## 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 ? :/

1. 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: $F = mg\sin \theta$

2. anonymous

:) yes

3. 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|

4. Michele_Laino

so the advantage is given by the ratio R/F, namely: $\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }}$

5. anonymous

ok:)

6. Michele_Laino

now, what is sin(\theta) ?

7. Michele_Laino

it is simple, here is its value: $H = L\sin \theta ,\quad \sin \theta = \frac{H}{L}$

8. Michele_Laino

where H is the height of the inclined plane, where L is its length

9. Michele_Laino

so, replacing that expression into the formula for R/F, we get: $\frac{R}{F} = \frac{{mg}}{{mg\sin \theta }} = \frac{1}{{\sin \theta }} = \frac{L}{H}$

10. anonymous

okay! so what do we plug in? :/

lenght and height

12. Michele_Laino

here is the next step: $\frac{R}{F} = \frac{L}{H} = \frac{5}{3}$

13. anonymous

ohh so we get 1.66666 so our solution is 1.67?

14. Michele_Laino

correct!

15. anonymous

yay! thank you!

16. Michele_Laino

:)