The Timpson Lyceum · University Physics I (2025)

Static Equilibrium, Center of Gravity, and Elasticity – Lecture 11 (University Physics I)

Overview

This lecture extends the idea of equilibrium from point particles to rigid bodies. For a rigid body to be in static equilibrium it is not enough that its center of mass not accelerate — it must also not rotate, so two conditions must hold at once: the net force is zero and the net torque is zero. We use the torque condition to extract extra information about real structures (bridges, ladders, vehicles) that ordinary force balance alone cannot give. The lecture also defines the center of gravity and proves it coincides with the center of mass in a uniform gravitational field. Finally it introduces stress, strain, and the elastic moduli — Young's modulus for stretching/compressing, the bulk modulus for pressure-driven volume change, and the shear modulus for parallel-force warping — which describe how real (non-rigid) materials deform under load.

Key Concepts

The two conditions for static equilibrium

A rigid body is in static equilibrium when both the net external force and the net external torque are zero:

$\sum \vec{F} = 0 \qquad \text{and} \qquad \sum \vec{\tau} = 0$

The force condition keeps the center of mass from accelerating; the torque condition keeps the body from rotating. A body can satisfy one without the other — equal up/down forces applied on the same side produce zero net force but a nonzero net torque, so it spins in place.
Because the body is not rotating, the net torque is zero about any point. You may choose the pivot freely; choosing the point where the most unknown forces act makes those forces drop out (a force applied at the pivot has zero moment arm and contributes zero torque).
Torque magnitude: $\tau = r F \sin\theta$ , where $\theta$ is the angle between the position vector and the force. Equivalently $\tau = r F_\perp$ , using only the force component perpendicular to $\vec{r}$ .

Center of gravity

The center of gravity is the point about which the net torque due to gravity is zero, no matter how the object is oriented.
Summing the gravitational torques over all the mass elements $m_i$ at positions $\vec{r}_i$ , and factoring out the constant $\vec g$ :

$\vec{\tau}_{\text{net}} = \sum \vec{r}_i \times (m_i \vec g) = \left(\sum m_i \vec{r}_i\right)\times \vec g = M\,\vec{r}_{\text{cm}} \times \vec g = \vec{r}_{\text{cm}} \times (M\vec g)$

The total gravitational torque acts as though the entire weight $M\vec g$ were applied at the center of mass. Hence in a uniform gravitational field the center of gravity and the center of mass are the same point. They differ only for objects large enough that $\vec g$ varies appreciably across them.
Practical consequence: dangle an irregular object from a point and the center of gravity hangs directly below the support line; doing this from two (or three, in 3-D) points and intersecting the lines locates it.
Tipping: an object topples when its center of gravity is no longer directly above its base of support, so the weight produces an unbalanced torque — why high-loaded vehicles tip more easily.

Stress, strain, and elastic moduli

Real bodies deform. The general (Hooke's-law) statement is that stress is proportional to strain, with the constant of proportionality called the elastic modulus:

$\text{modulus} = \frac{\text{stress}}{\text{strain}}$

Tensile/compressive (Young's modulus $Y$ ): a perpendicular force $F_\perp$ on cross-sectional area $A$ stretches (or compresses) a rod of initial length $L_0$ by $\Delta L$ :

$\frac{F_\perp}{A} = Y\,\frac{\Delta L}{L_0} \qquad\Longrightarrow\qquad \Delta L = \frac{F_\perp L_0}{A\,Y}$

Bigger force or longer rod → more stretch; larger area or larger $Y$ → less stretch.

Bulk modulus $B$ : a uniform pressure change $\Delta p$ changes volume $V_0$ by $\Delta V$ . The minus sign makes $B$ positive (increasing pressure shrinks volume):

$\Delta p = -B\,\frac{\Delta V}{V_0}$

Here pressure is force per unit area, $p = F/A$ .

Shear modulus $S$ : a force $F_\parallel$ applied parallel to area $A$ warps the body sideways by $x$ over height $h$ :

$\frac{F_\parallel}{A} = S\,\frac{x}{h}$

Units: strain is dimensionless (a length over a length, or volume over volume), so stress and every modulus carry the units of pressure — pascals (Pa $=$ N/m $^2$ ).
These relations assume the elastic regime: forces small enough that the material springs back when released, with no permanent deformation.

Worked Example

Ladder against a wall. A uniform ladder of length $L = 5.0\,\text{m}$ leans at $\theta = 53.1^\circ$ against a frictionless wall, its base on a floor with static friction. The ladder weighs $W_L = 180\,\text{N}$ ; a man of weight $W_m = 800\,\text{N}$ stands one-third of the way up. Find the normal and friction forces at the base.

Force balance gives the floor's normal force immediately, since the wall (frictionless) exerts only a horizontal force:

\begin{aligned} N_{\text{bottom}} &= W_L + W_m \\ &= 180 + 800 = 980\ \text{N} \end{aligned}

Taking torques about the base $O$ (where $N_{\text{bottom}}$ and friction act, so they drop out), the wall's normal force $N_{\text{top}}$ balances the two weights. Using $\beta = 90^\circ - \theta$ for the angle the ladder makes with each downward weight, and dividing through by $L$ :

\begin{aligned} N_{\text{top}}\,L\sin\theta &= \tfrac{1}{3}W_m\,L\sin\beta + \tfrac{1}{2}W_L\,L\sin\beta \\ N_{\text{top}} &= \frac{\left(\tfrac{1}{3}W_m + \tfrac{1}{2}W_L\right)\sin(90^\circ-\theta)}{\sin\theta} \\ &= \frac{\left(\tfrac{1}{3}(800) + \tfrac{1}{2}(180)\right)\sin(36.9^\circ)}{\sin(53.1^\circ)} \approx 268\ \text{N} \end{aligned}

Horizontal balance then makes friction equal the wall force, $f_s = N_{\text{top}} \approx 268\,\text{N}$ . The minimum coefficient of static friction follows from $f_s = \mu_s N_{\text{bottom}}$ :

\begin{aligned} \mu_s &= \frac{f_s}{N_{\text{bottom}}} \\ &= \frac{268}{980} \approx 0.273 \end{aligned}

Summary

Static equilibrium for a rigid body requires both $\sum\vec F = 0$ and $\sum\vec\tau = 0$ — no acceleration and no rotation.
Since the body does not rotate, the net torque is zero about any chosen pivot; pick the point where the most unknown forces act to simplify the algebra.
The center of gravity is the point where gravity exerts no net torque; in a uniform field it equals the center of mass, because the total weight acts as if applied there.
An object tips when its center of gravity passes outside its base of support, producing an unbalanced torque.
Stress $=$ modulus $\times$ strain: Young's modulus governs stretching ( $F/A = Y\,\Delta L/L_0$ ), the bulk modulus governs pressure-driven volume change ( $\Delta p = -B\,\Delta V/V_0$ ), and the shear modulus governs parallel-force warping ( $F/A = S\,x/h$ ).
Strain is dimensionless, so all moduli are measured in pascals; these elastic relations hold only for forces small enough that the material returns to its original shape.