Pressure inside a plastic bottle filled with water and squeezed by X weight?

[charles80]

Hi,

I'd like to know how it'd be possible to calculate the pressure inside a plastic bottle filled with water and squeezed by say 20 kg sitting on the bottle, which is lying on its side (so that resistance to deformation does't come to play as much as it'd if it was sitting on a standing bottle).

Thank you

Charles

 

[Chestermiller]

Hi,

I'd like to know how it'd be possible to calculate the pressure inside a plastic bottle filled with water and squeezed by say 20 kg sitting on the bottle, which is lying on its side (so that resistance to deformation does't come to play as much as it'd if it was sitting on a standing bottle).

Thank you

Charles

Hi Charles. Welcome to Physics Forums.

The approach to this problem depends on how accurate you want your answer. If you are willing to approximate the bottle as an infinitely long cylinder (neglecting the effects of the bottle ends), then you can get a pretty quick answer. The plastic would be treated as a flexible, inextensible membrane. You would also assume zero friction between the weight and the plastic and between the plastic and the table (i.e., butter on the outside of the bottle). The surface of the weight would have to be flat. You would neglect the weight of the liquid in the bottle, compared to the applied weight.

Chet

 

[Alfreds9]

I have this same question. Could the internal pressure be approximated to Force/Surface, so (mass*g)/surface area of a bottle?
How inaccurate would this be?

Thank you

 

[Chestermiller]

I have this same question. Could the internal pressure be approximated to Force/Surface, so (mass*g)/surface area of a bottle?
How inaccurate would this be?

Thank you

No. It couldn't be treated that way. What basic analysis have you done on this problem so far? Have you drawn a diagram?

 

[russ_watters]

The approach to this problem depends on how accurate you want your answer. If you are willing to approximate the bottle as an infinitely long cylinder (neglecting the effects of the bottle ends), then you can get a pretty quick answer. The plastic would be treated as a flexible, inextensible membrane. You would also assume zero friction between the weight and the plastic and between the plastic and the table (i.e., butter on the outside of the bottle). The surface of the weight would have to be flat. You would neglect the weight of the liquid in the bottle, compared to the applied weight.t

What am I missing here: it seems to me that similar to a car tire, the pressure could be literally anything (that the tire can withstand and enough to lift the wheel hub off the ground). Increasing the pressure reduces the size of the contact patch without affecting the supported weight.

 

[Chestermiller]

What am I missing here: it seems to me that similar to a car tire, the pressure could be literally anything (that the tire can withstand and enough to lift the wheel hub off the ground). Increasing the pressure reduces the size of the contact patch without affecting the supported weight.

You're not missing anything, although this is certainly much simpler to analyze than the mechanics of radial tires under contact loading with the ground (as I learned when I worked in that area).

 

[Alfreds9]

What am I missing here: it seems to me that similar to a car tire, the pressure could be literally anything (that the tire can withstand and enough to lift the wheel hub off the ground). Increasing the pressure reduces the size of the contact patch without affecting the supported weight.

I (perhaps wrongly) assumed that pressure inside a bottle with a weight sitting on top should be higher than pressure inside a bottle with no weight on?

Here's a sketch I did

 

[russ_watters]

The pressure is higher, but it isn't the total surface area that supports the weight it is just the contact area that supports the weight.

 

[Alfreds9]

The pressure is higher, but it isn't the total surface area that supports the weight it is just the contact area that supports the weight.

I can't get my head around this (or maybe I just explained my question poorly): if one considers only the weight-bearing area as you suggest, wouldn't a smaller supporting area yield a higher pressure even if weight stays the same, something which doesn't make sense to me since I guess that pressure experienced by an object inside the bottle would still be the same as long as the weights are equal?

Thank you

 

[Chestermiller]

I can't get my head around this (or maybe I just explained my question poorly): if one considers only the weight-bearing area as you suggest, wouldn't a smaller supporting area yield a higher pressure even if weight stays the same, something which doesn't make sense to me since I guess that pressure experienced by an object inside the bottle would still be the same as long as the weights are equal?

Thank you

When I have a chance, I will draw you a diagram.

 

[Chestermiller]

 

[Nidum]

I'm getting multiple repeats of the same posting - is there a glitch in PF software ?

 

[Chestermiller]

I'm getting multiple repeats of the same posting - is there a glitch in PF software ?

yes. I had some problems posting.

 

[russ_watters]

I can't get my head around this (or maybe I just explained my question poorly): if one considers only the weight-bearing area as you suggest, wouldn't a smaller supporting area yield a higher pressure even if weight stays the same...

That is exactly what happens. Think about the car tire example given above. You can inflate your car tires to any pressure until they pop, but the weight of your car doesn't change.

 

[CWatters]

Perhaps this version of the problem makes it easier to understand. The pressure in the bottle is the weight divided by the cross sectional area of the cork.

 

[Chestermiller]

Referring to post #11, if the total perimeter is $\pi D$ and the length of the cylinder is L, then the width of the footprint is $\frac{W}{pL}$ on both the weight side and the floor side, where p is the inside gauge pressure. So the combined length of the 2 sidewalls is $\pi D-\frac{2W}{pL}$. The sidewalls are circular, so the height of the weight above the floor is $D-\frac{2W}{\pi pL}$.

 

[russ_watters]

Perhaps this version of the problem makes it easier to understand. The pressure in the bottle is the weight divided by the cross sectional area of the cork.

View attachment 105384

My concern with this example is that the example has the potential to solidify the erroneous assumption that the cross sectional area would be fixed regardless of the mass sitting on the cork. With the bottle laying on its side, both the pressure and cross sectional area change with the mass and the pressure can also change independent of the other two (the other two being dependent).

 

[Cryston Waston]

As per my opinion it wouldn't increase the pressure on the bottle . A force on the outside of bottle would cause a deformation inside the bottle, hence air volume and pressure also increases.

 

[Chestermiller]

As per my opinion it wouldn't increase the pressure on the bottle . A force on the outside of bottle would cause a deformation inside the bottle, hence air volume and pressure also increases.

Can you please elaborate on this, perhaps with a diagram? It is unclear what you are saying.

Incidentally, you do realize that the last posting in this tread was almost 2 years ago, correct? And the last posting from the OP was 4 years ago.

 

[Jonathan1218]

I am trying to answer the original question: It is possible to calculate the pressure inside the bottle because if a weight is exerting a force upon the bottle (assuming the bottle is flexible, e.g plastic as opposed to steel) then there will be an equal and opposite force (which can be measures as pressure). ( Good old Newtons third law) Thus the downwards weight force is proportionate to the pressure that's pushing against the inside of the bottle. For a 20kg of mass, it will exert a weight of approx. 200Newtons of force on Earth and assuming the bottle has a surface area of one meter squared, the pressure resulting from the weight will be 200 Pascals as it means exerting 200N of force per meter square.

 

[Jonathan1218]

A correction to my previous post, in the last sentence i should say the assuming the contact area is one meter squared

 

[tallmonk]

Thanks @Jonathan1218 your explanation made the most sense