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Could someone help me with a problem involving units of lbm and lbf?

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I'd be glad to assist.
A bowling ball weighs 16 pounds-force (lbf) on earth, but only 7.6 lbf on Jupiter. What is the acceleration of gravity on Jupiter in meters/second? What is its mass on earth in lbm? What is the mass of the ball in kilograms on Jupiter?
I need help with the second and third questions.

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First off, I want to thank you for putting so much detail into your answer. This looks like I'm going to have to speak with my teacher immediately. B/c he said that the ball would be 16lbm (1kg/2.2lbm) = 7.25kg. This is why I kept getting so confused because I calculated 0.5 lbm on Jupiter, but then I keep hearing that 1lbf = 1lbm b/c you can convert from 1lbm to slugs, and then multiply that by the gravitational constant (whether it be on earth or some other planet).
Seriously... I HATE the English system. :-|
Who knows... maybe my instructor just screwed up. This is really bugging me. I have no problems with the metric system, but the English system is so *** backwards.
I hate it too. I should know it better (being an engineering student and all). I looked into it some more and my approach is wrong. \(1 lb_F = 1 lb_m\) on earth. So let's redefine \(lb_F\) as\[1 lb_F = 1 lb_m \cdot {g \over g_m}\]where \(g_m\) is the gravitational acceleration on earth. Therefore, the mass of the object on earth is, in-fact, \(16 lb_m\). To find the acceleration on Jupiter, let's solve the following equation for g. \[7.6 [lb_F] = 16 \cdot {g \over 32.17}\]g can readily be converted to m/s^2.
I guess what this means is the only lbm we can go off of is the one that we find on earth.
B/c you and I found two different lbms... one on earth and one on Jupiter which we know wouldn't be the case.
The mass will be the same. \[1 lb_f = 1 slug \cdot a\]\[1 slug = {1 lb_m \over 32.17}\]Therefore,\[7.6 lb_f = {16 \over 32.17} slug \cdot g_J\]which is equal to\[7.6 lb_f = 16 lb_m \cdot {g_J \over 32.17}\]
On earth this is expressed as\[16 lb_f = 16 lb_m \cdot {32.17 \over 32.17}\]
Therefore, your teachers conversion is correct. \[x kg = 16 \cdot 0.45\]
I'm still studying what you wrote...
I'll print out that 3rd to last thing your wrote and just keep looking it over. Thank you for really going over this with me. It's extremely confusing. But your explanation is making more and more sense as I let it sink in. And I'm sure it'll be even clearer tomorrow. One more medal for you if I could give it.
This is not easy to understand, it got me confused earlier. I have to study the principles every time I'm presented with an english units problem. Keep in mind that Newton's second law doesn't hold when we use pound-mass. \[F \ne [lb_m] \cdot a\]Instead, \[F = [slug] \cdot a\]Therefore, \[F = \left [ {lb_m \over 32.17} \right ] \cdot a\]since\[1 [slug] = \left [1 lb_m \over g_m \right] = \left [ 1 lb_m \over 32.17 \right]\] I might recommend just converting to slugs since we can use slugs just like we use kgs.
My apologies for the initial confusion.
Oh no, it's okay! My boyfriend's a mechanical engineer and took this class a few years ago... and he couldn't remember how to do stuff regarding British units. My dad couldn't and he's an engineer. So I wasn't expecting someone to just "get it" immediately. That's the nature of this system.

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