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In my physics class we currently have to do research about a motion and preform an experiment and write an essay on it. The idea I had was to drop two parachutes with different areas from the same height and use video measurement and a modeling program to graph the motion. I bought parachute soldiers off the internet of different sizes and at first I thought I was ready to go. Because the air density is the same, I chose to consider the drag coefficient to be the same, the areas of the parachutes were different (but that was the variable i chose to change, to compare the trajectory and the effect on the drag force.) The only variable that troubled me was the speed. Before I really started to think about it, I thought that the weight of an object didn’t affect the drag, since that’s what I had always heard. But after looking into it, it appears it might have an impact on it in away after all. Not directly, but via gravity and the acceleration. If this was the case it would skew the results a lot, since the sizes and weights of the two parachutes are really different. I got caught up in a whirlwind of contradicting information and read about terminal velocity, which made it 15 times as complicated as it was supposed to be. I’m still not sure on whether it does or doesn’t affect drag indirectly and if i should account for this. It would help tremendously if someone could help simplify this issue for me and pose a possible solution, since it largely involves theories we haven’t talked about in school. I’m a junior high school student from The Netherlands.

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  • $\begingroup$ Have you studied vectors. Have you had F=ma? $\endgroup$
    – Natsfan
    Commented Jun 2, 2020 at 19:43
  • $\begingroup$ How do you know that it is "15 times as complicated as it was supposed to be"? $\endgroup$ Commented Jun 4, 2020 at 7:19

3 Answers 3

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The drag force is given by:

$$F_D=\frac12 \rho v^2 C_d A\tag{1}$$

(for explanation of symbols, see this link)

The mass of the object doesn't enter the equation (no pun intended)

But in the case of an object of mass $m$ falling through air while dangling off a parachute, mass does play an important part.

When a parachute jump starts, the object's velocity is initially zero (or very low). As the object gains speed, $F_D$ increases, as per $(1)$.

Remember that $F_D$ is a force vector ($\mathbf{F_D}$) pointing vertically upward.

Terminal velocity is obtained when the drag force equals the vertically downward weight $\mathbf{W}$ vector of the object, so that:

$$\mathbf{F_D}=\mathbf{W}$$

In that case there's no net force acting on the object and per Newton's 2nd Law that means it must move at constant speed and zero acceleration $a$: $$F_{net}=ma=0 \Rightarrow a=0$$.

So:

$$\frac12 \rho v_t^2 C_d A=mg$$

where $v_t$ is the terminal velocity:

$$v_t=\sqrt{\frac{2mg}{\rho C_d A}}$$

So $v_t \propto \sqrt{m}$: the higher the mass, the higher the terminal velocity.

This explains why larger (or multiple) parachutes (larger $A$) are needed to safely parachute heavier objects like military material or food pallets.

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Assuming that the parachute reaches terminal velocity, the drag on the parachute equals the total weight of the toy soldier and the parachute, giving you a net force of zero (constant velocity = zero net force). So yes, the weight of the parachute does affect the drag force on that parachute.

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  • $\begingroup$ Strictly speaking mass doesn't affect drag force but it does affect the balance of forces. $\endgroup$
    – Gert
    Commented Jun 3, 2020 at 5:58
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After the parachute and the object it supports have reached terminal velocity, the drag exactly equals the weight of the object + parachute.

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