it's acceleration and this net force in this case is( mg-S) and not S .
Hope it helps 
Also the tension in the string can be easily calculated using the fbd of the system of B and C as using this we can easily ignore the internal tension between them and also  find the tension S  using the formulae S=(m1+m2)A
