Why does weight on Earth have a tangential component?

Question

I read in Pulsar's answer of this question that weight has a tangential component due to the rotation of the (assumed spherical) Earth but I don't understand this. Can someone expand on it please?

Answer 1

There are two reasons. One is that the Earth is not perfectly spherically symmetric and so at any point on the Earth's surface, the net gravitational force does not point directly towards the centre of the Earth.

Secondly, because the Earth is rotating, there is a pseudo-force, the centrifugal force, that appears to act directly away from the Earth's rotation axis. The combination of this with the gravitational force will have a component acting towards the centre and a component at right angles to this, tangential to the surface and acting along lines of longitude towards the equator.

Both of these effects are small; of order tenths of a percent, but the tangential force dueto the second disappears entirely at the poles (where the centrifugal force is zero) and at the equator (where the centrifugal force and gravity are in opposite directions, but act along the same line).

This is completely explained and demonstrated in How the apparent weight varies due to the rotational motion of Earth?

Answer 2

That question refers to apparent weight, as opposed to the true weight.

Your true weight, $W=mg$, is the force of gravity pulling on you.

Your apparent weight is how hard the ground has to push on you to keep you from falling. This is the weight you "feel". When you are accelerating up or down in an elevator while standing on a scale, the scale reads your apparent weight. It's actually measuring how hard it has to push to hold you up, not how hard the Earth pulls on you.

Pseudo-forces caused by accelerations can affect your apparent weight, but not your true weight. The accelerating elevator does it in the example above. In the rotating reference frame on the surface of the Earth you experience a pseudo-force from the acceleration that causes you to turn in a circle. In the rotating reference frame of a roller coaster loop-de-loop it's a bit more obvious.

As @Rob_Jeffries said, because the Earth is not a perfectly smooth, perfectly uniform sphere, there are local variations in gravity that could cause your true weight to have a horizontal component. In intro physics classes you probably don't need to worry about such things.

Answer 3

Just to expand a little bit... The centrifugal force on a particle of mass $m$ is given by $$\vec F_{ce}=-m\vec\omega\times(\vec\omega\times\vec r),$$ where $\vec\omega$ is the Earth's angular acceleration and $\vec r$ is the position of the particle. This force must be taken into account when using Newton's second law in the non inertial frame. Similarly to gravitational force, this term depend on the mass of the particle so it is useful to combine it with the weight $m\vec g$: $$\vec F=m\vec g-m\vec\omega\times(\vec\omega\times\vec r)=m\vec g_{eff},$$ where $$\vec g_{eff}\equiv \vec g-\vec\omega\times(\vec\omega\times\vec r)$$

This is the the acceleration feld by the particle. If the origin of the non inertial system is at the center of the Earth you can easily check that the vector $-\vec\omega\times(\vec\omega\times\vec r)$ is always pointing as in the white arrows below. The gravity $\vec g$ points radially inwards as the black arrows. Then the effective gravity $\vec g_{eff}$ points as in the red arrows. As you can see, for any place on Earth except over the poles and the equator line, $\vec g_{eff}$ will always have a tangential component (in a tangential plane).