This is a common physics exercise:
Suppose the earth is a sphere of radius $6370$ km. If a person stood on a scale at the north pole and observed the scale reading (his weight) to be $mg$, what would the scale read if he stood on it at a point on the equator?
I decided to try to solve it myself, first with intuition, and then a diagram.
My intuition tells me that weight (as seen on a scale) is the normal force $N$. On the North Pole, $N = W$, where $W$ is the weight. However, on the Equator, there is a centrifugal force. Like on a merry-go-round, one is pull away from the Earth. That means $N$ is "lightened" as such: $N = W – C$. From there, simple calculations are made to get a numerical answer. According to other sources, this is correct.
However, I tried solving this "without ever being on a merry-go-round". I draw a diagram. Draw the earth as a circle, with a person as a point. To the center, there is $W$, because $W$ is in the direction of the mass. In addition, centripetal force ($C$) is pointed to the center of a spinning object (i.e. same way as $W$). Opposing both is $N$. So, I get $N = W + C$. This is different than the above equation.
Where do I go wrong in my second solution?
Best Answer
The problem you are having is that you are thinking of "the centripetal force" as a force that you have to account for in drawing your diagrams. Not entirely you fault. The phrase does read that way.
Basic rule: there is no thing called "the centripetal force". Instead that phase is a label for some combination of real force to take on the role of causing the inward acceleration that must be present for curving motion to occur. What combination of forces varies from problem to problem and even from time to time in a single situation. What that combination is is something that you have to discover in each problem.
But you do have one hint. The simple kinematics of the motion tell you what the magnitude of acceleration must be.
In this case $F_c = |\text{gravity times mass}| - |\text{normal reaction force}|$. This is a specific case of the general rule that centripetal force is the sum of the inward pointing forces (or components of forces) minus the sum of the outward pointing ones.1
Secondly, you need to know that a bathroom scale measures the normal force between the person and the ground. This gives us a different understanding of "weight" than $mg$ (and it is one that works right when we say a astronaut in orbit is "weightless").
So, $$\text{weight} = \text{normal reaction force} = mg - m\frac{v^2}{r}\,,$$ is the expected reading on the scale at the equator.2
To expand a little on the "why" of this approach, I want to start by talking about equilibrium.
Once you have identified that some object is in equilibrium then you know something about the forces acting on it in aggregate: they add up to zero. That doesn't tell you about any single a priori, but if you know all the other force (or know that there is only one force) then it lets you finish.
The "centripetal force" is very much like that: because the object is in curving motion you know that it is not in equilibrium but accelerating, and furthermore you know that the component of its acceleration that points to the (instantaneous) center of curvature is $a_c v^2/r = r\omega^2$.
And that tells you that the total of all the radial forces acting on it must be $ma_c$ (and again, this is something you know about a collection of forces). But before you can use this knowledge to your advantage you must identify which forces (or components thereof) should be combined before being set to that total.
So an early step in working problems that involve "centripetal force" is asking "Which of these force point toward the center and which point toward the outside?" The former enter into the sum with a positive sign and the later with a negative sign.
1 You should note that because these are the only forces acting on the person at the equator and they do not cancel out because that person is not in equilibrium. But you knew that: they are moving in a circle, right?
2 Except that we have forgotten to account for the Earth's non-spherical mass distribution. But that is another story.