Proving Why the Center of Gravity is Used in Middle School Balance Problems via Integration

I took middle school entrance exams when I was in elementary school, and I wondered about something back then.

https://x.com/catatsuy/status/2007802987520115094

If I were to answer my elementary school self now, I would say:

"What you're saying is absolutely correct. To find this, you'd normally need integration, but since the calculation actually results in the same value, for now, please just remember it as useful knowledge."

I think that's how I'd answer.

If we consider the finer details, objects aren't necessarily uniform, and they can rotate, which brings in angular momentum and the need for cross products, making things complicated. However, since those complex issues don't come up within the scope of middle school entrance exams, I'll simplify the discussion and answer my childhood question with simple calculations.

Also, if we were to discuss physics strictly, gravitational acceleration would be necessary, but since it doesn't appear in middle school entrance exams, I'll simplify that part as well. For this article, I'll follow these guidelines:

• The object is in equilibrium, so we only consider the balanced state
• We'll think only in a two-dimensional world (ignoring balance in the depth direction)
• The density is uniform (to avoid complicating the integration)

Proof

Let's assume the whole system is in equilibrium, and here we will focus specifically on the right side (from 0 to $l$). Let the coordinate of the balance point be 0. The right side of the object extends from 0 to $l$, and we'll assume its thickness varies by position. Let the cross-sectional area at position $x$ be $S(x)$, and the density be a uniform $\rho$. If we cut out a thin slice of infinitesimal width $dx$, its "weight" (ignoring gravitational acceleration) is $\rho S(x)dx$.

The moment of force around 0 is "arm × weight," so the moment $T$ created by the entire right side around 0 is:

$$T=\int_{0}^{l} x\rho S(x)\,dx$$

Next, let the center of gravity of the right part (from 0 to $l$) be $g$. This can be defined as the point where the moments of force on the left and right are balanced. In other words, when point $g$ is used as the fulcrum:

$$\int_{0}^{g} (g-x)\rho S(x)\,dx=\int_{g}^{l}(x-g)\rho S(x)\,dx$$

holds true. Rearranging this equation, we get:

$$\int_{0}^{l}(x-g)\rho S(x)\,dx=0$$

which means

$$g\int_{0}^{l}\rho S(x)\,dx=\int_{0}^{l}x\rho S(x)\,dx$$

. Here, if we let the weight of the right side of the object be $M$, then

$$M=\int_{0}^{l}\rho S(x)\,dx$$

is nothing other than that. Therefore,

$$gM=T$$

, and we can see that $gM$ (the moment when the entire right-side weight $M$ is concentrated at point $g$) matches the moment $T$ of the right side that we first wrote as an integral. In other words, it has been shown that instead of integrating every time, you can calculate by assuming that all the weight is concentrated at the center of gravity (as it results in the same value).

Finally

As you can see from the calculation, the conclusion is not so much that it is okay to calculate assuming all the weight is concentrated at the center of gravity, but rather that the point where the result of such a calculation matches the actual moment of force is likely the very definition of the center of gravity.

Therefore, for my elementary school self, answering "The center of gravity is defined as the point that allows such a calculation" might have been a sufficient answer in its own way. That said, I don't think I would have been convinced at the time, so it's a mystery how meaningful this discussion really is.