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I have the following question based on a paragraph from my Physics textbook (I am translating, so the result might not be the most elegant):

It states that due to the centripetal force resulting from Planet Earth's movement, the Normal force becomes slanted away from a purely vertical line towards the centre of Planet Earth. This way the resultant force between gravity and the normal force gives us the centripetal force. It adds that this way Earth's rotation not only reduces weight and falling acceleration but also diverts them away from the vertical direction.

I am confused: First of all, on the previous page it stated that N = GmM/R^2-ma, with ma referring to the centripetal force. From this equation I had understood that the Normal Force is simply less around the equator, which was confirmed by several online sources who compare it to the reduced Normal Force in an elevator on its way down. So why would it become slanted?

Beyond that, I am lost as to the diversion of the direction of free fall…

Can somebody clear things up?

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  • $\begingroup$ Which Physics textbook? Page? Link? $\endgroup$
    – Qmechanic
    Aug 13 '18 at 12:57
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Image of obbject on Earth's surface

When you're not on the equator, you're travelling in a circle whose centre is not the centre of the earth. As you would know, gravity always acts towards the centre and centripetal force (which isn't actually a force but an effect) acts towards the centre of the latitudinal circle you're on(the green circle). So the contact force needs to balance out these two forces. If the contact force was acting only along the line joining centre to your location(along $r$), you would have a component of the centripetal force acting perpendicular to $r$ which won't be balanced out. So, the contact force has to be slanted and won't be along $r$ . A similar logic goes for free fall

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  • $\begingroup$ Thank you! That makes sense mathematically, but I am still doubting about the following: what is keeping me on Planet Earth, rather than moving in a tangential line off the planet due to my inertia, is /gravity/. I am not attached to the planet with a rope, as is the case with the other examples in my book that refer to the centripetal force. So how is the Centripetal Force not itself a component of gravity? I mean the centre of the circle at latitude x, parallel to the equator does not actually exert a force on me, as far as I understand. Only the centre of Planet Earth does. $\endgroup$
    – Pregunto
    Aug 13 '18 at 12:16
  • $\begingroup$ You're right.That's why is said centripetal force is not actually a force. The vector sum of gravitational force and normal force should be equal to $ma$ where $a$ points towards the centre of the circle. Centripetal force us the "net force" acting on the body so its the sum of the other two forces. I pretended like it was a force so that it would be easier to find the direction of the normal reaction. $\endgroup$
    – Skawang
    Aug 13 '18 at 12:20
  • $\begingroup$ Also, later on in the book they take a certain spot on Planet Earth (which evidently moves along with the planet) as a non-inertial framework, from which they analyse the balance of forces. The centripetal force is suddenly gone, but a fictitious centrifugal force has been added. I assume that the centripetal force is gone because it requires velocity in its equation (mv^2/R), and I understand why the centrifugal force is added to make things balance out. But is is correct to say that what was the centripetal force is now included in gravity? $\endgroup$
    – Pregunto
    Aug 13 '18 at 12:32
  • $\begingroup$ You've got to stop seeing centripetal force as an actual force. There is NO centripetal force. We call the term $ma$ as a force because when we're going in a circle, we have a physical sense of something pulling us towards the centre. This something can be one single force or a sum of a number of forces. The net force towards the centre is called the centripetal force but the net force, which is a sum of forces, need not be a force itself; Just like how you wouldn't call a collection of 10 chocolates a chocolate. $\endgroup$
    – Skawang
    Aug 13 '18 at 14:24
  • $\begingroup$ When you're in the frame of reference of the body, it's acceleration with respect to itself is gonna be zero. So we can say that the net force is zero. So it doesn't make sense to talk about a centripetal force since there is no net force towards the centre which we refer to as centripetal force. It doesn't make any sense to say that it's included in gravity since it's just not there in this frame. The centrifugal force is a pseudo force which arises due to the frame being non inertial. That is a whole different concept $\endgroup$
    – Skawang
    Aug 13 '18 at 14:28
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There are three real forces acting on the object: gravitation force towards the center of earth, normal force and friction. Normal force, as implied by its name, is normal to the surface. Assuming spherical earth, the normal direction is away from center of earth. Friction is perpendicular to the normal.

If earth and object were standing still, the gravitation and normal forces would equal each other. But the object is moving in a circle. So the vector sum of all three forces towards the center of circle should be equal the mass x radial acceleration. At the orthogonal direction, the body is not accelerating, so components of forces along that axis are summed to zero. These equations allow you to calculate the normal force and friction. The vector sum of normal force and friction is the "slanted normal".

The term mass x radial acceleration is coined centripetal force.

The above description is from an inertial frame.

In a non inertial frame, centripetal force is not there because it did not exist to begin with. The three true forces, gravitation, normal and friction do appear. Centrifugal force is added to "explain the physics" in the accelerating frame.

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