Drag in the x and y directions

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The discussion focuses on the application of drag forces in classical mechanics for a particle moving in both x and y directions. In the x-direction, the equation used is based on the premise that the acceleration is directly related to the drag force, expressed as ma_x = -kmv_x, which raises questions about the role of mass in deceleration. In contrast, the y-direction includes gravity as an additional force, leading to a different equation where the total force is the sum of gravity and drag. The participants debate the implications of these equations, particularly how mass influences drag and the behavior of objects in free fall. The conversation highlights the nuances of applying differential equations to understand motion in different contexts, emphasizing the importance of recognizing all acting forces.
Hermes Chirino
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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 ?
 
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Hermes Chirino said:
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).

Hermes Chirino said:
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}##
 
Nathanael said:
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?
 
Hermes Chirino said:
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?
 
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