Tension in the string of a pulley

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

• It would be $Mg+(M+m)g$ if the masses were not moving... Commented Feb 16, 2018 at 16:53
• 1) You need to specify if the pulley is masses and if the system is in motion if you want numerical help. 2) Conceptually, pulley "doesn't know" anything about the masses C and E. What it "does know" though is that there is a thing over it, which pulls it down with the force $2T$. The string (!) "knows" that there are two masses attached to it. Using Newton's 2nd law you can equate what pulley "knows" to what the string "knows". Commented Feb 16, 2018 at 17:02

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.

The way I always think of it is that as soon as an object is accelerating, it is "using up" some of the force for acceleration. In this case the heavier object that is falling is "using up" some of the $(m+M)g$ force, so it can not use that full force anymore to pull on the string.

For the smaller weight it is the other way around, since it is accelerating up the string needs to pull harder than $Mg$.

You can see it is not going to be the simple sum of the two weights.