Leverage and Forces on a Pivot System

[sophiecentaur]

I do see trouble for big strong guys: they can easily compress the stuff so tightly that it doesn't come out any more when the bin is collected...

This is a common problem and there's no sure fire solution. A slight taper on the container can help. The same solution that works to let moulds (molds) release their contents.
The arrangement in the diagram immediately above can produce a lateral force that can bend or bind the mechanism. Getting this just right in a design makes the difference between working and breaking. A cam mechanism can sometimes ensure that the vertical force stays vertical and better controlled. in these situations. But that puts the manufacturing cost up.

 

[Saints-94]

Firstly, thanks for your help.

The only unknown I have is the distance you have labelled Y. I have added 'F3' to the diagram to show the force acting against mechanism (representing the resistance of the waste).

How do I find the value of F2? Do I just substitute F2v for F3?

 

[sophiecentaur]

Firstly, thanks for your help.
View attachment 113820

The only unknown I have is the distance you have labelled Y. I have added 'F3' to the diagram to show the force acting against mechanism (representing the resistance of the waste).

How do I find the value of F2? Do I just substitute F2v for F3?

I notice that you now have the applied force acting at right angles to the lever. Would this be likely - as it would tend to tip the bin over. I would think that the natural direction to push would be vertically downward. The bottom line is that the force on the rubbish would be twice the applied force (more accurately [X+Y]/X plus or minus any variation in applied angel and friction). The angle of the bar to the horizontal will have no effect because the ratio of the perpendicular distances is the same.
Apart from any weight force from the press arms, the force on the rubbish is the same as the magnified force from the operator. F_2,v = F₃)

 

[Saints-94]

Decomposition of F2F2F_2:

F2,v=F2sin45∘F2,v=F2sin⁡45∘​

F_{2,\rm v} = F_2 \sin 45^\circ

So your formula should be

F2,v=F1X+YXsin45∘F2,v=F1X+YXsin⁡45∘​

F_{2,\rm v} = F_1 {X+Y\over X} \sin 45^\circ

So the equation above is not relevant as the angle does not need to be considered as the force is worked out using a ratio of X and Y?

 

[Nidum]

The compactor is going to be prone to tipping over anyway . Excessive lever length / lever ratio will make this problem worse .

Useful pictures here .

The picture with the lady operator and the handle at highest level is particularly interesting .

Lever ratio appears to be about 2.5 : 1

 

[Saints-94]

If the weight of the bin could be determined, could an equation be formed to work out the longest the lever could be without the bin tipping?

 

[sophiecentaur]

So the equation above is not relevant as the angle does not need to be considered as the force is worked out using a ratio of X and Y?

For a vertically applied force, you are dealing with similar triangles here. The ratio of lengths of the horizontal sides are the same as the ratio of measurements along the bar.
X and X+Y, actually

If the weight of the bin could be determined, could an equation be formed to work out the longest the lever could be without the bin tipping?

Yes. consider the edge of the bin as a fulcrum and the moments of the applied force and the total weight of bin plus rubbish plus lever system. Then equate the moments to give you the longest lever for equilibrium.
However, the user could probably exceed this value by adjusting the angle of the push and 'steadying' the system to avoid tipping.