[Physics] Tension in the string of a pulley

Question

In the diagram above why is the tension of the string attached to the pulley at "A"(the string attached to roof) equal to 2T?

Why is it not Mg+(M+m)g?(considering that the pulley is mass less)
I have trouble understanding

Answer 1

It would if the weights weren't accelerating but they are accelerating because they are not of equal mass. So to get the overall force on the pulley you have to take acceleration into account; subtract whatever force the acceleration creates from the force created by gravity

Tension is not going to equal to what the mass is at the ends of the strings when the masses are accelerating. The competing forces are Mg and (M+m)g so subtract those to get the overall force that drives acceleration. Acceleration is force over mass so $$A =\frac{F_a}{M_t}=\frac{Mg-(M+m)g}{M+(M+m)}=\frac{Mg-Mg+mg}{2M+m}=\frac{mg}{2M+m}$$

Mass is additive so we always add masses up. But forces can be subtractive as the case here. As you can see, the smaller the mass difference is (m) the less net force there is and the bigger M is the more mass the overall system has which means less acceleration created by the net force.

Now let's calculate the net force on the pulley by taking the overall force created by the masses and subtracting the force created by acceleration. $$T_t=F_g-F_a=M_tg-M_tA=M_t(g-A)$$

$$T_t=(2M+m)(g-\frac{mg}{2M+m}) = (2M+m)g - \frac{(2M+m)mg}{(2M+m)}$$

$$T_t=2Mg+mg-mg = 2Mg$$

So we can see that the pulley only has to support the smaller weight twice. Any differences in the two weights is in free fall. This also means that the wire holding onto the bigger weight only has to support the smaller weight and this makes intuitive sense since the bigger mass is falling and the wire is holding onto the smaller mass taking it along for the ride.