# Drag force of Parachute

1. Sep 13, 2016

### nysnacc

1. The problem statement, all variables and given/known data

2. Relevant equations
F=ma
Sum of force
3. The attempt at a solution

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

2. Sep 14, 2016

### Orodruin

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

3. Sep 14, 2016

### nysnacc

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

4. Sep 14, 2016

### Orodruin

Staff Emeritus
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.

5. Sep 14, 2016

### nysnacc

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

6. Sep 14, 2016

### Orodruin

Staff Emeritus
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.

7. Sep 14, 2016

### nysnacc

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

8. Sep 14, 2016

### Orodruin

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

9. Sep 14, 2016

### nysnacc

32 lbf?

10. Sep 14, 2016

### Orodruin

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

11. Sep 14, 2016

### Merlin3189

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

12. Sep 14, 2016

### Orodruin

Staff Emeritus
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.

13. Sep 14, 2016

### nysnacc

1 lbf = 32.174 lbm

W = 1lbf = 1 lbm * 32.174

14. Sep 14, 2016

### Orodruin

Staff Emeritus
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.

15. Sep 14, 2016

### nysnacc

Hard time finding it :(

16. Sep 14, 2016

### Orodruin

Staff Emeritus
17. Sep 14, 2016

### nysnacc

so 1 lbf = 32 lbm ?

18. Sep 14, 2016

### Orodruin

Staff Emeritus
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.

19. Sep 14, 2016

### nysnacc

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

20. Sep 14, 2016

### Orodruin

Staff Emeritus
Yes.