Let's break the question into two parts:
1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.
1. Ramp Scenario: In an incline, the only component of cart's weight(
mg) that is in the direction of motion is
. Therefore the effort force in this case must be equal or greater than .
Now we need to find
.
is the angle between the incline of the ramp and the ground.
Since the height is
5m and the length of the ramp is
8m, would be
5/8 or 0.625. Now that you have
, mutiple it with
mg.
=> m*g*
= 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is
125N.
2. Lever Scenario:
Just apply "moment action" in this case, which is:
= ?
= mg = 20 * 10 = 200N
= 10m
= 1m
Plug-in the values in the above equation:
= 200/10=
20NAs 20N << 125N, the best choice is to use lever.