Drag in the x and y directions

[Hermes Chirino]

I'm reading this book on Classical Mechanics and there are two examples in the book where we are asked to find one expression for the velocity $v$, and one for the position $x$, both as functions of time for a particle moving in the x-direction in a "medium" where the drag force is proportional to $v$. We are also asked to find velocity and position in the y-direction, same medium, drag force propotional to $v$. They use differential equations methods to solve it. I don't have any trouble understanding their methods or how they got the equations; my question is on their premises.

For the x-direction, they use as premise:
$$ma_x=m\frac{dv}{dt}=-kmv_x$$

I understand the drag force $-kmv$ is equal in magnitude to the force $ma$, but opposite direction.

For the y-direction, they use as premise:
$$F_T=m\frac{dv}{dt}=-mg-kmv_y$$

I understand the total force is equal to the force of gravity $mg$, minus the drag force pointing in the opposite direction.

My question here is: Why they don't use a similar premise for the x-direction, something like:
$$F_T=ma_x-kmv_x$$

What is the diference ? What I am missing here ?

 

[Nathanael]

For the x-direction, they use as premise:
$$ma_x=m\frac{dv}{dt}=-kmv_x$$

Shouldn't it be $ma_x=m\frac{dv_x}{dt}=-kv_x$ ?
Otherwise you're implying that deceleration due to drag would be independent of mass (but clearly feathers fall slower than heavy things).

My question here is: Why they don't use a similar premise for the x-direction, something like:
$$F_T=ma_x-kmv_x$$

That is not similar to what they did in the y-direction. $F_T$ is, by definition, equal to $ma_x$ right? So what sense does this equation make?

In the y-direction they equated $ma_y$ with $F_{net.y}$ and in the x-direction they equated $ma_x$ with $F_{net.x}$.
The y-direction just happens to have an extra force (gravity) that contributes to $F_{net.y}$

 

[Hermes Chirino]

Shouldn't it be $ma_x=m\frac{dv_x}{dt}=-kv_x$ ?
Otherwise you're implying that deceleration due to drag would be independent of mass (but clearly feathers fall slower than heavy things).That is not similar to what they did in the y-direction. $F_T$ is, by definition, equal to $ma_x$ right? So what sense does this equation make?

In the y-direction they equated $ma_y$ with $F_{net.y}$ and in the x-direction they equated $ma_x$ with $F_{net.x}$.
The y-direction just happens to have an extra force (gravity) that contributes to $F_{net.y}$

They use m, so when solving for the differential equation, the math comes a little bit easier, they mention in the book that we shouln't really too much on the mass, they use it only for mathematical purposes. I understand your point in the y direction, that gravity comes as an extra force, but when an object is in free fall, isn't gravity the only force?

 

[SteamKing]

I understand your point in the y direction, that gravity comes as an extra force, but when an object is in free fall, isn't gravity the only force?

Only when things are falling in a vacuum.

Why do you think that a cannon ball and a feather dropped from the same height in air hit the ground at different times?