Why is $F = mg - T$ in this case?

Question

The situation is as follows:

I am told that $F_{net} = mg - T$ in this case, but doesn't that not take into account that $T$ isn't applied to the center of mass? Newton's second law is defined for discrete particles, but this is a continuous body... If a force is applied to the center of mass of the body, it can be treated as a particle. But $T$ isn't applied to the center of mass - does this change anything? If it does, how would you find the translational acceleration in this case (given $r$, $m$, and $T$)?

Answer 1

$T$ must be accounted for when you sum the forces. The application of $T$ away from the center of mass is equivalent to moving $T$ to the center of mass and adding an appropriately sized couple.

Answer 2

You can think of ${\vec F}_{\rm net}$ as consisting of all of the external forces acting on a system of particles, irrespective of where those forces are applied. The only place where the center of mass comes into play is on the other side of Newton's second law: the ${\vec a}$ in $\sum {\vec F}_{i} = m{\vec a}$ is most assuredly the acceleration of the center of mass of the system of particles, not of any one of the constituents.

Now, you ask, what about the INTERNAL forces in the system? Like all of the interatomic forces holding our wheel together above? Well, here we have the magic of newton's third law, telling us that they can't change the overall momentum of the system, and therefore, they can't contribute to the acceleration of the center of mass of our system!

Answer 3

F=ma but, If object is falling or going downwars the acceleration will be considered as "acceleration due to gravity" hence "g" instead of "a". If "T" is the opposing force, It would be expressed with a negative sign.

F=ma 

Since, a=g

Therefore,
F=mg 

Since, T is oposing force Therefor,

F=mg + (-T)

Hence,

 F=mg-T