Calculating Force and Torque for Linear motion (image provided)

[Cadormare]

Homework Statement

This isn't a homework problem. This is for a project I am working on. I am trying to understand how to calculate the force needed to compress the spring in the mechanism shown in the image.

Relevant Equations

A = the distance from the axis of rotation to the pivot joint that connects the handle (shown in black) to the rod (show in blue) and spring (shown in gray) assembly.

B = the distance from the location force will be applied to the handle (shown in black) to the pivot joint that connects the handle (shown in black) to the rod (shown in blue) and spring (shown in gray) assembly.

C = the distance the spring (shown in gray) needs to be compressed.

As the handle (shown in black) has force applied to it the spring (show in gray) will be compressed.

This is how I understand the torque (T1) created at the axis of rotation due to the spring would be calculated.
This is considering the spring is starting from a fully extended state and is not already compressed.
This is considering that (A) has a distance of 2 in. and the force from the spring is 20 lbs/In.
T1 = (2 in.) x (20 lbs/In)
T1 = 40 lbs/In

If the torque (T2) needed to compress the spring is 40 lbs/In then this is how I understand to calculate how much force (F1) is needed at the end of the handle (show in black).
This is considering that (A) has a distance of 2 in. and (B) has a distance of 4 in.
T2 = (A + B) x F1
T2 = 6 in. x F1
F1 = T2 / 6in.
F1 = (40 lbs/In) / 6 In
F1 = 6.67 lbs.

So to compress the spring an inch I would need 6.67 lbs of force for every inch I wanted to compress the spring?
If I wanted to compress the spring half an inch would it be ( 6.67 lbs. ) x (0.5) = 3.33 lbs. ?If the spring has a rate of 20 lbs./In and is already compressed an inch, would T1 then be (40 lbs/In) x 2 = 80 lbs/In ?

 

[Lnewqban]

Welcome!
The force with which the spring pushes back on the point of connection with the handle depends on how much distance or deflection it is compressed.
The rate or constant of this spring has a magnitude of 20 lbf/in.
If the handle deforms it 0.5 inch, it will push back with 10 lbf.

If you can actually apply the center of your hand's force at 6 inches from the pivot (A+B distances), then you have a mechanical advantage of 2.
That means that the rod-handle joint will "feel" twice the force that you hand applies (on a perpendicular direction) on the end of the handle.

Please, see:
https://en.wikipedia.org/wiki/Mechanical_advantage

If we consider the angles, the spring is "feeling" magnitudes of force that differ from the above for significant deviations from vertical of the handle.
In practical terms, there should be some way to decouple the angular movement of the lever and the rectilinear movement of the rod.