The force of gravity is $F_g=+mg$ or $F_g=-mg$? I have noticed that in my classical mechanics course and in the textbook I read for it, seem to ignore the gravitational force's position. For example, if we were dealing with a system with a ball of mass $m$ tied to a ceiling and were asked to find the forces, in an introductory textbook (as least the ones I read) would state the summation of forces as $F_{net}=T−mg$. However, in the mechanics textbook it is written as $F_{net}=T+mg$. I'm wondering why it seems that the downward aspect of the gravitational force is ignored.
 A: $\boldsymbol g$ is a vector. You define it as a scalar only when you have mentioned a clear reference frame in which the sign of $\boldsymbol g$ makes sense.
You could define a vector field from the law of universal gravitation for two bodies $A$ and $B$:
$$\boldsymbol F_{AB}(\boldsymbol r) = -G\frac{m_Am_B}{|\boldsymbol r_{AB}|^2}\boldsymbol {\hat{r}}_{AB}.$$
A: If you do not define the direction of the force, you need to do the math explicitly. Usually, a letter like $g$ identifies a scalar quantity. If you want to show a vector, there are a number of typographical conventions. I have seen $\mathbf{g}$, $\vec{g}$, $\overline{g}$, $\underline{g}$ ...
So when I have tension up and gravity down, I know that the two are opposing each other, and I need to make that clear in how I describe the math. Vectors are unambiguous - and you will sometimes see them written as $\mathbf{\hat{i},\hat{j},\hat{k}}$ which indicate unit vectors along three orthogonal directions (the $\hat{ }$ symbol describes "unit" something). Then you could write your example as
$$F = T\mathbf{\hat{k}} - m\cdot g\cdot \mathbf{\hat{k}}$$
But it would be just as easy to describe the situation where the tension is at 45 degrees to the vertical:
$$F = T\frac{\mathbf{\hat{i}}+\mathbf{\hat{k}}}{\sqrt{2}} - m\cdot g\cdot \mathbf{\hat{k}}$$
and there is no ambiguity...
It is also possible that your author uses $g=-9.81 m/s^2$ as the convention, in which case you write the sum of the forces in the usual way:
$$F_{total} = \Sigma{F_i} + \Sigma{m_i\cdot a_i}$$
where the direction has to be accounted for by the sign of the accelerations. Again, the usual approach (in anything that isn't a 1D problem) would be to write the whole thing as a vector equation.
