Hydraulic pressure dropping bottle

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    Hydraulic Pressure
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Calculating the hydraulic pressure at the bottom of a bottle upon impact involves understanding the force exerted by both the bottle and the liquid inside. The impact force can be estimated by dividing the total force by the area of the bottle's bottom, assuming even distribution. Pascal's principle indicates that the pressure at the bottom consists of the pressure from the liquid above and any additional pressure from the impact. While the hydrostatic pressure is present, it may be negligible compared to the pressure generated by the impact force. Overall, the impact force is relevant for determining the hydraulic pressure at the bottom of the bottle.
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I am attempting to calculate the pressure at the bottom of a bottle at the moment of impact. I have calculated the force of impact ,but now I am unsure if this will be the same force (approx) applied to the top of the liquid to determine the hydraulic pressure at the bottom of the bottle.
In other words I need to find the hydraulic pressure exerted in a bottle, ( a full bottle that has fallen a certain distance).
If i have a bottle full of liquid of given mass falling a given distance.I can calculate the force at which it impacts. But does this help me find the answer to the above question?
 
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onebad1968 said:
I am attempting to calculate the pressure at the bottom of a bottle at the moment of impact. I have calculated the force of impact ,but now I am unsure if this will be the same force (approx) applied to the top of the liquid to determine the hydraulic pressure at the bottom of the bottle.
In other words I need to find the hydraulic pressure exerted in a bottle, ( a full bottle that has fallen a certain distance).
If i have a bottle full of liquid of given mass falling a given distance.I can calculate the force at which it impacts. But does this help me find the answer to the above question?

I suppose you could estimate it by simply dividing the force by the area of the bottom of the bottle. This makes the assumption that the force is distributed evenly across the bottom and the fluid is in contact with the bottom of the bottle at impact.

CS
 
It would seem to me that the sum of both masses, (bottle and liquid) will make up the total impact at point of outside contact. The liquid inside will gain energy in the fall, but being a liquid will spread it's energy at impact to the inside bottom and all around the inside walls of the bottle, also the gas portion above the liquid would receive some of the energy through compression.

Just don't ask me to do any math:biggrin:

Ron
 
According to pascals principle.. The pressure applied to the top of the fluid can be calculated by using... (P1 being a force pushing down on the top of the liquid)

p1= Force1/area1... The pressure on the bottom of the bottle p2=p1+static fluid pressure.. and finally the force on the bottom of the bottle F2 = P2 A2. so I still don't know if the calculation of the force at which the bottle hits is relevant to calculating the above or is there another way to do this.
 
onebad1968 said:
According to pascals principle.. The pressure applied to the top of the fluid can be calculated by using... (P1 being a force pushing down on the top of the liquid)

p1= Force1/area1... The pressure on the bottom of the bottle p2=p1+static fluid pressure.. and finally the force on the bottom of the bottle F2 = P2 A2. so I still don't know if the calculation of the force at which the bottle hits is relevant to calculating the above or is there another way to do this.

Draw a FBD and sum the forces. Then calculated the pressure. There will be a small hydrostatic head pressure plus whatever pressure is generated by the impact force. The hydrostatic head is probably negligible compared to the equivalent impact pressure so you can probably just ignore it. P1 is on the top of the bottle (atmospheric) and on the bottom so it cancels out...ignore it too.

CS
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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