Calculating Drag Force for Parachute: F=ma and Sum of Forces

[nysnacc]

When 200 lb is considered to be lbf, I don't have to multiply by g?
when it is considered as lbm, I need to?

 

[nysnacc]

The relationship between lbf and lbm is:
1 lbf = g* lbm??

 

[Orodruin]

The relationship between lbf and lbm is:
1 lbf = g* lbm??

Yes, I have already said this in this thread. Note that g is a dimensionful quantity. Therefore g is not equal to 9.82 or 32. It is 9.82 m/s^2 = 32 ft/s^2.

Do not take this the wrong way, but it seems to me that you really need to read up on and understand units and physical dimensions. If these are not second nature, you will struggle more than necessary when studying physics.

 

[nysnacc]

Yes, I have already said this in this thread. Note that g is a dimensionful quantity. Therefore g is not equal to 9.82 or 32. It is 9.82 m/s^2 = 32 ft/s^2.

Do not take this the wrong way, but it seems to me that you really need to read up on and understand units and physical dimensions. If these are not second nature, you will struggle more than necessary when studying physics.

Yes, I'll take that :) so when do I know 200 is lbf and lbm? is it the problem in this exercises not stating the units clear?

 

[Orodruin]

Yes, I'll take that :) so when do I know 200 is lbf and lbm? is it the problem in this exercises not stating the units clear?

The 200 in the problem is a mass. Masses cannot be measured in lbf since it is a unit of force. Masses can be measured in lbm since it is a unit of mass.

 

[nysnacc]

But then in my expression of sum of force, F = lbf +F_drag

lbf = lbm *g why isn't it 200*32?

 

[nysnacc]

-sign before lbf, should be

 

[Orodruin]

But then in my expression of sum of force, F = lbf +F_drag

lbf = lbm *g why isn't it 200*32?

The gravitational force is $-mg$. The mass is $m = 200$ lbm making the gravitational force $-200\ {\rm lbm} \cdot g = -200\ {\rm lbf}$. Where do you have a problem with this computation?

 

[Orodruin]

You also seem to be mixing units and the physical quantities. You cannot just write "lbf" instead of $mg$ in the force balance equation. Again, it seems as if you need to take some steps back and understand units and physical dimensions before you tackle this type of problems.

A physical quantity consists of a suitable unit to describe it and a numerical value. For example, a suitable unit for measuring lengths is the meter. In order to know how long a distance is, you give a measured value with reference to the unit used. You can use any length unit, but the numerical value will depend on your choice.

When you construct new physical quantities by multiplying others, the units will also be multiplied. For example, you can construct an area by multiplying two lengths together. If you measure the lengths in m, then the resulting area will have units m^2. You cannot use length units to describe an area, you must use units of length^2.

The pound mass (lbm) is a unit of mass, you can only use it as a unit of mass. The pound force (lbf) is a unit of force, you can only use as a unit of force. The pound force is defined such that $1\ {\rm lbf} = 1\ {\rm lbm} \cdot g$, which is mass * acceleration and therefore a unit of force.

Edit: Note that lbm is not "equal to 200". It is a unit and will be the same regardless of the problem (and approximately 0.45 kg). A unit in itself does not have a numerical value. A physical quantity has a numerical value given an appropriate unit.

 

[vela]

The 200 in the problem is a mass.

I didn't read it that way. The problem statement does say the drag force is in units of pounds, which suggests when it says the man weighs 200 pounds, it's referring to the man's weight, not mass.

To the OP: You're not recognizing that the numerical value of $g$ depends on the system of units you're using. I'm sure you're not confused with the fact that $g=9.8\text{ m/s}^2$ and $g=32.2\text{ ft/s}^2$ correspond to the same physical quantity even though they're numerically different.

Your book appears to be using the old English system, where force is measured in pounds, acceleration has units of ft/s², and mass is in units of slugs. They're related by $$1\text{ lb} = 1\text{ slug} \times 1\text{ ft/s}^2.$$ Since acceleration is in ft/s^2, you'd have g=32.2 ft/s^2, and a 1-slug object would weigh 32.2 lb. This usage of the word pound is still reflected in units like the ft-lb for energy and lb-ft for torque.

In another system of units, the pound is not a unit of force but a unit of mass. If you specify accelerations in ft/s², the unit of force would be pound ft/s², which, as far as I know, doesn't have a name. (But see below.)

To reduce the confusion between the two uses of the word pound, we use the pound-mass (lbm) and pound-force (lbf). The pound-force is the weight of a pound-mass, and in this set of units, we can say $g=1 \text{ lbf/lbm}$. Note that $1\text{ lbf/lbm} = g = 32.2\text{ ft/s}^2$ so that $\frac{1}{32.2} \text{lbf} = 1\text{ lbm ft/s}^2.$ So a pound(-mass) ft/s² doesn't have a special name, but it's equal to 1/32.2 of a pound-force.

 

[Orodruin]

I didn't read it that way. The problem statement does say the drag force is in units of pounds, which suggests when it says the man weighs 200 pounds, it's referring to the man's weight, not mass.

Well, the problem is using a unit system in which this does not matter in a gravitational field of standard strength ... By definition, 1 lbf is the gravitational force on a mass of 1 lbm in a standard gravitational field.

In another system of units, the pound is not a unit of force but a unit of mass.

Pound force and pound mass are (sometimes) used in the same system of units, see e.g. https://en.wikipedia.org/wiki/English_Engineering_units

 

[vela]

The moral of the story here is Avoid the use of English units.