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I know that all objects, regardless of their density, feel buoyancy force. But do partially submerged, meaning an object that has lesser density than the liquid,lose weight.

For example and object with a density of 200kg/m3 and volume of v and mass of m is submerged in water.

Only one fifth of it would be submerged. So the buoyant force would be= (v/5)(1000)(g)=200vg.

Now according to archimedes's rule the object should lose 200vg amount of weight.

But the entire weight of the object in air is = v(200)g= 200vg.

So this means the weight of the object while being partially submerged should be 200vg-200vg= 0. Is this correct? If correct, is it because it has an net acceleration of 0 which causes it to stay afloat?

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2 Answers 2

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I know that all objects, regardless of their density, feel buoyancy force. But do partially submerged, meaning an object that has lesser density than the liquid, lose weight.

There is an "apparent" weight loss equal to the weight of the liquid displaced by the object. But there is no change in the object's gravitational definition of weight, which is simply the force exerted on the object due to gravity, without regard to the presence of an upward buoyant force.

For example and object with a density of 200kg/m3 and volume of v and mass of m is submerged in water. Only one fifth of it would be submerged. So the buoyant force would be= (v/5)(1000)(g)=200vg.

Correct.

Now according to archimedes's rule the object should lose 200vg amount of weight.

Again, that is the "apparent" weight loss.

But the entire weight of the object in air is = v(200)g= 200vg.

Correct, given the weight of the displaced air is considered negligible.

So this means the weight of the object while being partially submerged should be 200vg-200vg= 0. Is this correct? If correct, is it because it has an net acceleration of 0 which causes it to stay afloat?

All this means is an object that has a density equal to or less than the liquid experiences a net force of zero (the upward buoyant force equals the downward force of gravity), and thus zero acceleration.

Hope this helps.

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  • $\begingroup$ Thanks. I feel like I was neglecting the concept "apparent weight". $\endgroup$
    – safwan05
    Dec 21, 2023 at 16:40
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If you partially submerge an object in a liquid, say water, the weight of volume of liquid displaced by it is in upward direction and reduces the observed weight of the object that is directed downward.

The reason that partial submergence would at first glance not appear to be causing the apparent weight lose is that the submerged volume of object is not enough to draw out enough amount of water whose weight would completely nullify the weight of the object. There is a certain minimum volume of water that will have to be displaced or equivalently, a certain height upto which the object must be submerged that the weight of displaced water just exactly cancel it out.

For an object with $\rho<\rho_w$, this 'critical volume' can be reached pretty easily and even partial submergence would cause it to float. But for objects with $\rho>\rho_w$, even complete submergence could not draw enough volume of water to completely counter the weight of the object, only partially cancelling it out.

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