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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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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here is my reasoning: |dw:1433092666869:dw| in order to lift the box, if I use the inclined plane, I apply the subsequent force: \[F = mg\sin \theta \]
:) yes
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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so the advantage is given by the ratio R/F, namely: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }}\]
ok:)
now, what is sin(\theta) ?
it is simple, here is its value: \[H = L\sin \theta ,\quad \sin \theta = \frac{H}{L}\]
where H is the height of the inclined plane, where L is its length
so, replacing that expression into the formula for R/F, we get: \[\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }} = \frac{L}{H}\]
okay! so what do we plug in? :/
lenght and height
here is the next step: \[\frac{R}{F} = \frac{L}{H} = \frac{5}{3}\]
ohh so we get 1.66666 so our solution is 1.67?
correct!
yay! thank you!
:)

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