Should a force be positive or negative when using $F=ma$? When finding the equations of motion of an object under a constant force, the sign of the force does not matter to much of an extent, only affecting the final sign of the result. 
For example, solving $F=ma$ for $ma=-mg$ or $ma=mg$ only affects the sign of the final displacement equation, either $s=ut-0.5gt^2$ or $s=ut+0.5gt^2$. In these cases, it is the orientation of the axis you pick for your calculation which affects the result.
However, solving $F=ma$ for forces which are velocity or dispacement dependent, like for air resistance, the sign of the term matters more than just the axis orientation. 
$ma=bv-mg$ and $ma=-bv+mg$ yield very different results when solved, exponentially increasing velocity and exponentially decaying velocity respectively, even though they seem to only vary in how you label you positives and negatives on your axis.
Is there a way to determine the correct sign of a force? How do you solve $F=ma$ for air resistance when your axis labels downwards as negative?
 A: There are two kinds of forces : those who are conservative (associated with a potential energy $V$) and those who are not.
For the latter, it is often obvious which sign to affect. For example a friction force propotional to $v$ or $v^2$ will always decrease the acceleration therefore they come with a sign $-$, e.g.
$$
\vec{F} = -\gamma m \vec{v}
$$
For the former, what is really intrinsic is that the force will lead you to minimize the potential
$$
\vec{F} =-\vec{\nabla} V
$$
Think about a particle in a parabolic potential $V(x) = \frac 1 2 k x^2$ (e.g. attached to a spring). The corresponding force will lead the particle to move towards $x=0$ so the sign of the force $\vec{F} = - k x \vec{u_x}$ is opposite to te sign of $x$.
The case of gravity leads you to a minus sign
$$
\vec{F} = -m g \vec{u_z}
$$
because going higher would cost you some energy and therefore decelerates your object, assuming higher means "more away" from the massive object considered, like earth.
A: This is a good question.  Normally when solving Newton's Laws we define the forces, draw a free body diagram, etc, and there's not much else to think about.
With a velocity-dependent force, we have to think, and add more information.
The force always serves to slow down the object.  When objects slow down, the acceleration is in a direction opposite to the velocity.  So our velocity-dependent force must be in the direction opposite to the actual acceleration not opposite to the force of gravity, always being careful, as has been mentioned, to properly account for whatever coordinate system you set up, e.g, the force of gravity is negative.
A: The way to determine the correct sign of a force is to use your coordinate system. A force which points in the direction you've chosen to be negative is represented by a negative number, and a force which points in the direction you've chosen to be positive is represented by a positive number. That's all there is to it.
The reason you get different results from $ma=bv-mg$ and $ma=-bv+mg$ is that they represent different physical situations. If you have $+bv$ as in the first equation ($b > 0$), that means the air resistance is acting in the same direction as the velocity, i.e. acting to speed the object up. This should be fairly clear: the air resistance force $F_d = bv$ has the same sign as $v$. But if you have $-bv$ as in the second equation, that means the air resistance is acting to slow the object down: $F_d = -bv$ has the opposite sign, and thus opposite direction, as $v$. Clearly, only one of these corresponds to reality.
A: If writing $F = ma$ and bothering with the direction of axes confuses you, then I suggest writing 
$$F=m~\frac{\mathrm dv}{\mathrm dt}$$
Air resistance always tend to act in the direction opposite to that of the increasing velocity, meaning it serves to decrease velocity with time whereas gravity tends to increase its velocity with time (assuming of course that you are thinking about a body falling towards the earth...in the direction of gravity).
It should be clear to you now as to what sign has to be used with the '$bv$' term and what sign should be used with the '$mg$' term.
A: You are dealing with vector quantities here.  To never go wrong in getting the correct signs in your equations, all you need to do is include the unit vector for each force or acceleration.  Try it and see how it simplifies things.  For example, if the force is in the negative x direction, just multiply the magnitude of the force by $(-i_x)$ to get the vector representation of the force.
