# Question on finding force needed to compress system

1. Jun 28, 2012

### TokerM

I have attached a pdf file showing the calculation in question.

The 2 views in the pdf are section in the same line. When the operator presses the thumb lever, the pivot shaft rotates the pusher lever which in turn rotates the tooling.

The tooling is forced open via the compression spring.

My question is, how much force will the operator need to exert to compress the system, and does anyone know if/what OSHA has to say on how much a human thumb can/should push?

Thanks in advance, and if any further dimensions/info is needed, let me know, I hope I gave way more info than needed to calculate this.

Edit: For ease in calculations, assume 0 friction and ignore the mass of the levers and tooling adding to the gravitational torque of the system.

#### Attached Files:

• ###### torque force.pdf
File size:
16.8 KB
Views:
96
Last edited: Jun 28, 2012
2. Jun 29, 2012

### TokerM

95 views and nobody knows how to set up this calculation?

3. Jul 1, 2012

### nvn

TokerM: We currently cannot relate the right-hand diagram to the left-hand diagram, except for the second linkage bar (the pusher lever), which I recognize in both diagrams. Therefore, we currently cannot understand the mechanism, due to insufficient data. And the third linkage bar (the tooling) image is cut off, and seemingly does not show all of its boundary conditions; as currently drawn, it would just rotate on its pivot, without compressing the spring. You also do not show how the thumb lever linkage bar is connected to the pusher lever linkage bar, nor how the pusher lever linkage bar is connected to the tooling linkage bar.

4. Jul 2, 2012

### TokerM

One view is to be superimposed over the other. The thumb lever, and pusher lever are fixed to the same shaft. In plan view, the pusher lever is "L" shaped to "hook" on the tooling.

The tooling is similar to a ladder runglock. Due to NDAs, that's all I can say.

I'm really only looking for the final pushing force this system creates, not the detailed attachments of the mechanism. That part is good. This is one of those "works on paper" ideas.

I am getting confused by how to set up the torque calculation.

Push the thumb lever, it rotates the shaft 8 deg. The center of the thumb pad is 2.13" away from the CL of the pivot shaft.

Rotating the shaft 8 deg rotates the pusher lever which, due to the L shape, pushes on the tooling face. By pushing through the 8 deg, the length of travel on the tooling is 0.82" along the arc. This action causes the tooling to rotate on a separate shaft, in the position shown.

In order for the tooling to rotate clear of obstructions, it must be rotated 15 deg.

With the spring that has been selected, in the position shown, there is 8 lbs of force needed to compress the spring the 15 deg. The pusher lever is pressing against the tooling at a position 3.09" from the tooling CL.

I'm attaching a dwg view of everything overlapped, you'll see why I broke things into 2 sections, the pic is a bit busy!

Thanks again.

File size:
15.9 KB
Views:
110
5. Jul 2, 2012

### nvn

TokerM: P1 = [6.00*5.54/(2.13*3.09)](8 lbf) = 5.050*8 = 40.4 lbf (180 N), where P1 = required force exerted on thumb lever by operator thumb.

I did not look for an OSHA limit. However, P1 = 180 N is far too high. Even 90 N is rather painful. And 110 N is quite painful. Therefore, you should not exceed roughly 60 % of the rather painful value; i.e., do not exceed P1 = 55 N. Or preferably, do not exceed P1 = 50 N (or even better, 45 N).

Last edited: Jul 2, 2012
6. Jul 3, 2012

### TokerM

That's what I was afraid of... As the old adage goes... "Back to the board!"

Thank you for helping with the calc. From that I can find a workable setup.