Centre of Gravity

The figure shows an irregularly shaped cardboard sheet balanced perfectly on the tip of a pencil. The pencil tip is placed at a point labelled G, and the cardboard remains horizontal and stationary — it does not tilt or fall off. Two forces are drawn acting on the cardboard: its total weight $Mg$ (where $M$ is the mass of the cardboard and $g$ is the acceleration due to gravity) acts vertically downward through G, and the upward normal reaction $R$ from the pencil tip acts vertically upward, also through G. In addition, the figure schematically indicates several small masses $m_1, m_2, \dots$ distributed across the cardboard, each with its own weight $m_i g$ acting at its own location.

The physical idea is straightforward: the cardboard is in translational equilibrium because the net external force on it is zero — the upward reaction $R$ exactly balances the downward weight $Mg$. But more importantly, it is also in rotational equilibrium: because the pencil tip is placed exactly at the centre of gravity G, the line of action of every individual weight $m_i g$ passes through G, so none of them produces a torque about G. The cardboard therefore does not rotate.

The key formula the textbook develops from this figure is the condition for a rigid body to be in static equilibrium:

The first condition says the vector sum of all external forces must be zero — for the cardboard, $R - Mg = 0$. The second condition says the sum of all external torques about any point must be zero. When the support is at the centre of gravity, the torque due to each particle weight about G is $m_i g \times 0 = 0$, so the net torque is automatically zero. If the pencil tip were placed anywhere else, the weights would produce a net torque and the cardboard would tilt.

The figure shows an irregularly shaped lamina hanging freely from a ceiling. A hook or pin at point S on the ceiling holds the body at a point on its edge labelled A. The body is free to rotate about A. A dashed vertical line, labelled AA₁, runs downward from A through the interior of the body. The centre of gravity G is marked on this dashed line, inside the body. Two other suspension points, B and C, are also marked on the edge of the body, with corresponding vertical dashed lines BB₁ and CC₁ that would appear if the body were hung from those points instead.

The physical idea is straightforward: a freely suspended body always comes to rest with its centre of gravity directly below the point of suspension. Gravity pulls straight down, so the line from the suspension point to the centre of gravity must be vertical. By hanging the body from three different points (A, B, C) and drawing the vertical line through each suspension point, you get three lines that all intersect at the same point — the centre of gravity G. The figure captures the first of these three steps: the body hung from A, with the vertical AA₁ drawn, and G lying somewhere on that line.

The key formula is not a new equation but the condition for rotational equilibrium that this method exploits. For a rigid body suspended at a point, the torque due to gravity about the suspension point must be zero when the body is at rest. The torque is the vector cross product of the position vector of the centre of gravity relative to the suspension point and the weight vector. Mathematically, if $\vec{r}_{G/A}$ is the position of G relative to A, and $M\vec{g}$ is the weight, equilibrium requires:

Since $M\vec{g}$ is vertically downward, the cross product is zero only when $\vec{r}_{G/A}$ is parallel to $\vec{g}$ — that is, when G lies directly below A along the vertical line. The figure shows this vertical line as AA₁.