What is the tangential component? Taylor p.347

[SebastianRM]

I do not understand why the tan component for a gravity affected by the centrifugal force:
g = Ω^2 * R * sinθ * cosθ

So I tried to draw this: using a "big" X-shaped axis where the / component goes along the main gravity direction while \ points normal to / this direction. Then the centrifugal force points out of the center of the X into the x direction, with and angle θ between the y-axis and the / direction. Thus the angle between the centrifugal force and \, should be 90 - θ, and so its projection along that direction should be cos(90-θ).
So i did not arrive to the conclusion of the book, where am I going wrong?

Thank you in advance.

 

[kuruman]

I do not understand why the tan component for a gravity affected by the centrifugal force:
g = Ω^2 * R * sinθ * cosθ

So I tried to draw this: using a "big" X-shaped axis where the / component goes along the main gravity direction while \ points normal to / this direction. Then the centrifugal force points out of the center of the X into the x direction, with and angle θ between the y-axis and the / direction. Thus the angle between the centrifugal force and \, should be 90 - θ, and so its projection along that direction should be cos(90-θ).
So i did not arrive to the conclusion of the book, where am I going wrong?

Thank you in advance.

I cannot tell you where you went wrong because your description of what you did is not clear to me. Here is my suggestion of how to derive what you need.

The centrifugal part of the acceleration according to equation 9.44 is $~~\vec a_{cf}=\Omega^2~R~\sin\theta~\hat \rho$. Now look at figure 9.10. It shows the centrifugal acceleration horizontally out. The component you are seeking is tangent to the surface of the circle representing the Earth at that point. To get that component you need to multiply the magnitude $a_{cf}$ by the appropriate trigonometric function. What might that be?