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Can underwater pressure be equated with weight? (after conversions)

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user100
#1
Jul15-14, 11:18 AM
P: 6
It is a established fact that pressure i.e ρgh will be same at a given depth. Suppose in a tank filled with water the gauge pressure works out to 1 bar, then the total pressure at the depth of tank would be 1 bar + 1 bar atmospheric pressure = 2 bar pressure.

2 bar pressure equals 2 kgf/ cm2

Question

Suppose the base of the tank is 200 cm2, can it be said that "water in tank exerts 400 kg WEIGHT on the base of tank" [i.e 200 cm2 * 2kgf/cm2]
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msrlasky
#2
Jul15-14, 03:04 PM
P: 1
The weight exerted on the base of the tank, is due to the volume of water, & it's density (sea or fresh). I assume fresh water. The height of the tank, to exert 1 atmosphere of pressure, would be ~ 10 meters. So, 1,000 cm ht. x 200 cm2 base = 200 liters tank volume.
At 1 kg per liter, the total weight of water, exerted on the base, would be 200 kg. The atmosphere would exert a pressure equivalent to the weight of an additional 200 kg of water, on the base of the tank. However, the atmospheric pressure, being exerted in all directions, is negated. So, weight and pressure, are not the same. A fresh water filled 1 inch diameter tube, 10 meters tall, would still exert 1 atmosphere of pressure, at it's base, but weigh only about 5 kg.
sophiecentaur
#3
Jul15-14, 05:59 PM
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Pressure and weight are different quantities with different units. Sometimes, their numerical value can be the same but that is not significant. So the weight is independent of the pressure, and they would not be the same numerical value if the shape or orientation of the tank changed. Your example works because you happen to have chosen a particular situation where the pressure happens to be the same over the whole area of the bottom of the tank and the sides are vertical so there is no downwards or upwards force on them - just horizontal forces. If the sides were sloping either way than the contributions of pressure on the sides would need to be added in (or subtracted).
So the expression should rather be 'the force on the bottom' - unless the tank is cuboidal with the sides vertical.
The weight of water has a definite value and the pressure at a given depth is also well defined (atmospheric plus hydrostatic).

Note, in SI units, we do not have a 'weight' in kg. Weight is in units of Newtons because it is a force. In the dreaded Imperial system, there is always a problem if you are using pounds for mass and force.

user100
#4
Jul16-14, 04:26 AM
P: 6
Can underwater pressure be equated with weight? (after conversions)

Quote Quote by msrlasky View Post
The weight exerted on the base of the tank, is due to the volume of water, & it's density (sea or fresh). I assume fresh water. The height of the tank, to exert 1 atmosphere of pressure, would be ~ 10 meters. So, 1,000 cm ht. x 200 cm2 base = 200 liters tank volume.
At 1 kg per liter, the total weight of water, exerted on the base, would be 200 kg. The atmosphere would exert a pressure equivalent to the weight of an additional 200 kg of water, on the base of the tank. However, the atmospheric pressure, being exerted in all directions, is negated. So, weight and pressure, are not the same. A fresh water filled 1 inch diameter tube, 10 meters tall, would still exert 1 atmosphere of pressure, at it's base, but weigh only about 5 kg.
ok. But assume i have the base made of glass which can withstand only 395 kg of weight So if i fill the tube (with base of 200cm2) to a height of 10 metres. As u had explained the capacity will be 200 litres fresh water, if it's a vertical tube. Then, when the water reaches 10 metres height will the pressure be enough to break the glass?. i.e will 2 bar pressure acting on 200 cm2 be enough to break the glass?
user100
#5
Jul16-14, 04:37 AM
P: 6
Quote Quote by sophiecentaur View Post
Pressure and weight are different quantities with different units. Sometimes, their numerical value can be the same but that is not significant. So the weight is independent of the pressure, and they would not be the same numerical value if the shape or orientation of the tank changed. Your example works because you happen to have chosen a particular situation where the pressure happens to be the same over the whole area of the bottom of the tank and the sides are vertical so there is no downwards or upwards force on them - just horizontal forces. If the sides were sloping either way than the contributions of pressure on the sides would need to be added in (or subtracted).
So the expression should rather be 'the force on the bottom' - unless the tank is cuboidal with the sides vertical.
The weight of water has a definite value and the pressure at a given depth is also well defined (atmospheric plus hydrostatic).

Note, in SI units, we do not have a 'weight' in kg. Weight is in units of Newtons because it is a force. In the dreaded Imperial system, there is always a problem if you are using pounds for mass and force.
Thanks for the reply. But can u explain further regarding "If the sides were sloping either way than the contributions of pressure on the sides would need to be added in (or subtracted)."

i.e the instances where weight would be higher than pressure or vice versa, when sopes are slanting either way. A link to any material will be very useful. thank you.
CWatters
#6
Jul16-14, 04:58 AM
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Quote Quote by user100 View Post
ok. But assume i have the base made of glass which can withstand only 395 kg of weight
Presumably you mean it can only withstand 395kg distributed uniformly over the 200cm^2 area of the base? or do you mean a 395kg point load in the middle of the base?

If it will withstand only 395kg distributed uniformly then clearly it will fail if you apply 400kg distributed uniformly.

If it will withstand a 395kg point load then it might or might not withstand 400kg distributed. Point loads are generally worse than distributed loads.
sophiecentaur
#7
Jul16-14, 05:25 AM
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Quote Quote by user100 View Post
Thanks for the reply. But can u explain further regarding "If the sides were sloping either way than the contributions of pressure on the sides would need to be added in (or subtracted)."

i.e the instances where weight would be higher than pressure or vice versa, when slopes are slanting either way. A link to any material will be very useful. thank you.
I thought that might get you going!
What I meant was that the pressure on any surface will be ρgh, at any point of the surface where the depth is h. If the container is a truncated conical funnel shape (a frustum, point down), the downward force on the cone sides near the water surface will be less than near the bottom but will still add to the force at the bottom. Likewise, with the funned inverted, there will be a downward force over all the bottom (plane) surface but pressure acting upwards on the inverted sloping surface will cause an upwards force, subtracting from the total force acting on the flat bottom. This sort of argument will apply to all shapes container - always giving you the weight force as the same for the same mass of water.
The point is that the 'weight' has contributions from any non-vertical surface.

Weight cannot be 'higher or lower' than pressure because they are two different quantities in two different units. The force on the bottom (pressure times area) can obviously be different from the weight, which is the sum of all vertical forces.


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