[Physics] How the apparent weight varies due to the rotational motion of Earth

Question

I learned that as the earth rotates about its axis, the bodies on the earth also follow a circular path. In most books I read, they give the example of a person standing on a weight balance at the equator… and I did understand that. However, by doing the following calculation, I am seeing that the apparent weight at other points on the earth (apart from the poles) is the same

This is the picture on my mind:

At B,

$$W-N=m{\omega}^2 R$$

At A, a component of weight will provides the centripetal force to rotate around the circle with the radius $r$,
$$W\cos{\theta} – N\cos\theta =m{\omega}^2r $$
as $r=R\cos\theta$,
$$W\cos{\theta} – N\cos\theta =m{\omega}^2R\cos\theta$$

The equation eventually ends up as…
$$W – N =m{\omega}^2R $$

So from this, I think that the normal reaction force which is the apparent weight remains the same as to the apparent weight at the equator.

However the book states that the apparent weight varies along $A$ and $B$.

Also, we assume for simplicity that the Earth is spherical. (I know this is not a realistic assumption, cf. e.g. this, this and this Phys.SE posts, but that's another story.)

I am really sorry for making the question so long. Could someone please tell me which part of my concept is wrong?

Answer 1

If we look at the $x$- and $y$-components of the forces, we see that $$ \begin{align} W_x - N_x &= m\omega^2 R\cos\theta,\\ W_y - N_y &= 0. \end{align} $$ The normal force $\vec{N}$ is a radial force, so if we assume that the Earth is spherically symmetric, we indeed have $N_x = N\cos\theta$ and $N_y = N\sin\theta$, so that $N=N_R$. But the weight force $\vec{W}$ is not purely radial. It also has a tangential component, precisely because of the rotation of the Earth: $$ \begin{align} W_R &= W_x\cos\theta + W_y\sin\theta,\\ W_\theta &= -W_x\sin\theta + W_y\cos\theta. \end{align} $$ So in general $W_x\neq W\cos\theta$. For the normal force, we have $$ \begin{align} N_R &= N_x\cos\theta + N_y\sin\theta = N\cos^2\theta + N\sin^2\theta = N,\\ N_\theta &= -N_x\sin\theta + N_y\cos\theta = 0. \end{align} $$ From $$ \begin{align} W_x\cos\theta - N_x\cos\theta &= m\omega^2 R\cos^2\theta,\\ W_y\sin\theta - N_y\sin\theta &= 0, \end{align} $$ we find the centripetal force $$ W_R - N = m\omega^2 R\cos^2\theta, $$ but note that there's also a tangential force $$ W_\theta = -m\omega^2 R\sin\theta\cos\theta. $$