# Drag Force Acting on an Object with Respect to Velocity

#### Marcell

1. The problem statement, all variables and given/known data
The final velocity of an object falling through air from various heights is given. From this, can you derive an equation for the drag force acting on the object with respect to velocity?

2. Relevant equations
Maybe relevant?
Wno drag$$=mgh,$$
Wreal$$=\frac{1}{2}mv^2,$$
Wdrag$$=mgh-\frac{1}{2}mv^2,$$

3. The attempt at a solution
Wdrag$$=F(drag~average)s$$
This allows me to solve for the average drag, but I don't know how to go from there to equating the actual drag force acting on the object at a specific velocity.

Thanks for any and all help! :)

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#### Ray Vickson

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1. The problem statement, all variables and given/known data
The final velocity of an object falling through air from various heights is given. From this, can you derive an equation for the drag force acting on the object with respect to velocity?

2. Relevant equations
Maybe relevant?
Wno drag$$=mgh,$$
Wreal$$=\frac{1}{2}mv^2,$$
Wdrag$$=mgh-\frac{1}{2}mv^2,$$

3. The attempt at a solution
Wdrag$$=F(drag~average)s$$
This allows me to solve for the average drag, but I don't know how to go from there to equating the actual drag force acting on the object at a specific velocity.

Thanks for any and all help! :)
If you have also measured the time of the fall along with the terminal velocity, you can set up a mathematical problem: instead of writing the differential equation for velocity as a function of time in the form $dv/dt = g - f(v)$ for your drag force function $f(v)$, let's re-write it as
$$\frac{dt}{dv} = \frac{1}{g - f(v)}$$
Thus, the fall-time $T$ for terminal velocity $V$ is given by
$$T = \int_0^V \frac{1}{g - f(v)} \, dv \hspace{1.5cm} (1)$$
If you have measurements of lots of $(V,T)$ pairs, you can try various functional forms in (1) to see if you can get a fit. For example, if you try
$$f(v) = a v + b v^2$$ in (1), the result for $T$ vs. $V$ will contain the unknown constants $a$ and $b$. You can then try a data-fitting program to find values that are consistent with your measured values.

If you have not measured the fall-times, but only the distance $X$ and the terminal velocity $V$ you will have a much harder problem.

#### Marcell

If you have also measured the time of the fall along with the terminal velocity, you can set up a mathematical problem: instead of writing the differential equation for velocity as a function of time in the form $dv/dt = g - f(v)$ for your drag force function $f(v)$, let's re-write it as
$$\frac{dt}{dv} = \frac{1}{g - f(v)}$$
Thus, the fall-time $T$ for terminal velocity $V$ is given by
$$T = \int_0^V \frac{1}{g - f(v)} \, dv \hspace{1.5cm} (1)$$
If you have measurements of lots of $(V,T)$ pairs, you can try various functional forms in (1) to see if you can get a fit. For example, if you try
$$f(v) = a v + b v^2$$ in (1), the result for $T$ vs. $V$ will contain the unknown constants $a$ and $b$. You can then try a data-fitting program to find values that are consistent with your measured values.

If you have not measured the fall-times, but only the distance $X$ and the terminal velocity $V$ you will have a much harder problem.
Thanks for the help! But I don't quite see how solving for the time taken to reach terminal velocity $T$ would be useful in finding the drag force acting on the object at a specific velocity...

#### haruspex

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Wdrag$$=mgh-\frac{1}{2}mv^2,$$
Try differentiating that with respect to ...

#### Marcell

Try differentiating that with respect to ...
distance? Would that give the instantaneous work being done on the falling object and allow me to find the force of drag? If so I'm really not sure how to differentiate something like that, and my go to 'derivate-calculator.com' is lost here as well.

#### Marcell

Try differentiating that with respect to ...
Wait, even if I do differentiate it with respect to something, wouldn't here only be one component on the left hand side? and would said component then be equal to zero?

#### haruspex

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distance? Would that give the instantaneous work being done on the falling object
If you differentiate the work done by a force with respect to the distance the force advances, what do you get? Or to put that in reverse, what, when integrated wrt distance, gives the work a force does?

#### Marcell

Try differentiating that with respect to ...
Distance.
If you differentiate the work done by a force with respect to the distance the force advances, what do you get? Or to put that in reverse, what, when integrated wrt distance, gives the work a force does?
I get the force done on the body.

However, differentiating $W(air~drag)=mgh-\frac{1}{2}mv^2,$ with respect to s (s=h), got me $0=mg$

#### Apashanka

instantaneous acceleration equation for falling object gives,
mdv/dt=mg-Fdrag.
Now given the form of Fdrag ,v(t) can be determined.

#### Marcell

instantaneous acceleration equation for falling object gives,
mdv/dt=mg-Fdrag.
Now given the form of Fdrag ,v(t) can be determined.
I'm not sure if I follow, but I don't want to find v(t), I'm looking for $F_{drag}$ and I don't have the form of $F_{drag}$...

#### haruspex

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I get the force done on the body.
Right. Specifically, you get the force at any particular distance.
However, differentiating
$W(air~drag)=mgh-\frac{1}{2}mv^2$ with respect to s (s=h), got me 0=mg
No, you get $F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)$.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?

#### Marcell

Right. Specifically, you get the force at any particular distance.

No, you get $F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)$.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?
Wait I'm quite confused, wouldn't the right hand side just equal $mg$? The only way I see getting a non-constant force of drag would be if the right hand side were differentiated with respect to v (though that doesn't make too much sense to me...)

