On the edge of a wall, we build a brick tower that only holds together because of the bricks' own weight. Our goal is to build a stable tower whose overhang $d$ is greater than the length $\ell$ of a single brick. What is the minimum number of bricks you need? [Hint: Find the center of mass for each brick and add.]

On the edge of a wall, we build a brick tower that only holds together because of the bricks' own weight. Our

V. Numerical Problems · Q7

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Concept understanding — Center of Mass

What is the Center of Mass?

Imagine you pick up a broom by the handle and try to balance it horizontally on one finger. You instinctively slide your finger along the handle until the broom stays level. That point — the one where the broom doesn't tip — is its center of mass.

Now think about throwing a cricket bat. It spins and wobbles in the air, but there is one point on the bat that follows a smooth, parabolic path, as if all the bat's mass were concentrated there. That point is also the center of mass.

The core idea is simple: the center of mass is the average position of all the mass in an object. It's the point where you could imagine the entire mass of the object being concentrated, and the object would behave the same way under the influence of external forces.

Why does this matter?

When you push an object at its center of mass, it moves in a straight line without rotating. Push it anywhere else, and it will both move and spin. This is why:

A car's stability depends on where its center of mass is (lower = safer).
A tightrope walker holds a long pole — moving the pole shifts their combined center of mass back over the rope.
In projectile motion, the center of mass of a system (like an exploding firework) continues along the original parabolic path, even though the fragments scatter.

The precise definition

For a system of particles, the center of mass is the weighted average of their positions, where the weight is the mass of each particle.

\vec{R}_{\text{CM}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_n \vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

Here:

$\vec{R}_{\text{CM}}$ is the position vector of the center of mass
$m_i$ is the mass of the $i$ -th particle
$\vec{r}_i$ is the position vector of that particle

For a continuous object (like a rod or a sphere), the sum becomes an integral:

\vec{R}_{\text{CM}} = \frac{1}{M} \int \vec{r} \, dm

where $M$ is the total mass and $dm$ is an infinitesimal mass element.

Breaking it down with an example

Take two masses on a light rod: $m_1 = 2 \text{ kg}$ at $x = 0$ , and $m_2 = 3 \text{ kg}$ at $x = 5 \text{ m}$ .

The center of mass is:

x_{\text{CM}} = \frac{(2)(0) + (3)(5)}{2 + 3} = \frac{0 + 15}{5} = 3 \text{ m}

So the center of mass is at $x = 3 \text{ m}$ , closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.

Note

The center of mass does not have to be inside the object. A ring or a hollow sphere has its center of mass at the geometric center, which is empty space.

Key properties to remember

External forces only — Internal forces (like collisions between parts of the system) do not affect the motion of the center of mass. Only external forces can change its velocity.
If no external force acts, the center of mass moves with constant velocity (or stays at rest). This is the law of conservation of momentum applied to the whole system.
For symmetric objects with uniform density, the center of mass coincides with the geometric center. For irregular shapes, it shifts toward the region with more mass. …

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Step 1. Set up the stacking rule. Build the tower from the top brick downward. For any group of the topmost $k$ bricks (each of length $\ell$ ), treated as a single rigid unit, the maximum they can overhang the brick immediately below them — without the whole group toppling — is exactly reached when the combined center of mass of the top $k$ bricks sits directly above the very edge of the $(k+1)$ -th brick. Any further overhang and the group topples; this is the greatest stable overhang, since the problem asks for the minimum number of bricks for a given required overhang, using the best (maximum-overhang) stacking at every stage.

Step 2. Find the extra overhang contributed when going from $k$ bricks to $k+1$ bricks. Adding the $(k+1)$ -th brick beneath a stable stack of $k$ bricks (whose combined center of mass is known), the new combined center of mass (of all $k+1$ bricks) shifts by an amount inversely proportional to $k$ ; working through the center-of-mass balance (treating the top $k$ bricks, of total mass $km_0$ for single-brick mass $m_0$ , as concentrated at their own combined center of mass, positioned at the very edge of brick $k+1$ ), the additional overhang gained at this step is

\Delta_k=\frac{\ell}{2k}.

Step 3. Sum the overhangs to get the total overhang after $n$ bricks. Starting from the top (brick 1, contributing nothing extra since it is the very first) and adding each subsequent brick below it, the total overhang beyond the edge of the bottom (table or wall) support, after $n$ bricks, is

d_n=\sum_{k=1}^{n-1}\frac{\ell}{2k}=\frac{\ell}{2}\left(1+\frac12+\frac13+\cdots+\frac{1}{n-1}\right)=\frac{\ell}{2}H_{n-1},

where $H_{n-1}$ is the $(n-1)$ -th harmonic number.

Step 4. Apply the required condition, $d_n>\ell$ . …

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