Parachute drag force calculation in imperial units

[nysnacc]

Homework Statement

Homework Equations

F=ma
Sum of force

The Attempt at a Solution

Not sure if my a is correct, but actually, I need to have it in imperial (US) unit,[/B]

 

[Orodruin]

You cannot assume a constant force in the second part. The force depends on velocity, which is changing.

 

[nysnacc]

but first part is good? what should I do with the force then?

 

[Orodruin]

No, the first part is not ok either. According to your computation, the sky diver will crash into the ground with a velocity larger than 30 m/s. This does not seem like a very good parachute. Hint: Units are important. Always convert values of the same type of quantity to the same units!

Edit: The problem is also not well stated. It states that $F = 0.5 v^2$ (in pounds) but fails to tell you what units this requires $v$ to be specified in.

 

[nysnacc]

this is my updated value, as I was supposed to use US units thanks for pointing that out :)

 

[Orodruin]

No this is not correct either. Look up the definition of lbf vs the definition of lbm. Your sky diver would still accelerate from a velocity of 30 m/s and crash into the ground at a higher velocity. (This hurts!)

Also, please stop using images in your posts. It is impossible to quote parts of them and they do not always appear well on mobile devices etc. There is a perfectly viable way of writing equations in LaTeX in this forum if you want more advanced formulas than what can be written in plain text.

 

[nysnacc]

Umm... what's wrong with my initial equation?

 

[Orodruin]

Umm... what's wrong with my initial equation?

The mass is given in lbm. What is the gravitational force on an object with mass 1 lbm?

 

[nysnacc]

32 lbf?

 

[Orodruin]

32 lbf?

No. Again, look up the definition of lbf vs the definition of lbm.

 

[Merlin3189]

Edit: The problem is also not well stated. It states that $F = 0.5 v^2$ (in pounds) but fails to tell you what units this requires $v$ to be specified in.

Absolutely! It seems crazy to give a formula where the units are defined for only one of the variables. Of course I'm not accustomed to USC units, so perhaps ft and sec are implicit if F is in pounds?

Surely the proper way to give such a formula, is to say F=kv² then give the value of k with units.
Then k = 0.5 lbf sec² ft^-2 for honest Americans, or for us foreigners k = 24 N sec² m^-2

 

[Orodruin]

Of course I'm not accustomed to USC units, so perhaps ft and sec are implicit if F is in pounds?

I would not bet my money on this. Without giving too much away, using this would give a rather large g-force on the sky diver (higher than any I could find referenced for sky diving in a quick Google search). If the velocity $v$ is intended to be in m/s, you instead get a rather low value.

On the other hand, the option k = 0.5 lbf s^2/m^2 also gives a large terminal velocity (roughly 20 m/s) and k = 0.5 lbf s^2/ft^2 gives roughy 7 m/s. Both are high velocities. 7 m/s would correspond to free-falling from roughly 2.5 m height.

 

[nysnacc]

32 lbf?

1 lbf = 32.174 lbm

W = 1lbf = 1 lbm * 32.174

 

[Orodruin]

1 lbf = 32.174 lbm

W = 1lbf = 1 lbm * 32.174

No, 1 lbf cannot be equal to 1 lbm since the former is a unit of force and the latter a unit of mass. They simply have different physical dimension.

 

[nysnacc]

No, 1 lbf cannot be equal to 1 lbm since the former is a unit of force and the latter a unit of mass. They simply have different physical dimension.

Hard time finding it :(

 

[Orodruin]

Hard time finding it :(

"The pound-force is equal to the gravitational force exerted on a mass of one avoirdupois pound on the surface of Earth."
https://en.wikipedia.org/wiki/Pound_(force)

 

[nysnacc]

so 1 lbf = 32 lbm ?

 

[Orodruin]

so 1 lbf = 32 lbm ?

No. Again, they are units of different physical dimension. One is a measure of force and the other is a measure of mass.

1 lbf is the magnitude of the gravitational force on an object of mass 1 lbm.

This means that, calling the standard Earth gravity g, 1 lbf = (1 lbm) g. This relationship is given in the first equation on the Wikipedia page.

 

[nysnacc]

so for my sum of force in y, I should have put 200 lbf instead of 200*32??

 

[Orodruin]

so for my sum of force in y, I should have put 200 lbf instead of 200*32??

Yes.

 

[nysnacc]

so F_y = - 200 +0.5* ⁽¹⁰⁰⁾ <-- instantaneous

 

[nysnacc]

oh typed it wrong,
F_y = - 200 +0.5* (100)²

 

[nysnacc]

F_y = 4800 (pointing up)

a = 4800 lbf / 200 lbm = 24 ft/s²

 

[Orodruin]

oh typed it wrong,
F_y = - 200 +0.5* (100)²

Yes, with the unit lbf (always give the units!) - and assuming that v in the task is supposed to be inserted in ft/s, which (as mentioned above) is not at all clear from the problem.

F_y = 4800 (pointing up)

a = 4800 lbf / 200 lbm = 24 ft/s²

No. 1 lbf = (1 lbm) g and therefore ...

 

[nysnacc]

a = 4800 lbf / 200 lbm*32 ft/s² = 24 ft/s²

 

[nysnacc]

not 24 but 0.75

 

[Orodruin]

4800 lbf / 200 lbm = 4800 g (1 lbm)/(200 lbm) = (4800/200) g = 24 g

 

[nysnacc]

so it is roughly 768 ft / s²?

 

[nysnacc]

4800 lbf / 200 lbm = 4800 g (1 lbm)/(200 lbm) = (4800/200) g = 24 g

768 ft /s^2 not making sense. 24g meant to be 24 ft/s² ??

 

[Orodruin]

768 ft /s^2 not making sense.

Why not? Because it is a large acceleration? I said that already in this thread. No, it is not supposed to be 24 ft/s^2. That computation had obvious fallacies in terms of units.