Download The equation for a Lever System is: F1D1=F2D2 That same

The equation for a Lever System is:
F1D1=F2D2
That same equation works for all 3 class levers.
The force directly vertical is the only force the Lever system will take. To get vertical forces on a wheel
at any angle, one must use the following equation:
(mg)COS(angle)
[1]
This can be visually checked if one place a load on top of a post standing 90*. The movement force is
zero because the post absorbs all the force. Now the post is 0* and the load is hanging from the top of
it, then the force is the total weight.
This was done to find the force at every angle between 90-0.
Each force from the above at a given angle is then plugged into the Lever equation as F2.
A balanced Lever has equal distance and force between the fulcrum. This is common in gravity wheels
as the fulcrum is the axis of the wheel.
In order to find the USEFUL force, one needs to find F1. The equation for F1:
F1=F2D2/D1
[2]
where F2 is the force found by [1], D2 is the distance between the axis and the force, and D2 is the
distance between the axis and the force we are trying to find.
Substitute F2 for [1]:
F1=(mg)COS(degree)(D2)/D1 [3]
Example 1: At 90*
F2 = (5)COS(90) = 0
D2 = 40
D1 = 40
F1 = (0)(40)/40 = 0
Example 2: At 0*
F2 = (5)COS(0) = 5
D2 = 45
D1 = 40
F1 = (5)(45)/40 = 5.6
In Example 2, F1 is 5.6 while F2 is 5. Since this is on a wheel and is horizontal to the axis, then there is
a required addition of 0.6 force to the current 5 force to balance the lever at D1. The addition is the
USEFUL force at D2 as if the only weight on the entire wheel. The equation to find the USEFUL force:
F1=(mg)COS(degree)(D2)/D1 – (mg)COS(degree) [4]
Now the useful force is defined. We now want to find the force at the stationary rack in which we call
the Small Wheel. This time, we are using the 2nd class lever system under the same equation [2]:
F1=F2D2/D1
[2]
where F2 = [4], D2 is D1 from [3], and D1 is the radius of the Small Wheel.
Example 1: At 0*
F2 = (5)COS(0)(45)/40 – (5)COS(0) = 0.6
D2 = 40
D1 = 4
F1 = (0.6)(40)/(4) = 6.00
The reason for D2 to be D1 from [3] is because the distance moved of D1. D1 becomes the distance of
the Small Wheel which is the stationary rack.