Pressure on block immersed in water?

  • #1
i was solving a problem based on archimedes principle. but got confused whilr determinig the pressure.
well the situation was that a block is immersed in water. the height of water column above it is .05m. the blocks side is .02 m and its base area is 25cm^2.and the height of water below the block is not given.
the presure on top of the block was simply the application of formula 'pgh' = 500pascal.
but how do i determine the pressure on the lower surface ?

also in general while determinig pressure on lower surface i THINK that we consider the height of water column from the upper edge of block to the bottom of the water container.
if this is true then why?
explain
 

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Answers and Replies

  • #4
Spinnor
Gold Member
2,176
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From your picture it looks to be .07m
 
  • #5
From your picture it looks to be .07m

yes the question was solved in my book using .07m
but why? i mean we have to find the pressure acting on the lower surface. so why not consider the height of water under it?
i dont understand this point.
 
  • #6
gneill
Mentor
20,925
2,866
yes the question was solved in my book using .07m
but why? i mean we have to find the pressure acting on the lower surface. so why not consider the height of water under it?
i dont understand this point.

It's the weight of the water above (and the weight of the atmosphere above that!) that is is creating the pressure at any given depth. The water below only provides a "surface" upon which the weight acts to manifest the pressure. If this were not the case the oceans would leap from their beds, there being no water above them to hold them down against the pressure of all that water below the surface level :smile:
 
  • #7
It's the weight of the water above (and the weight of the atmosphere above that!) that is is creating the pressure at any given depth. The water below only provides a "surface" upon which the weight acts to manifest the pressure. If this were not the case the oceans would leap from their beds, there being no water above them to hold them down against the pressure of all that water below the surface level :smile:

ok...you have a point
but then according to what you are saying , the pressure on the lower surface should also be in the lower direction and thus there will be no upthrust. pressure on both surfaces will be directed downwards.....
also i remeber when i was in 8th grade there was a problem similar to this one and the book authors while computing pressure on lower surface considered the water column to be = height of block+water under it
so i am all confused...explain a bit more
 
  • #8
gneill
Mentor
20,925
2,866
ok...you have a point
but then according to what you are saying , the pressure on the lower surface should also be in the lower direction and thus there will be no upthrust. pressure on both surfaces will be directed downwards.....
also i remeber when i was in 8th grade there was a problem similar to this one and the book authors while computing pressure on lower surface considered the water column to be = height of block+water under it
so i am all confused...explain a bit more

Pressure is caused by the weight of the material above, but it isn't weight itself. Gases and liquids are composed of highly mobile particles that distribute the force in all directions. The weight of the material above sets the magnitude of the distributed force at any given depth.

Pressure on an immersed object is always directed inwards, normal to the surface of the object. Upward facing, downward facing, sideways, or slanted, it makes no difference.

The problem you're remembering must have used a coordinate system that had its origin at the bottom of the container, and likely determined the depth as the difference between a given height in the column and the height of the surface.
 
  • #9
Pressure is caused by the weight of the material above, but it isn't weight itself. Gases and liquids are composed of highly mobile particles that distribute the force in all directions. The weight of the material above sets the magnitude of the distributed force at any given depth.

Pressure on an immersed object is always directed inwards, normal to the surface of the object. Upward facing, downward facing, sideways, or slanted, it makes no difference.

The problem you're remembering must have used a coordinate system that had its origin at the bottom of the container, and likely determined the depth as the difference between a given height in the column and the height of the surface.

ok ...thanks
i never really knew that pressure was always directed normally inwards
now i understand that pressure in other words is a distributive force due to weight of material above it
so by above understanding the pressure on lower surface is also directed inwards or in other words:it is directed uipwards and its magnitude is more than at the top surface
thanks
 
  • #10
olivermsun
Science Advisor
1,244
118
ok ...thanks
i never really knew that pressure was always directed normally inwards
now i understand that pressure in other words is a distributive force due to weight of material above it
It's just the way fluids work (and hence don't hold their shape) -- they try to distribute forces in every direction they can, on any surface available.
 

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