Effect of gravity on a rigid body

The problem gravity poses. Gravity acts on every particle of a rigid body — a billion tiny downward pulls. Adding the moment of each one separately would be impossible. The rescue is a single beautiful fact: for the purpose of resultant force and moments, all of that weight behaves as one force acting at a single point — the centre of mass.

The centre of mass (c.o.m.) is the point at which the entire weight of the body may be taken to act. Write the weight as

$W = mg$

where $m$ is the mass (in kg) and $g = 10\ \text{m s}^{-2}$ is the acceleration due to gravity. This force acts vertically downward, and its line of action passes through the centre of mass.

Where is it? For a uniform body the centre of mass is at the geometric centre (the centre of a uniform rod is its midpoint; of a uniform rectangular lamina, the intersection of the diagonals). The actual calculation of the centre of mass belongs to other subtopics — here we simply use the fact that one exists and that the weight acts there.

Cambridge tip. Whenever a problem mentions the weight of a rod, lamina or block, draw the single weight arrow $mg$ acting vertically downward through the centre of mass. That one arrow carries all of gravity's effect.

Because the weight acts as a single force $mg$ vertically downward through the centre of mass $G$, its moment about a point $O$ is found exactly like any other force: force times perpendicular distance from $O$ to the line of action.

The line of action of the weight is vertical. So the perpendicular distance from $O$ to it is the horizontal distance between $O$ and $G$. If $G$ is a horizontal distance $\bar{x}$ from $O$, then

$\text{moment of weight about } O = mg \times \bar{x}$

The vertical position of $G$ does not matter for this moment — only how far $G$ sits horizontally from $O$.

Sense. A weight acting downward with $G$ to the right of $O$ turns the body clockwise about $O$; with $G$ to the left of $O$, anticlockwise. As always, fix a sense convention (anticlockwise positive) and sign accordingly.

Cambridge tip. A common slip is to multiply $mg$ by the straight-line distance $OG$, or by the vertical height of $G$. The weight is vertical, so its moment arm is the horizontal distance only — drop a perpendicular and read off the horizontal gap.

This is the single most important fact of the subtopic. When a rigid body is suspended freely from a fixed point $P$ and hangs in equilibrium, it always settles so that

the centre of mass $G$ lies vertically below the point of suspension $P$.

Why. Hanging from $P$, the body feels only two forces: the upward pull of the support at $P$, and the weight $mg$ acting downward through $G$. Take moments about $P$: the support force passes through $P$ (zero moment), so for equilibrium the weight must also have zero moment about $P$. The weight is vertical, so its moment about $P$ is zero exactly when $G$ is on the vertical line through $P$ — i.e. $G$ is directly below $P$. If $G$ were off to one side, the weight would have a non-zero moment and swing the body until $G$ came underneath.

Two ways examiners use this:

• Finding an angle of tilt. A body is suspended from one of its points. You know where $G$ sits relative to the body. The body rotates until $PG$ is vertical; the angle the body has turned (or the angle a particular edge now makes with the vertical/horizontal) is found from the geometry of the triangle formed by $P$, $G$ and the horizontal/vertical offsets.
• Locating the centre of mass. Suspend the body from two different points in turn; $G$ lies on the vertical through each suspension point, so it is at their intersection (the "plumb-line" method).

Getting the tilt angle. Set up the right-angled triangle whose sides are the horizontal and vertical offsets of $G$ from $P$ measured in the body's own frame, then

$\tan\theta = \frac{\text{horizontal offset of } G \text{ from } P}{\text{vertical offset of } G \text{ from } P}$

gives the angle through which the chosen body-line has tilted from the vertical. (Which lengths are "horizontal" and "vertical" depends on how the question defines the angle — always sketch $P$, $G$ and the offsets first.)

Cambridge tip. The whole method rests on one sentence: "In equilibrium $G$ is vertically below $P$." Write that sentence down, draw the vertical line through $P$, mark $G$ on it, and the required angle drops out of basic trigonometry.

The suspension fact can be run in reverse to find the centre of mass of an irregular body — the classic plumb-line experiment, and a favourite "show that" setting in 9231.

The plumb-line method.

1. Suspend the body from a point $A$. In equilibrium, $G$ lies on the vertical line through $A$ — call it line $\ell_A$.
2. Suspend it again from a different point $B$. Now $G$ lies on the vertical through $B$ — line $\ell_B$.
3. The centre of mass is the intersection of $\ell_A$ and $\ell_B$, since $G$ must lie on both lines.

Using it numerically. If a question gives the position of $G$ and asks you to verify an angle, you reverse the tilt calculation: impose that $G$ is on the vertical through the suspension point, write the geometric condition, and solve for the unknown (an angle, a length, or a mass that makes $G$ land where required).

A frequent variant: a uniform shape has a portion removed, or an extra mass attached, shifting $G$. You first find the new $G$, then apply the suspension condition to get the resulting tilt. The mechanics here is always the same — equilibrium $\Rightarrow$ $G$ vertically below the pivot — only the geometry of finding $G$ changes.

Cambridge tip. In a "hence find the angle the edge $AB$ makes with the vertical" question, identify clearly (i) the suspension point, (ii) the centre of mass, and (iii) the line whose angle is asked. The angle between $PG$ (vertical) and that line is what trigonometry then delivers.