Is it that in water apparent weight is just net force which is 0 when floating like when we are standing on our floor than our net force is also 0 because normal force cancel outs gravity so we can say that in water weight may differ if we measure weight with a weighing machine as it calculates normal force but not the net force unlike spring balance right?
Forces – Is Apparent Weight Really Apparent Weight or Just Net Force?
forces
Related Solutions
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
The force of gravity acts throughout our body and doesn't stretch or compress us.
When we stand on the ground, the normal force from the ground acts at one place, our feet and so we feel it as it tends to try to compress us.
In this example the net force is zero, but we experience the force on our feet.
If you were in a lift moving downwards that suddenly stopped, you'd feel heavier. The force of gravity has been constant, it's the force at our feet that has changed and increased.
Best Answer
What is called apparent weight depends on your definition of weight as explained in the post What is actually weight?.
Taking the definition of (true) weight as the gravitational attraction on a body then you can say that the apparent weight of the body is the reading which would be obtained if the body was placed on some scales.
In the case of a body with a fluid around it the apparent weight of the body would be the true weight (ie no fluid around the body) minus the buoyancy force (upthrust / Archimedean force) produced by the fluid surrounding the body.
This means that if you weigh a body in vacuum (no buoyancy force - true weight) using some scales then the reading would be different from a reading taken when the body is in air, water etc, (apparent weight).
With a floating body the net force on the body is zero, ie its apparent weight (reading on scales) is zero, because its true weight (gravitational attraction of the Earth) has the same magnitude and is opposite in direction the the buoyancy force. Often the effect of air is neglected because the density of air is so much less than that of the body but with a hot-air balloon you can think of the apparent weight as actually being negative with the buoyancy force being greater than the true weight.
There are also examples where a scale reading (apparent weight) would alter when the body is accelerating.
For example a body in a accelerating lift or a body in free fall / orbiting the Earth where the body appears to be weightless.
For humans that sensation of a change in "weight" or "weightlessness" is due to the change in the force or lack of force pushing up due to the floor of the lift (normal force) or lack of any sensation that something is pushing up (weightlessness).