A car drives 40 km due east and then 50 km due west .what is the cars overall displacement?
2 answers:
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Answer:
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= 20N
As 20N << 125N, the best choice is to use lever.
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
Now we need to find
Since the height is 5m and the length of the ramp is 8m,
=> m*g*
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:
Plug-in the values in the above equation:
As 20N << 125N, the best choice is to use lever.