mjcoon
Well-known member
Oh dear, metrication won't be complete until all those shop weigh-scales are re-labelled in Newtons! ;-)
How much does my 40Kg anchor weigh when in sea water?
- Thread starter tudorsailor
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boomerangben
Well-known member
I see what you are trying say (I think). If you put a flat plate on the seabed, it would weigh the mass of plate x g, reduced by the buoyancy. Then you have the weight of seawater above it pushing down on it. But if the seabed is permiable, there will be water beneath the plate (between the grains of sand) “connected” to the water above the plate thereby pushing up on the under side of plate. The thickness of the plate means the pressure acting beneath the plate is slightly higher than the pressure acting on the top of the plate (density x g x thickness). That difference in pressure is what we call buoyancy.
If the seabed is less permiable, say sticky mud, trying to lift the plate will involve trying to lift some of the water above it too because there won’t be so much water pressure beneath the plate. If you go a step further, and put a metal skirt around the plate, put that on the seabed and pump the water out between the plate and seabed, the plate will be pressed down onto the seabed and the skirt will provide a barrier to water permeation and you have a suction anchor which as you say will require you to lift much of the water column above it. But an anchor on the seabed will generate very little suction and therefore only need a force that just exceeds the mass x g - buoyancy force to lift. Mid water, as others have said, you have to rationalise the water pressure acting all over the surface of the anchor
Bouba
Well-known member
Yes...in the real world there are variables...mine was like a perfect environment for experimentation purposes only...perhaps we should have said swimming pool instead of sea.
The answer everyone seems to want is what is the buoyancy of steel?...which is a known factor and all variables like salinity, temperature and depth are probably also available online... but the question was weight...and weight is the gravitational attraction from the Earth on an object, including anything on that object, so for you and me it would include everything in our pockets, for Charles it would include the weight of his gold hat...and for an anchor the weight of water above it....of course in a real world none of this is possible to ascertain because of all the forces involved...
Bouba
Well-known member
Of course I could be wrong
boomerangben
Well-known member
Ok let’s be clear. Mass is the measure of the amount substance within an object and does not change regardless of its place in the universe (notwithstanding the edges of physics where black holes and speeds approaching that of light muddy the waters so to speak)
Weight and mass are directly proportional on the surface of the earth and whilst in a scientific different, they are interchangeable by a factor of just under 10.
Seawater also has gravity acting on it and therefore has weight, but it is a fluid and that allows it to transmit a pressure force in all directions (up, down, sideways, diagonally etc)
If you put something solid into water, like your boat, it feels pressure on its external hull surface. If you were to drill holes in your hull with the same drill size but at different depths with in the hull, water will be forced in by water pressure, most enthusiastically from the holes deeper in the hull. The water pressure is higher the deeper you go, why because gravity is acting on a deeper (more massive) water column. Fill the holes in and the boat floats. Why because the water pressure acting on the hull tries to push it up which is matched by the gravity force acting on the mass of your boat.
So you have a 40kg anchor on deck. Weighs 40 x 9.81 newtons. Let’s round g to 10 for simplicity. So it weighs 400newtons.
Deploy it to mid water on a neutrally buoyant line so we can ignore the effect of rode.
Mid water it has a mass of 40kg, a weight of 400N (40kg x g) and hydrostatic pressure forces acting all over each and every surface. Because the anchor is solid (made up of solid components) and has a top surface that experiences a lower hydrostatic pressure than its bottom surface, there is a net upward pressure force which we call buoyancy. Determining the net pressure force is a pain unless it’s a simple shape. Thankfully Archimedes came up with a short cut that’s says the buoyancy force is equal to the mass of displaced water x g, ie the product of displaced volume, displaced fluid density and g .
The volume of the anchor is equal to mass/density of steel, the mass of displaced water is density x volume and since the volume of the anchor remains constant, the buoyancy force is 40 x g x 1026 (density of seawater) / 7850 (density of steel). Which is about 52 newtons. So the tension in the anchor rode is about 350 newtons. Which is equivalent to about 35 kg.
In short, water pressure acts all over the surface of the anchor and indeed the water at depth is supporting the water above it and is itself supported by the water below it.
Note if anchoring near black holes these numbers will not be correct, the holding will be poor and the rate of drift huge. Moreover your new generation anchor will not be suitable and you would need to look to The Next Generation of skyhooks. Ask Neeves for test results
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KevinV
Well-known member
Note if anchoring near black holes these numbers will not be correct, the holding will be poor and the rate of drift huge. Moreover your new generation anchor will not be suitable and you would need to look to The Next Generation of skyhooks. Ask Neeves for test results
