Resultant force on a parachutist

[Quantum55151]

Homework Statement

A parachutist is in free fall before opening her chute. The net force on her has magnitude F and is
directed down; this net force is somewhat less than her weight W because of air friction. Then
she opens her chute. At the instant after her chute fully inflates the net force on her is

(A) greater than F and still directed down.
(B) less than F and still directed down.
(C) zero.
(D) directed upward, but could be more or less than F.

Relevant Equations

F = ma

I would say B, C, and D. If the upward force of the parachute is greater in magnitude than F, then the new resultant force must point upward (and may be more or less than F). If the upward force of the parachute is equal and opposite F, then they must cancel out, and if the upward force is smaller in magnitude than F, then the new resultant force must point downward.

 

[kuruman]

Don't think in terms of forces but in terms of how the motion changes before and after the parachute opens.
Remember that

• The acceleration is in the same direction as the net force.
• When the velocity and the acceleration are in the same direction, the speed increases.
• When the velocity and the acceleration are in opposite directions, the speed decreases.

What is the speed doing before the parachute opens? Increase or decrease?
What is the speed doing after the parachute opens? Increase or decrease?

 

[Quantum55151]

What is the speed doing before the parachute opens?
What is the speed doing after the parachute opens?

Well, in a typical skydiving scenario, the parachute would slow down the fall of the person, in which case the resultant force must be upward. However, this is not explicitly mentioned in the question, so I would assume that all possible scenarios should be considered...

 

[kuruman]

Note that there is also no explicit mention that the parachutist falls towards the Earth. This doesn't mean that you should also consider the scenario that she falls away from the Earth.

 

[Mister T]

Well, in a typical skydiving scenario, the parachute would slow down the fall of the person, in which case the resultant force must be upward. However, this is not explicitly mentioned in the question, so I would assume that all possible scenarios should be considered...

It seems to me that before you wrote the first sentence quoted above you did consider all possible scenarios.

 

[haruspex]

Well, in a typical skydiving scenario, the parachute would slow down the fall of the person, in which case the resultant force must be upward. However, this is not explicitly mentioned in the question, so I would assume that all possible scenarios should be considered...

It is reasonable to assume that the parachutist waited until clear of the aircraft/balloon, in which case her speed would already exceed a safe landing speed.

 

[Quantum55151]

Note that there is also no explicit mention that the parachutist falls towards the Earth. This doesn't mean that you should also consider the scenario that she falls away from the Earth.

Fair. Yet I've found a solutions document, according to which the answers are B and D (I would also add C):

The author arrives at this conclusion precisely by considering different scenarios. Are they wrong in doing so?

 

[jbriggs444]

Fair. Yet I've found a solutions document, according to which the answers are B and D (I would also add C):

View attachment 347531
The author arrives at this conclusion precisely by considering different scenarios. Are they wrong in doing so?

It is going to take an extremely unusual circumstance to get a downward acceleration at the moment of full inflation. Maybe a downward microburst or a suicidal choice of equipment. In my judgement, B is incorrect.

Engineering choices dictate a range of plausible drag forces and drop distances for full inflation -- we do have information to go on.

 

[Quantum55151]

It is going to take an extremely unusual circumstance to get a downward acceleration at the moment of full inflation.

Could the acceleration be zero though?

 

[kuruman]

Could the acceleration be zero though?

Let's see what we know.

1. Before the parachute opens, the velocity is down and we are told that the net force is also down. This means that the down speed is increasing. That's given.
2. The parachute needs some time to be fully deployed, however we are not concerned about it because the question wants an answer when "the chute fully inflates".
3. When the chute fully inflates, the velocity is still down and remains down until the parachutist lands. That's how parachutes work. Furthermore, the down velocity reaches some terminal landing speed that is considerably less than the speed before the chute is deployed.
4. When terminal speed is reached, the acceleration is zero and so is the net force on the parachutist. However, this does not occur immediately at the moment the chute opens. One wants the time derivative of the acceleration, a.k.a. the jerk, not to be too large lest the shoulders of the parachutist be dislocated or worse.
5. Therefore, when the chute is fully deployed, the velocity is down and the acceleration is non-zero because the speed starts decreasing on its way to reaching its terminal value.

Which is the correct answer?

 

[PeroK]

Homework Statement: A parachutist is in free fall before opening her chute. The net force on her has magnitude F and is
directed down; this net force is somewhat less than her weight W because of air friction. Then
she opens her chute. At the instant after her chute fully inflates the net force on her is

(A) greater than F and still directed down.
(B) less than F and still directed down.
(C) zero.
(D) directed upward, but could be more or less than F.
Relevant Equations: F = ma

I would say B, C, and D. If the upward force of the parachute is greater in magnitude than F, then the new resultant force must point upward (and may be more or less than F). If the upward force of the parachute is equal and opposite F, then they must cancel out, and if the upward force is smaller in magnitude than F, then the new resultant force must point downward.

