The strength of the buoyant force acting on an object in a fluid depends on the objects's?
156716
in reality, it is the surface tension (adherance of molecules to one another) and density of the fluid\nbecause the question does not state that the object is suspened in the fluid\n\notherwise the force acting to keep an object buoyant would need to equate and thus counteract the objects weight (which is the force of gravity pulling said object down)\n\nmass X volume = density which is a factor in determining a force acting upon the object; a force working to cause the object to disperse within the less dense fluid or conversely the fluid to compress and colapse the object\n\nSo the author of the question given is expecting the answer: weight
63147
Archimedes' principle states that the buoyant force experienced by a submerged object is equal to the weight of the liquid displaced by the object. Experimentally this appears in the fact that the submerged object apparently weighs less by an amount equal to the weight of the liquid displaced. The buoyant force can be expressed as:\n\nFb = W(air) - W(liquid) = dgv \n\nwhere d is the density of the liquid, g is the acceleration of gravity and v is the volume of the immersed object (or the immersed part of the body if it floats).
238622
it is the normal force per area...\nyou can visit this site:\nhttp://en.wikipedia.org/wiki/Pressure
1026719
Buoyancy\n\nThe buoyancy force is a force exerted by a fluid on an immersed object and is equivalent to the weight of the volume of such fluid displaced by that object. Water serves as a particular fluid in this case.  The water buoyancy provides large mammals such as such as whales an environment were they can survive our planet’s gravitational pull. As we know, beached whales get suffocated as their lungs compress under the weight of their own body. \nOceans waters also provide abundance of food to support their huge frames, but this is another story :) .
182931
Lift under a liquid (or BOUYANCY FORCE) is defined as the weight of the volume of liquid displaced.  In other words, say a ball displaces 1 cubic meter of water.  The ball will be pushed back up with the same amount of force as the weight of that 1 cubic meter of water.\n\nSince water's density is 999 kg/cubic meter, we're looking at 999 kg of water (water is really heavy!).\n\nNow if we want the force of that water, we'd simply multiply it by the gravitational acceleration, 9.8 m/s/s:\n\n999 * 9.8m/s/s = 9790 Newtons (1 Newton = 1 (kg * m)/(s^2))\n\nSo the ball would be pushed back up with a force of 9790 Newtons.\n\nNow, the hard shell ball will (theoretically) not deform when subjected to the pressure of 10 feet of water.  So the *bouyance force* placed on the hard shell ball is easy to calculate, given the formula above.\n\nThe latex balloon, however, will shrink as it is exposed to increasing pressure, so the volume of water it displaces will be smaller. (I will not solve that problem here as it takes up quite a bit of space to show how much a balloon shrinks as it is subjected to increasing pressure.  Essentially it comes down to writing a spherical force balance and integrating over volume)\n\nConsequently, the bouyancy force exerted on the latex balloon will be smaller, and the ping pong ball would be forced back up through the water harder (and faster) than the balloon.\n\nAs an extreme, ever wondered what happens to helium filled balloons when they are released into the air and they float up until you can't see them anymore?  Since the atmospheric pressure is lower and lower the higher you go, the balloon expands until eventually it pops.  At that point, the popped balloon either falls back to earth or gets swept up into atmospheric currents, which can take it basically anywhere on earth.
1144696
gravity and air resistance.
17865
There are a lot of sites that might help you calculate buoyancy. Click the link below to get started.
257654
Say the anchor weighs 15 pounds in air and displaces 2 pounds of water.\nBefore the drop, the boat displaces its weight plus the 15 lb weight in air of the anchor. This is the maximum displacement situation, so the water is at maximum height.\nAfter the drop, if the anchor hits bottom, the boat displaces its weight only and the anchor displaces 2 lb of water, the minimum displacement situation, so the water is at minimum height.\nAfter the drop, if the anchor doesn't reach the bottom, the boat displaces its weight plus the 13 lb weight in water of the anchor. The anchor displaces 2 lb water. The total displacement is the same as the maximum displacement situation, so the water is at maximum height.
70735
The upthrust force equals the weight of the water displaced.\n\nIf d = density of the water,\nV = volume displaced,\ng = acceleration due to gravity,\n\nThe force of upthrust is,\nF = d*V*g
177575
If the fluid is brought to a rest by the beam then the drag coeff Cd is one.\nhttp://en.wikipedia.org/wiki/Drag_coefficient\n\nThis paper seems relevant to your calculations.\nhttp://www.artificialreefs.org/ScientificReports/CancunHotelRBstability.htm
152879
No, it will not.  The other answers to this question are on the right track, in that they point out that an object's density is the key thing in most situations, since if it is less dense than the fluid it is in, it will displace fluid equal to its weight before it is completely submerged, and hence float.\n\nBut in *this* problem, we are told from the outset that there is not enough water to be displaced.  Therefore, the object will displace all the water, all 1 lb. of it, and still not have displaced enough water to equal its weight.  Therefore, it will rest on the bottom, i.e. not float.
118752
The fricional force between an object and a surface is dependent on two things,\nthe coefficient of friction between the two surfaces and the normal force exerted by the surface on the object.\nIf you increase either of these, the frictional force is increased, likewise, if you decrease (or make zero) any of these, the frictional force is less (or zero).
967963
Either air pressure, or fluid pressure on a fluid present in the cylinder.
823708
Weight of the body is always less than the weight of the liquid it displaces which is the buoyant force , if it has to float \n\nIt is the law of bouyancy ,\n\nIt states that:\n " any body completely or partially submerged in a fluid (gas or liquid) at rest is acted upon by an upward, or buoyant, force the magnitude of which is equal to the weight of the fluid displaced by the body."\n\nThe buoyant force is equal to the weight of the displaced fluid
340597
there's not too much difference, other than  the density which causes the drag coefficient. however, water has surface tension, which may be slightly important in causing greater drag
952124
1. The bouyant force of the water on the object is the difference in the reading, i.e. BF=(.0554-.0370)g\n                                   =0.0184gN\n\nBased on Archimedes' Principle, BF =Volume*Density of water displaced.\n \n0.018g=Volume*1000g\nVolume=.018/1000\n            =1.8*10^-5m3\n\n2. Density of Object=Mass/Volume\n                                =.0554/1.8*10^-5\n                                =3078kg/m3
187607
The density of water is about 100 times the density of any gas .\nSo the gas will be lifted. (Archimedes law).\nBasicly the gravity energy decreases.