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anonymous

  • one year ago

An inclined plane is 5.00 m long and 3.00 m high. What is the ideal mechanical advantage of this machine? 1.00 , 2.00, 1.67 , or 2.50 ? :/

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  1. Michele_Laino
    • one year ago
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    here is my reasoning: |dw:1433092666869:dw| in order to lift the box, if I use the inclined plane, I apply the subsequent force: \[F = mg\sin \theta \]

  2. anonymous
    • one year ago
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    :) yes

  3. Michele_Laino
    • one year ago
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    whereas if I don't use the inclibed plane I apply this force: \[R = mg\] namely the weight of the box |dw:1433092864019:dw|

  4. Michele_Laino
    • one year ago
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    so the advantage is given by the ratio R/F, namely: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }}\]

  5. anonymous
    • one year ago
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    ok:)

  6. Michele_Laino
    • one year ago
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    now, what is sin(\theta) ?

  7. Michele_Laino
    • one year ago
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    it is simple, here is its value: \[H = L\sin \theta ,\quad \sin \theta = \frac{H}{L}\]

  8. Michele_Laino
    • one year ago
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    where H is the height of the inclined plane, where L is its length

  9. Michele_Laino
    • one year ago
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    so, replacing that expression into the formula for R/F, we get: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }} = \frac{L}{H}\]

  10. anonymous
    • one year ago
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    okay! so what do we plug in? :/

  11. yadu123
    • one year ago
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    lenght and height

  12. Michele_Laino
    • one year ago
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    here is the next step: \[\frac{R}{F} = \frac{L}{H} = \frac{5}{3}\]

  13. anonymous
    • one year ago
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    ohh so we get 1.66666 so our solution is 1.67?

  14. Michele_Laino
    • one year ago
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    correct!

  15. anonymous
    • one year ago
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    yay! thank you!

  16. Michele_Laino
    • one year ago
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    :)

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