# Missing Resistance Component

Assume an object travelling in a horizontal plane with no friction and in a resistive medium. For resistive motion, I'm being taught that $$ma = -R$$, where $$F=ma$$ would be force exerted by the object.

The force $$R$$ is the resistive force acting in the opposite direction of motion.

My problem is that this formula seemingly implies that net force acting on the object is zero, since those are the only two forces affecting the object's horizontal motion. However, the object still moves. Therefore my first question is why the object moves while the formula implies that it shouldn't?

Also, regarding resistive motion for an object in freefall: we know that drag in fluid or air is a function of (square) of velocity, so it changes with time as object accelerates. Supposing an object is released in the air and enters freefall in a resistive medium (i.e. air). The two forces affecting its motion would then be $$mg$$ and $$R$$, where $$R$$ is air resistance/drag. Since $$R$$ changes over time, while $$mg$$ remains constant, what is the other force(s) component in addition to air resistance if net forces are zero?

$$\vec F=m\vec a$$ is Newton's second law.
With a resistive force $$R$$ in the opposite direction to the motion of the object, which is assumed to be in the $$\hat i$$ direction, the force on the body $$\vec F = -R \hat i$$.
$$\vec F = m \vec a \Rightarrow -R \hat i = m a \hat i \Rightarrow a = - \frac R m$$.
In free fall again there is a net force on the object as it falls in the $$\hat y$$ direction made up of two forces, the gravitational force $$mg\hat y$$ and a resistive force $$-R(v)\hat y$$.
In this case $$\vec F = m \vec a \Rightarrow mg\hat y-R(v)\hat y = ma \hat y \Rightarrow a =\frac {mg-R(v)}{m}$$