If you take a giant whale out of the water and put it on land for long enough, it will crush itself under its own weight. Why doesn't the animal get crushed under its own weight when it's in water?
[Physics] Why don’t massive water-borne animals crush under their own weight when they’re in water
biophysicsforcesnewtonian-mechanics
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
Where did I err?
The equation $x=\frac{F}{A}\frac{L}{Y}$ applies to each infinitesimally-thin horizontal layer of the wire.
If the cross-sectional area $A$ of the wire were constant then $F$ (= weight of wire below any layer) would increase linearly with $y$, the height from bottom to top. Because the only variation (that in $F(y)$) is linear, the total extension $x$ for the whole wire can be found by using the average value of $F$ at the ends. This is just the same as finding the area under a straight-line graph from the range and the average height at the ends.
However, for non-uniform wire area $A(y)$ decreases (linearly) from bottom to top while $F(y)$ increases but no longer linearly. The overall variation is not likely to be linear. So taking the average between the end values no longer works. This is the same as the reason why the area under any non-linear curve is not necessarily equal to the range times the average height at the two ends.
See also the worked solution to this problem on the Mathalino website.
Best Answer
One place this issue is discussed is here. The key paragraph is
which is pretty cool. But then this leads to another problem addressed here:
At the same time, as the article points out, the lack of gas exchange is useful in the sense that it circumvents the problem of nitrogen getting absorbed into the blood that would then bubble out at lower depths.