#### haruspex

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wouldn't the right hand side just equal mg?
No, v is a function of h.
You are given a set of pairs of values for h and v. How can you estimate $\frac d{dh}v^2$ from those?

#### Marcell

No, v is a function of h.
You are given a set of pairs of values for h and v. How can you estimate $\frac d{dh}v^2$ from those?
Eyyyy, thank you so much, I got it! Squared the line of best fit's equation and then just took its derivative.

#### haruspex

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Eyyyy, thank you so much, I got it! Squared the line of best fit's equation and then just took its derivative.
Sounds good to me. Or you could square first and take best fit second. The result won't be quite the same... not sure which is the more justified.

#### Marcell

Right. Specifically, you get the force at any particular distance.

No, you get $F_{drag}= \frac d{dh}(mgh-\frac{1}{2}mv^2)$.
So now you just need a way to estimate the right hand side of that at various values of h. How can you do that?
Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...

#### Ray Vickson

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Thanks for the help! But I don't quite see how solving for the time taken to reach terminal velocity $T$ would be useful in finding the drag force acting on the object at a specific velocity...
I mean that you need to do the integral to find $T = F(a,b,V),$ where
$$F(a,b,V) = \int_0^V \frac{1}{9.81 - a v - b v^2} \, dv$$
is some function of the terminal velocity $V$ that contains the unknown constants $a$ and $b$. You can then try to find numerical values of the parameters $a$ and $b$ so as to fit the $(T,V)$ data as well as possible---that is, to make $F(a,b,V_i)$ come out close to the measured value of $T_i$ for all pairs $(T_1,V_1), (T_2,V_2), (T_3,V_3), \ldots, (T_n,V_n)$ in the experimental data set.

An exact fit cannot be expected, because of experimental errors (and possibly because of an incorrect $f(v)$ expression), but some form of "least-squares" fit might work well. Once you have found appropriate "best-fit" numerical values of $a$ and $b$, you can then substitute those values into the formula $f_{\text{fit}}(v) = a v + b v^2$, so you can calculate your best estimate $f_{\text{fit}}(v)$ of the drag force $f(v)$ for any input value of $v.$ Of course, for all this to work, you must know how to do integration, and have access to a nonlinear "least-squares" computer package.

Since I do not know your background I cannot judge whether this answer is meaningful to you and if so, how much additional explanatory material to post, if any.

Last edited:

#### vela

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Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...
You have to be careful with the signs of the various quantities. With your sign convention, is v positive or negative? What about dv/dh?

#### haruspex

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Wait, doesn't this equation suggest that as velocity increases the force of drag acting on the object decreases? There's no way that can be right...
No, why?
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum. Sketch the difference. What can you say about the slope as h increases?

#### Marcell

I mean that you need to do the integral to find $T = F(a,b,V),$ where
$$F(a,b,V) = \int_0^V \frac{1}{9.81 - a v - b v^2} \, dv$$
is some function of the terminal velocity $V$ that contains the unknown constants $a$ and $b$. You can then try to find numerical values of the parameters $a$ and $b$ so as to fit the $(T,V)$ data as well as possible---that is, to make $F(a,b,V_i)$ come out close to the measured value of $T_i$ for all pairs $(T_1,V_1), (T_2,V_2), (T_3,V_3), \ldots$
That is quite interesting, but I don't think I could do it since I only have one pair of $(T_1,V_1)$, after all the whole experiment is done with a single object.
You have to be careful with the signs of the various quantities. With your sign convention, is v positive or negative? What about dv/dh?
Down is positive, so both v and dv/dh are positive

#### Marcell

No, why?
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum. Sketch the difference. What can you say about the slope as h increases?
Where does gh come from? $\frac d{dh}(mgh)=mg$ no? please correct me if I'm wrong.

#### haruspex

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Where does gh come from? $\frac d{dh}(mgh)=mg$ no? please correct me if I'm wrong.
I'm suggesting to sketch mgh, ½mv2 and mgh-½mv2, but with the m omitted throughout.
You can then consider how the slope, $\frac d{dh}(gh-\frac 12v^2)$, changes with h.

#### Marcell

I'm suggesting to sketch mgh, ½mv2 and mgh-½mv2, but with the m omitted throughout.
You can then consider how the slope, $\frac d{dh}(gh-\frac 12v^2)$, changes with h.
Well the slope doesn't change too much if it all, it appears to be linear. Also I don't quite see why differentiating $\frac d{dh}(gh-\frac 12v^2)$ _wouldn't_ lead to a drag force which decreases with velocity.

#### haruspex

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Well the slope doesn't change too much if it all, it appears to be linear. Also I don't quite see why differentiating $\frac d{dh}(gh-\frac 12v^2)$ _wouldn't_ lead to a drag force which decreases with velocity.
Then you are sketching it wrongly.
As I wrote in post #19:
Sketch gh and ½v2 as functions of h. The first is a straight line, the second starts with the same slope but rises towards a maximum.
This implies the second rises ever more slowly, so the gap between the two increases faster than linearly. The difference therefore rises faster than linearly, so the slope (the drag force) increases.

#### Marcell

Then you are sketching it wrongly.
As I wrote in post #19:

This implies the second rises ever more slowly, so the gap between the two increases faster than linearly. The difference therefore rises faster than linearly, so the slope (the drag force) increases.
taking $$v=ah+b,$$
$$\frac{d}{dh}\left(mgh-\frac{1}{2}m(ah^2+2\operatorname{abh}+b^2\right))$$
then $$F=\left(mg-\frac{1}{2}m\left(2ah+2ab\right)\right)$$

And from that, as h increases, F decreases which doesn't make sense.

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