anonymous
  • anonymous
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 ? :/
Physics
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
here is my reasoning: |dw:1433092666869:dw| in order to lift the box, if I use the inclined plane, I apply the subsequent force: \[F = mg\sin \theta \]
anonymous
  • anonymous
:) yes
Michele_Laino
  • Michele_Laino
whereas if I don't use the inclibed plane I apply this force: \[R = mg\] namely the weight of the box |dw:1433092864019:dw|

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Michele_Laino
  • Michele_Laino
so the advantage is given by the ratio R/F, namely: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }}\]
anonymous
  • anonymous
ok:)
Michele_Laino
  • Michele_Laino
now, what is sin(\theta) ?
Michele_Laino
  • Michele_Laino
it is simple, here is its value: \[H = L\sin \theta ,\quad \sin \theta = \frac{H}{L}\]
Michele_Laino
  • Michele_Laino
where H is the height of the inclined plane, where L is its length
Michele_Laino
  • Michele_Laino
so, replacing that expression into the formula for R/F, we get: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }} = \frac{L}{H}\]
anonymous
  • anonymous
okay! so what do we plug in? :/
yadu123
  • yadu123
lenght and height
Michele_Laino
  • Michele_Laino
here is the next step: \[\frac{R}{F} = \frac{L}{H} = \frac{5}{3}\]
anonymous
  • anonymous
ohh so we get 1.66666 so our solution is 1.67?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
yay! thank you!
Michele_Laino
  • Michele_Laino
:)

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