Centripetal acceleration problem (box sitting on a table)

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The discussion centers on understanding the forces acting on a box resting on a table in both inertial and non-inertial frames. In an inertial frame, the box experiences zero net force due to the balance of gravitational and normal forces. However, when considering the Earth's rotation, the box is subject to centripetal acceleration, which alters the force balance, resulting in a slight excess of gravitational force over the normal force. The key point is that the rotating frame of the Earth is non-inertial, requiring the introduction of fictitious forces like centrifugal force to accurately describe the situation. This distinction clarifies the apparent contradiction in the force analysis.
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Let me imagine a box placed on a table. It has got no acceleration. If I were a person who trusted Newton's laws then I would argue that the net force on the box should be zero. Now in another situation I am an observer outside the Earth and I see that the box is rotating along with the earth, so it should have a centripetal acceleration and gravity provides it...but in the previous case, gravity was canceled by the normal reaction. So shouldn't the centripetal force also be zero. Please explain where I went wrong.
 
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You've got the right intuition, but let's iron a few things out. :biggrin:

For a minute, forget about the Earth. Just imagine an idealised flat, rigid surface of infinite extent. Impose a uniform gravitational field ##-g \hat{\mathbf{y}}##. The book sitting on the table has its weight exactly canceled by the normal force exerted by the table, to satisfy the equilibrium condition.

If the Earth wasn't rotating, that would be a pretty good local model. In fact, even though the Earth is rotating, the rate of rotation is sufficiently slow that this is an approximately correct model.

But we might want to think about a different, maybe more accurate, model in which the Earth has non-zero rotational speed. To simplify the description, imagine the table is at the equator. As you correctly deduced, the net force on the book must now be non-zero and pointing toward's the Earth's centre, in order to provide the necessary acceleration ##\rho \omega^2## toward the centre. In other words, the magnitude of the weight will slightly exceed the magnitude of the normal force.

These two descriptions aren't in contradiction, simply because they're not describing the same scenario. The assumptions that underly the two models are different. :smile:
 
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Physics guy said:
Let me imagine a box placed on a table. It has got no acceleration. If I were a person who trusted Newton's laws then I would argue that the net force on the box should be zero. Now in another situation I am an observer outside the Earth and I see that the box is rotating along with the earth, so it should have a centripetal acceleration and gravity provides it...but in the previous case, gravity was canceled by the normal reaction. So shouldn't the centripetal force also be zero. Please explain where I went wrong.
Newton's Laws apply to inertial frames. The rotating rest frame of the Earth is not exactly inertial. To make Newton's 2nd Law applicable there, you have to introduce inertial forces (here Centrifugal force):

Rotating rest frame of the Earth:
Gravity + Centrifugal + Normal = 0

Inertial frame where the Earth rotates:
Gravity + Normal = Centripetal

See also:
https://en.wikipedia.org/wiki/Rotating_reference_frame
 
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To go along with what @A.T. said, in the non-inertial local frame you have the fictitious centrifugal force and the gravitational force. Since they are both proportional to mass you can simply add them together to get an overall “effective” g. This effective g will change from place to place over the globe.
 
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Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
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