How should the sign of acceleration be chosen when I apply Newton's second Law

Question

While I was playing around with the equations of motion and velocity of a free falling object experiencing drag and gravity, I started wondering how the choice of the coordinate system affects the equations. I know that the choice of coordinate system doesn't affect our answers, but in this case, I get different answers depending on which system I use. For example, let's say that an object is in free fall experiencing only the force of gravity. I choose upwards as my positive direction. Thus $ma=-mg$ or $a=-g$.

Now, what if I choose down as my positive direction? Then, $mg=ma$ or $a=g$. Now, I start incorporating drag in my equation. Let's say that $F_d\propto v$. If I choose down as my negative direction, I get $ma=-mg+bv$ since drag is opposite to the direction of motion. However, after solving this differential equation for velocity, I end up with $$v(t)=\frac{1}{b}\left(C\exp\left(\frac{bt}{m}\right)v+mg\right)$$ This means that the velocity keeps growing exponentially, which we know is not the case. But when I choose down as my positive direction, the solution is $$v(t)=\frac{1}{b}\left(mg-C\exp\left(-\frac{bt}{m}\right)\right)$$ This one makes more sense since the velocity approches a constant value: the terminal velocity.

My question is: what did I do wrong in the first way of solving the problem. Did have I have to put a negative sign on the acceleration? If so, why did I not have to do that with the constant acceleration equation?

Answer 1

Here's a way to imagine the problem. Forget about friction and gravity for a second. The force on a spring will be of the form:

$$ F = -kx$$

So $m\ddot{x} = -kx$. This means that as you increase $x$, the force tries to bring it back to the equilibrium position. If instead, the force was $F=kx$, the more you stretch the more it expands, leading to an infinitely accelerating mass.

Equally, the force of friction should be a restoring force. Imagine that the force of friction is such that:

$$m\dot{v} = bv$$ What does this mean? If it's not moving, no problem. v=0, $\dot{v}$=0. But if it starts to move, it accelerates more, the more it accelerates, the more $v$ grows, and so on. So if it's going to be a restoring force, we need the acceleration and the speed to have opposite signs: $$ m\dot{v} = -bv$$ Now it behaves like the real world: The faster you go, the more the drag tries to make you decelerate. We are now ready to add gravity to it: $$ m\dot{v} = -bv+mg$$ or $$ -m\dot{v} = +bv-mg$$ Note that your equation should also describe the situation where the object is thrown upwards. In that case, both $mg$ and $bv$ are trying to slow down the object.

Answer 2

I want to show you that the choice of up or down as the positive direction does not produce conflicting results.

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Suppose you choose up $\hat z$ to be positive.

Think of all your vector quantities as components in the $\hat z$ direction so any downward vector would be in the direction $-\hat z$ or have a negative component in the $\hat z$ direction.

$\vec F = m\vec a\, \Rightarrow \, -mg \hat z = ma \hat z \, \Rightarrow \, a = -g$.

The component of the acceleration $\vec a$ in the $\hat z$ direction is $-g$ and so the direction of the acceleration is downwards.

The frictional force is $-bv \,\hat z$, with the sign of the component of velocity $v$ in the $\hat z$ direction switching sign when the direction of motion of the particle changes and this in turn switches the direction of the frictional force.
You must not prejudge the direction of motion of the particle rather set up the differential equation and solve it to find the component of velocity in the $\hat z$ direction as a function of time.

The equation of motion is the same for the particle going up and for the particle coming down.

$\vec v = v \hat z \, \Rightarrow \, \vec F_{\rm friction} = -bv \hat z \,\Rightarrow \, -mg \hat z -bv \hat z = ma \hat z\Rightarrow - mg - bv = ma$

Note that when the particle is going up and the component of velocity in the $\hat z$ direction is positive the frictional force is in the downward direction and when the particle is going down and the component of velocity in the $\hat z$ direction is negative the frictional force is in the upward direction.
This is exactly what you want to happen?

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Now suppose you choose down $\hat z$ to be positive.

As before $\vec v = v \hat z \, \Rightarrow \, \vec F_{\rm friction} = -bv \hat z \,\Rightarrow \, +mg \hat z -bv \hat z = ma \hat z\Rightarrow + mg - bv = ma$

Note that the direction of motion of the particle has not been chosen ie the component of velocity of the particle $v$ in the $\hat z$ direction might be positive or it might be negative.

If the component of $v$ is positive it means that the particle is moving downwards and the frictional force is upwards.
If the component of $v$ is negative it means that the particle is moving upwards and the frictional force is downwards.

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The final solution for the component of velocity in the $\hat z$ direction only differs by a sign as to whether $\hat z$ is chosen to be up or down.

The component of velocity in the $\hat z$ direction at a time $t$ is given by

$v(t) = A \exp \left (-\dfrac {bt}{m}\right )\pm \dfrac{mg}{b}$

the constant $A$ being determined by the initial velocity of the particle.

When it comes to the initial velocity $\vec v(0)$ it is easier to have it as either $v_0\,\hat z$ or $-v_0\,\hat z$, depending on the initial direction of the particle and the direction of $\hat z$, and with $v_0$ positive.

Note that as time $t \rightarrow \infty$ then the component of velocity in the $\hat z$ direction tends to $+\dfrac{mg}{b}$ if $\hat z$ is down and $-\dfrac{mg}{b}$ if $\hat z$ is up.