Answer:
The efficiency of the ramp11%
Explanation:
Let AC be the length of inclined plane or ramp = 3m (given)
BC be the height through which the load is lifted = 1m.
Load or Box (L) = 20kg
= 20×9.8 N = 196N
Effort (E) = [tex]\mathrm{L} \times \sin (\theta)[/tex]
= 196 ×0.33 =64.68
where θ is the angle between AC and AB.
Therefore,
Mechanical Advantage (MA) = [tex]\frac{\mathrm{L}}{\mathrm{L} \sin (\theta)}[/tex]
MA = 0.33.
Now, Velocity Ratio (VR) =[tex]\frac{1}{\mathrm{h}}[/tex]
= [tex]\frac{\mathrm{AC}}{\mathrm{BC}}[/tex]
= 3/1 = 3
Efficiency =[tex]\frac{M A}{\mathrm{VR}}[/tex]
[tex]\frac{M A}{\mathrm{VR}}[/tex]
= [tex]\frac{0.33}{3}[/tex]
=0.11 = 11% which is the efficiency of the ramp
