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Topic 13 of 19

Static Equilibrium & Elasticity

A structure is in static equilibrium when it is not accelerating — neither translating nor rotating. Two conditions must be simultaneously satisfied: the net force must be zero and the net torque (about any point) must be zero. This topic also introduces the elastic response of solid materials.

Key Concepts

Conditions for Equilibrium

(1) Translational:

\sum\vec{F} = 0

(no net force). (2) Rotational:

\sum\vec{\tau} = 0

(no net torque about any point). Both must hold simultaneously.

Choosing a Pivot

For the torque condition, any point can be used as the pivot. A strategic choice (e.g., at the location of an unknown force) eliminates that force from the torque equation, reducing algebra.

Center of Gravity

The point where gravity effectively acts on an extended body. Coincides with the center of mass in a uniform gravitational field. An object topples when its center of gravity falls outside its support base.

Stable, Unstable, Neutral Equilibrium

Stable: displaced object returns (lowest potential energy). Unstable: displaced object moves further away (highest PE). Neutral: displaced object stays displaced (constant PE).

Hooke's Law (materials)

A material under stress

\sigma = F/A

deforms with strain

\epsilon = \Delta L/L

. Young's modulus

E = \sigma/\epsilon

characterizes stiffness. Hooke's Law applies in the elastic (linear) regime.

Shear and Bulk Modulus

Shear modulus

G

relates shear stress to shear strain (shape change at constant volume). Bulk modulus

B

relates pressure change to volume change (compression/expansion).

Key Equations

Equilibrium conditions

\sum F_x = 0, \quad \sum F_y = 0, \quad \sum \tau = 0

All three must hold; the torque equation can be written about any convenient pivot.

Young's modulus

E = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}

Ratio of tensile stress to tensile strain. Units: Pa (N/m²). High E → stiff material.

Tensile deformation

\Delta L = \frac{FL}{EA}

Elongation of a rod under axial load F; A is cross-sectional area, L is original length.

Shear modulus

G = \frac{F/A}{\Delta x / L}

Shear stress over shear strain; determines resistance to shape-changing deformation.

Worked Example

Beam Supported at Two Points

Problem

A uniform 10 kg beam 4 m long is supported at its left end and at a point 1 m from the right end. Find the support forces ( $g = 10$ m/s²).

Solution

Label left support $N_L$ at $x=0$ , right support $N_R$ at $x=3$ m. Weight $W = 100$ N at center $x=2$ m.

Force equation: $N_L + N_R = W = 100$ N.

Torque about left end (eliminates $N_L$ ):

N_R\times 3 - 100\times 2 = 0 \implies N_R = \frac{200}{3} \approx 66.7 \text{ N}

N_L = 100 - 66.7 = 33.3 \text{ N}

Answer Left support ≈ 33.3 N; right support ≈ 66.7 N.

Practice

Exercises

7 problems

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2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

The beam is in static equilibrium. Two downward forces act on it (shown in red). A wall reaction force R acts upward at the left. Using ΣF = 0, find R.

R = N

Exercise 2 / 7 Free

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A sign weighing 80 N hangs from a wall bracket supported by a cable at 45° from horizontal. For static equilibrium ΣFy = 0: T·sin(45°) = 80 N. Find the cable tension T.

T = N

3 of 7

A ladder of mass $12$ kg and length $5$ m leans against a frictionless wall, making a $60°$ angle with the floor. A person of mass $70$ kg stands $3/5$ of the way up the ladder. Find the horizontal friction force (in N) at the base. Use $g = 9.8$ m/s².

Unlock Exercise 3