First let us examine a big kid and a small kid on a teeter-totter. The big kid has to slide toward the middle to balance the small kid. This is because the force times distance (from the pivot) on one side has to balance the force times distance on the other side.
With a rotating arm, the horizontal distance from the pivot point depends on the sine of the angle from vertical. A table will be provided at the end of this discussion.
Let us assume for convenience that the spring and string forces are constant on the other side of the pull.
Actually these spring forces vary as the square of the distance pulled from zero. That means if a force Fs is required to pull the spring ¼ inch, say, then the force to pull ½ inch is 4xFs, the force to pull ¾ inch is 9xFs, and to pull 1 inch is 16xFs. This calculation is complicated further by the sine of the angle of rotation, which is also non-linear. However, we assume both parties pull the same angles, so the total spring/string forces winds up the same for both.
Jumping ahead to avoid confusion with math derivations, it turns out that the amount of downward rod force, Fr, varies inversely with the sine of the angle of rotation.
That means Fr = K / sine(a) where (a) is the angle of rotation from vertical in degrees, and K is a constant as assumed above (which it really is not).
A crude table of 1/sine(a) follows:
Code: Select all
(a) deg 1/sine(a)
0 infinity
2 28.6
5 11.47
10 5.88
20 2.94
30 2.00
40 1.52
45 1.41
47 1.37
50 1.30
60 1.15
70 1.06
80 1.01
90 1.00
The inverse sine at 2 degrees is 28.6, which differs a lot from 2.82.
Note that at 90 degrees, the rod force is equal to 1x the spring/string force, the minimum pull force required.
