Here is a diagram to show the the force on a point mass $m$ on the surface of an ideal (spherical, uniform density etc) Earth of mass $M$, radius $R$ and angular speed $\omega$.
The force acting on the mass $m$ is $\dfrac{GMm}{R^2}$ at all positions on the surface of the Earth.
Except at the poles the gravitational force of attraction can be thought of as providing two accelerations on the point mass.
One is the centripetal acceleration $r \omega^2 = \dfrac {v^2}{r}$ where $r$ is the radius of the "orbit" and $v$ is the tangential speed of the mass.
At the poles $m g_{\rm p} = \dfrac{GMm}{R^2}$ where $g_{\rm p}$ is the acceleration of free fall at the poles and $m g_{\rm p}$ is the reading on a spring balance at the poles.
At the Equator $m (g_{\rm e} + m R \omega^2) = \dfrac{GMm}{R^2}$ where $ R \omega^2 = \dfrac{v^2}{R}$ is the centripetal acceleration of the mass and $g_{\rm e}$ is the acceleration of free fall at the Equator which will be less than it is at the Poles or anywhere else on the Earth.
At a general position with latitude $\lambda$ on has to include the directions of the force and the accelerations as they are not collinear.
The vector triangle is shown on the diagram.
In this case the centripetal acceleration is $R \cos \lambda \omega^2$ and the acceleration of free fall $g$ is between the value at the Poles and at the Equator.
