How can a pulley work? 
Assuming $g=10m/s^2$, the $2M$ mass would exert a $20M$ force on $M$, and by newton's third law, the $M$ mass would exert an equal and opposite force of $20M$ on the $2M$ mass.

How would this move? Sorry if this is a really stupid question.
EDIT: Thanks a lot guys! I had forgotten that the two had the same acceleration, and since the masses are different, the net force acting on either had to be different.
 A: Firstly, we have to assume that the string is light and inextensible. That way, the tension is constant in the string, and the weight of the string is negligible. Also, the pulley must be smooth to avoid friction. Therefore, we can use $F=ma$ to calculate the acceleration. Both masses will be modelled as connected particles, so must have the same acceleration, $a$.
$F_{1}=m_{1}a$
With $m_1$ being the mass of m. Calculating tension, $T$,
$T-m_1g=F=m_1a$
Doing the same for $m_2$,
$m_2g-T=F=m_2a$
As $m_1=m$, and $m_2=2m_1$:
$T-mg=ma$ and $2mg-T=2ma$
Thus;
$T=ma+mg=2mg-2ma$
$ma+mg=2mg-2ma$
$a+g=2g-2a$
$3a=g$
$a=\frac{g}{3}$
Therefore, the system is moving, with the second ball moving down with acceleration $\frac{g}{3}$ and the first ball moving up with acceleration $\frac{g}{3}$ also.
A: Do a free body diagram individually to each block: 
To the 2M block there is 2 forces acting on it: T-W=m1*a1
To the M block there is 2 forces as well: T-W=m2*a2
Notice that m1=2M ; m2=M ; and a1=-a2, since they have equal intensities of accelerations but opposite signs; In 1 rope the tension is equal along the whole rope.
Solve the system of equations:
T-2M*10=2M*(-a2) 
T-M*10=M*a2 
T=-a2*2M + 20M
-a2*2M +20M -M*10=M*a2 
10M= 3M*a2 
a2=10/3 m/s^2.
Tension is just an internal force of the system That connects the 2 bodies and transmits an action from one to another. So, if the bodies wouldnt be attached to each other they would fall with acceleration 9.8 m/s^2, since they are connected and 1 is heavier then the other, 1 will fall and the other will lift, with an acceleration of 10/3 m/s^2, because the 2M block acelerates the system, and the M block retards the system.
A: Your question is not stupid. It's basic and important to know.
You're forgetting about the tension. At the point where the string is attached to the 2M block, two forces are acting: weight 2Mg straight down and an unknown force T straight up. These are not action-reaction pairs. Same goes for block M: weight Mg straight down, an unknown tension force T, up. 
As pointed in the previous answer, T turns out to be a little bigger than Mg and a little smaller than 2Mg. You can calculate this if by replacing some number for M.
