Today I did HRK Chapter 14 (Gravitation). I just started reading the theory today, so I'm not too far into the chapter yet. However, I did come across an interesting thing that we've probably never thought about before:
If you have a block of mass $m$ at rest on Earth, the normal force on it from the ground is surprisingly not equal to the gravitational force on the block.
Explanation:
For this explanation, let $r_E$ be the radius of the Earth and let $\omega$ be the angular velocity.
Let's consider the simple case first, if the Earth wasn't rotating. Then, the downward force due to gravity is $mg_0$, where $g_0$ is the acceleration due to gravity at that point on Earth. The net force on the block is $0$, so the normal force is also $mg_0$. In this case, the normal force from the ground is equivalent to the gravitational force.
Now, in reality, the Earth is rotating. Note that the block is rotating with it. Hence, some centripetal force must be applied on the block, which means that the net force on it is not $0$. The gravitational force will still be the same, namely $mg_0$. But some of this force is used up for centripetal force, which isn't "felt" by an observer on Earth (the reason this isn't "felt" is because for an observer on Earth, the block is not rotating). Only the remaining portion of $mg_0$ is "felt" as normal force. So, the normal force will now be $mg$ (also known as weight), where $g \approx 9.81$ is the gravitational acceleration measured by an observer on Earth. The difference in the real gravitational force and the normal force (which is the perceived gravitational force on Earth) is thus the centripetal force, which is $m\omega^2 R$. $\square$
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