I would take this approach to this problem. Physics and mathematics is all about being precise, and systematic.

We have two cases: a) before the shute is open; and, b) after the shute is open. We need to establish all forces on the parachutist in both cases. And, in the absence of numbers, we must estimate them quantitatively.

Let's take downwards as positive:

a) Force of gravity: $mg$, where $m$ is the combined mass of the parachutist and her equipment. Air resistance: $-R_a$, where we can assume that $0 < R < mg$. Net force $F = mg - R > 0$.

b) Force of gravity: $mg$, where $m$ is the combined mass of the parachutist and her equipment. Air resistance: $-R_b$, where we can assume that $R_a < R_b$ (as the purpose of a parachute is to increase air resistance). Net $F_b = mg - R_b$.

What can we say about $R_b$ other than $R_a < R_b$? The answer is nothing. So, we have three sub-cases:

b.1) $R_a < R_b < mg$

b.2) $R_a < R_b = mg$

b.3) $mg < R_b$

I'll let you describe the consequences of each of these.

The book answer is woolly and imprecise. We are already told the parachutist is not at terminal velocity.

Note that just because you have a "word" problem and a "word" answer, does not mean your analysis is only words. Physics is about calculation, equations and quantities. It's not a word puzzle.

 

[kuruman]

I would take this approach to this problem. Physics and mathematics is all about being precise, and systematic.

I'm all for that, so let's do some math and, while we're a it, put in some (realistic) numbers. I will model the air resistance as proportional to the velocity. With "down" taken to be the positive direction, Newton's second law gives $$\frac{dv}{dt}=-\frac{b}{m}v+g ~~~~(b>0).$$The general solution of this is $$v(t)=v(0)e^{-(b/m)t}+\frac{mg}{b}\left[1-e^{-(b/m)t}\right].\tag{1}$$Let $t=t_o$ be the deployment time when the parachute opens. We assume that deployment is instantaneous. We also assume that we have two air resistance coefficients, $b_1$ and $b_2$ before and after deployment respectively. Finally, assuming that the parachutist starts with zero vertical velocity and falls straight down, we write
$$v_1(t)=\frac{mg}{b_1}\left[1-e^{-(b_1/m)t}\right]~~~~~~(t\leq t_o).$$ The net force $F$ at deployment mentioned in the problem is $$F=m \left . \frac{dv}{dt}\right |_{t=to}=mg~e^{-(b_1/m)t_o}\implies e^{-(b_1/m)t_o}=\frac{F}{mg} .$$ At the time of deployment, $$v(t_o)=\frac{mg}{b_1}\left[1-e^{-(b_1/m)t_o}\right]=\frac{mg}{b_1}\left[1-\frac{F}{mg}\right].$$ Now the velocity is continuous through the deployment and we use the expression above as the initial velocity in equation (1) to write the velocity after deployment: $$v_2(t) =\frac{mg}{b_1}\left[1-\frac{F}{mg}\right]e^{-(b_2/m)t}+\frac{mg}{b_2}\left[1-e^{-(b_2/m)t}\right]~~~~~~(t>t_o)$$ Of interest is the acceleration after deployment which, after some algebra, is given by$$a_2(t)=\frac{dv_2}{dt}=g\left[1+\frac{F}{mg}-\frac{b_2}{b_1}\right]e^{-(b_2/m)t}.$$According to our adopted convention, if the factor multiplying the exponential is positive the acceleration is down, else it is up. We are told by the problem that $F$ is "somewhat less" than the weight. This means that $1< (1+\frac{F}{mg})< 2.$ We also know that $b_2 > b_1$ because the air resistance increases when the parachute opens. The terminal velocity is given by $v_{\text{ter}}=\dfrac{mg}{b}$, therefore $$\frac{b_2}{b_1}=\frac{v_{\text{ter,1}}}{v_{\text{ter,2}}}.$$ The web tells me here that in the stable belly-to-earth position the terminal velocity (parachute not deployed) is $v_{\text{ter,1}}=~$120 mph. It also tells me here that the landing velocity of an open parachute is $v_{\text{ter,2}}=~$15 mph. This means that the ratio $b_2/b_1=v_{\text{ter,1}}/v_{\text{ter,2}}=8$ which makes the factor multiplying the exponential negative. Thus, the acceleration and hence the net force are directed "up" after deployment. In order for the net force to be down, one would need a landing speed that is in excess of 60 mph. Ouch!