How do write the force diagram for the following situation?

[eprparadox]

Homework Statement

We have a crate sitting on a scale that is on the surface of the Earth. We want to come up with the value of the acceleration due to gravity, $g$, when we take into consideration the rotation of the Earth.

Homework Equations

In the book, here's how they go about this:

There is an upward normal force from the scale, $F_N$. There is a downward gravitational force that is given by $m a_g$. These two forces sum and cause the centripetal acceleration $-\omega^2 R$. That is,

$$F_N - m a_g = -m \omega^2 R$$

But then they say that $F_N$ equals the force $mg$ from the scale and they write

$$mg - m a_g = -m \omega^2 R$$

They solve for $g$ to get

$$g = a_g - \omega^2 R$$

Note that in the book, $a_g$ is the gravitational acceleration given by $a_g = \frac{MG}{R^2}$

Ultimately, in the book they write

$$g = a_g - \omega^2 R$$

The Attempt at a Solution

[/B]
This is confusing me. I don't know how you can justify putting in $mg$ for $F_N$.

In the case where we're not rotating, I know that we can write

$$F_N - mg = 0$$

and so $F_N = mg$

but I just don't get how the book justifies their expression for $g$.

Any thoughts on how I can think about this more clearly? I know the problem isn't a difficult one but I want to make sure I'm crystal clear on what's going on.

 

[tnich]

Homework Statement

We have a crate sitting on a scale that is on the surface of the Earth. We want to come up with the value of the acceleration due to gravity, $g$, when we take into consideration the rotation of the Earth.

Homework Equations

In the book, here's how they go about this:

There is an upward normal force from the scale, $F_N$. There is a downward gravitational force that is given by $m a_g$. These two forces sum and cause the centripetal acceleration $-\omega^2 R$. That is,

$$F_N - m a_g = -m \omega^2 R$$

But then they say that $F_N$ equals the force $mg$ from the scale and they write

$$mg - m a_g = -m \omega^2 R$$

They solve for $g$ to get

$$g = a_g - \omega^2 R$$

Note that in the book, $a_g$ is the gravitational acceleration given by $a_g = \frac{MG}{R^2}$

Ultimately, in the book they write

$$g = a_g - \omega^2 R$$

The Attempt at a Solution

[/B]
This is confusing me. I don't know how you can justify putting in $mg$ for $F_N$.

In the case where we're not rotating, I know that we can write

$$F_N - mg = 0$$

and so $F_N = mg$

but I just don't get how the book justifies their expression for $g$.

Any thoughts on how I can think about this more clearly? I know the problem isn't a difficult one but I want to make sure I'm crystal clear on what's going on.

It seems to me that the book is using $a_g$ for the acceleration of gravity and $g$ for the normal acceleration at Earth's surface due to gravity and centrifugal force. I use the word centrifugal (instead of centripetal) intentionally because the Earth's rotation does cause a very small apparent upward force in a rotating reference frame fixed at a point on the Earth's surface. I think your confusion stems from the use of $g$ to represent the normal force. The point is, I think, that the normal force measured by a scale on the surface of the planet is not just the force of gravity. For practical purposes we commonly ignore the centrifugal force, which is negligibly small compared to the gravitational force, and assume that the normal force and the gravitational force are the same. So there is a little ambiguity in the use of $g$.

 

[Chestermiller]

The equation $mg - m a_g = -m \omega^2 R$ in your book is the source of your confusion. They should have (more appropriately) written: $$mg_{app} - m a_g = -m \omega^2 R$$where $g_{app}$ is the "apparent gravitational acceleration" at the surface.