# Why isn't the apparent weight of a body in a fluid equal to the buoyant force? Why does buoyancy reduce it instead?

I have been very confused about this for some time now. We know that when an elevator accelerates upward, the apparent weight of a body placed inside it equals the normal force exerted by the elevator and this force acts in the upward direction. But why doesn't this logic hold for the buoyant forces in fluids, even though they are the normal forces present in fluids and act in the upward direction? For example, suppose a person is submerged in a swimming pool, then the water will exert a buoyant force on the person and that force will be equal to the weight of the displaced water. But why does this force, reduce the apparent weight of the body, unlike the force in the elevator? Because apparent weight of a body is - by definition - equal to the normal force on the body and I've read that in the case of fluids this normal force is the buoyant force, so why isn't the apparent weight equal to the buoyant force? For example, if a body weighs 10 N and it is completely submerged in water such that the buoyant force(which is the Normal Force here) = 8 N, so shouldn't the apparent weight = 8 N also? Are some other forces, which are unknown to me, playing a role here? ( Interestingly, in a similar question asked here, an answer said that the apparent weight of a completely submerged body Is equal to the buoyant force; while another source said that the apparent weight will be equal to the difference between the true weight and buoyant force. So, which one is right?). Also, what would happen if an object is placed on a weighing machine that is atop water shooting out of a fountain (and accelerating)? I think because it's analogous to the elevator case the weighing machine would register an increased mass (i.e. the apparent weight would be greater), am I correct?

Bonus question: I read that once a body gets completely submerged under a fluid, the buoyant force experienced by it doesn't change. Then, why does it get increasingly hard for us to push the object farther down to the bottom?

• Where did you read that the buoyant force is a normal force? Jun 20, 2017 at 5:26
• @probably_someone, I don't remember the actual links. But some of them were from this forum and others from some other sources. What I got from them was: fluids have various stresses (I don't remember the exact term); the tangential component of which is the viscous force and the Normal component is the buoyant force and that this force is the only normal force present in fluids (at least static), and is the counterpart of The normal force in solids. A link that I remember is: physics.stackexchange.com/questions/254221/… Jun 20, 2017 at 6:27
• You are asking why the apparent weight of the body isn't equal to the buoyant force. Where would actual weight of the body go? The apparent weight of the body would have to be somewhat related to the actual weight of the body, no? Jun 20, 2017 at 6:46

In the elevator scenario, the elevator frame is getting accelerated; hence, the when you draw the free-body diagram, with respect to the elevator, the pseudo force acts downwards (opposite to the direction in which the frame is getting accelerated). Hence, the apparent weight increases as the pseudo force gets added up with the weight of the person.

Suppose the acceleration of the elevator is $a$ and the mass of the body is $m$, then the apparent weight of the body in the elevator frame is - $$N = m(a + g)$$

In the second scenario, the buoyant force acts in the upward direction, because the buoyant force is always directed against the pressure gradient i.e, the direction in which pressure decreases. (Much like an electric field directed in the direction where the potential decreases) Of course, the buoyant force exerted is equal to the weight of the fluid displaced by the body (Which is the Archimedes principle); but -

Drawing the FBD in the second case yields the weight of the body acting downwards, and the buoyant force acting upwards. This results in the weight decreasing (since the buoyant force is subtracted from the weight, not added up with it), and not increasing.

Say, the buoyant force acting on the body is $B$ and the actual weight is $W$, the net weight of the body (acting in the downward direction) then would be - $$W' = W - B$$

Which is why the apparent weight of the body in the liquid decreases.

(This is considering that the density of the body is greater than the density of the liquid, in the case where it is opposite (the body doesn't sink; but floats partially), the signs of $W$ and $B$ are swapped and the net force is acting in the upward direction. In another scenario where the the weight of the the body is equal to the buoyant force, the net force on the body then is zero, hence it floats being completely submerged)

Keep in mind that a body loses weight in a liquid which is equal to the weight of the liquid displaced by it/equal to the buoyant force.

As for the bonus question, look into the answer to this question - https://physics.stackexchange.com/a/296537/134658

• > (since the buoyant force is subtracted from the weight, not added up with it), vs_292 could you please explain this part Dec 9, 2021 at 14:14

In the case of a body submerged in a liquid the body accelerates but the frame doesnt, hence the resultant acceleration due to gravity does not change. We can simply write the equation for motion, but in the case of a body accelerated in a lift, the frame accelerates and hence the resultant acceleration due to gravity changes and as the frame is non-inertial, we will have to include a pseudo force in the free body diagram.