Find how the weight of bodies on the Earth's surface depends…
Find how the weight of bodies on the Earth's surface depends on the geographic latitude (the effect of the Earth's rotation on the gravitational force).
Explanation of Method
Weight against latitude.
Step 1: Idea
On a rotating Earth a body at rest on the ground actually moves on a circle around the Earth's axis. It is easiest to describe this in the rotating frame of the Earth, where we add a centrifugal force directed away from the axis. The weight (what a scale shows) is the combination of the gravitational force $mg$, pointing to the centre, and this centrifugal force.
Step 2: Centrifugal force at latitude $\varphi$
A body at latitude $\varphi$ moves on a circle of radius $R\cos\varphi$ (Problem 9), so
\[ F_c = m\omega^2R\cos\varphi. \]
It points horizontally away from the axis, not vertically upward (except on the equator).
Step 3: Component along the vertical
The local vertical runs along the radius, which makes the angle $\varphi$ with the direction of $F_c$. Resolve $F_c$ into two parts: $F_{c,\text{vert}} = F_c\cos\varphi = m\omega^2R\cos^2\varphi$.
\[ F_{c,\text{hor}} = F_c\sin\varphi = m\omega^2R\sin\varphi\cos\varphi. \]
The vertical part acts directly against gravity and reduces the weight. The horizontal part is tiny (at most $0.017\ \mathrm{m/s^2}$, at 45°); it only tilts a plumb line by about 0.1° and hardly changes the magnitude. Therefore
\[ W(\varphi) \approx mg - m\omega^2R\cos^2\varphi. \]
\[g_{\text{eff}}(\varphi) \approx g - \omega^2R\cos^2\varphi\]
Step 4: Size of the effect
\[ \omega^2R = (7.27 \cdot 10^{-5})^2 \cdot 6.37 \cdot 10^6 = 0.0337\ \mathrm{m/s^2}. \]
So the reduction is $0.034\ \mathrm{m/s^2}$ (0.34%) at the equator, $0.017\ \mathrm{m/s^2}$ at 45°, and zero at the poles. Bodies are lightest at the equator and heaviest at the poles.
Check
Equator ($\varphi = 0$): reduction $\omega^2R$, the full centrifugal force, pointing straight up $\checkmark$
Pole ($\varphi = 90^\circ$): reduction 0 $\checkmark$
Measured $g$ differs by about $0.052\ \mathrm{m/s^2}$ between pole and equator; about $0.034$ of that is rotation, the rest comes from the Earth's flattening
Result
\[\boxed{\,g_{\text{eff}}(\varphi) \approx g - \omega^2R\cos^2\varphi\,}\]
Remark
Note. A common shortcut subtracts the whole centrifugal force $m\omega^2R\cos\varphi$ from the weight. That is only right on the equator; elsewhere the force is not vertical, and only its vertical part, with $\cos^2\varphi$, counts.