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brinethery

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

  • 2 years ago
  • 2 years ago

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  1. eashmore
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    I'd be glad to assist.

    • 2 years ago
  2. brinethery
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    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?

    • 2 years ago
  3. brinethery
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    I need help with the second and third questions.

    • 2 years ago
  4. brinethery
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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).

    • 2 years ago
  5. brinethery
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    Seriously... I HATE the English system. :-|

    • 2 years ago
  6. brinethery
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    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.

    • 2 years ago
  7. eashmore
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    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.

    • 2 years ago
  8. brinethery
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    I guess what this means is the only lbm we can go off of is the one that we find on earth.

    • 2 years ago
  9. brinethery
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    B/c you and I found two different lbms... one on earth and one on Jupiter which we know wouldn't be the case.

    • 2 years ago
  10. eashmore
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    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\[7.6 lb_f = 16 lb_m \cdot {g_J \over 32.17}\]

    • 2 years ago
  11. eashmore
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    On earth this is expressed as\[16 lb_f = 16 lb_m \cdot {32.17 \over 32.17}\]

    • 2 years ago
  12. eashmore
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    Therefore, your teachers conversion is correct. \[x kg = 16 \cdot 0.45\]

    • 2 years ago
  13. brinethery
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    I'm still studying what you wrote...

    • 2 years ago
  14. brinethery
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    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.

    • 2 years ago
  15. eashmore
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    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.

    • 2 years ago
  16. eashmore
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    My apologies for the initial confusion.

    • 2 years ago
  17. brinethery
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    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.

    • 2 years ago
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