# Silly Question on Binding for Piston

1. Sep 1, 2015

### dsurfer21

Hi,

Another fellow engineer told me that for a dual connected piston guided inside a tube (as shown in the diagram below), the distance D1 has to be chosen correctly in order to avoid binding (I'm guessing from friction only) since there may be a side force present. This is ignoring gravity for now. In the example discussed there is a force F applied with some side force component Fn. I couldn't make sense as to why D1 is relevant. Can anyone explain it easily?

The only thing that comes to mind is that I see a normal force at point A and point B as a result of Fn. I can see how maybe there is a moment created at point B by the normal force at point A times the distance D1 and then a moment at point B from the force F times D2/2. Would the balance of those moments result in the value for D1?

2. Sep 1, 2015

### Bystander

What happens to the moments you've described as D1 goes to zero?

3. Sep 1, 2015

### dsurfer21

The moments at point B would be

F times D2/2

minus

Fn/2 times D1

As D1 goes to zero the moment at point B would be F times D2/2. I'm not sure how this relates to binding unless a positive moment is contributing to a force normal to the tube and friction comes into play. Is there another correct explanation?

4. Sep 2, 2015

### Nidum

As stated the question is not very meaningful . Piston stability depends on many different things .

5. Sep 2, 2015

### dsurfer21

Thank you, I thought the same and I know there are other factors to consider for piston stability.

However I was asked what distance D1 would eliminate binding ignoring all other effects. Is this a matter of solving for the moments about point B and making sure they sum to zero?