Chapter 11: Equilibrium and Elasticity – Study Notes

Equilibrium and Elasticity

Introduction

Many structures, such as bridges and ladders, are designed to remain stationary and not accelerate. However, real materials are not perfectly rigid; they are elastic and can deform under force. This chapter introduces the concepts of equilibrium, center of gravity, and the mechanical properties of materials, including stress and strain, to understand how bodies maintain stability and respond to forces.

Conditions for Equilibrium

Static Equilibrium

For a body to be in static equilibrium, two essential conditions must be satisfied:

• First Condition (Translational Equilibrium): The sum of all external forces acting on the body must be zero. $\sum \vec{F} = 0$ $\sum F_x = 0$, $\sum F_y = 0$

• Second Condition (Rotational Equilibrium): The sum of all torques about any axis must be zero. $\sum \tau = 0$

These conditions ensure that the body does not translate or rotate.

Reference Point for Torque

The reference point for calculating torque can be chosen arbitrarily for convenience, as long as the sum of forces is zero. The total torque about point O can be related to the torque about another point P:

• $\vec{\tau}_O = \sum_{i=1}^N \vec{r}_i \times \vec{F}_i$

• $\vec{\tau}_P = \sum_{i=1}^N (\vec{r}_i - \vec{r}_P) \times \vec{F}_i$

This flexibility allows for easier problem solving.

Center of Mass and Center of Gravity

Definitions and Properties

The center of mass (CM) is the point at which the mass of a body can be considered to be concentrated for translational motion. The center of gravity (CG) is the point where the gravitational force can be considered to act. If gravity is uniform, CM and CG coincide.

• Translational Equilibrium: $\vec{F} + \vec{g} \sum_{i=1}^N m_i = 0$

• Rotational Equilibrium: $\sum_{i=1}^N \vec{r}_i \times m_i \vec{g} = (\sum_{i=1}^N m_i \vec{r}_i) \times \vec{g} = 0$

For balance, an object will hang with its CM directly below the suspension point. If not, gravity creates a torque that rotates the object until equilibrium is achieved.

Examples of Equilibrium

Equilibrium for a Bar

When analyzing a bar in equilibrium, all its weight is considered to act at the CG. The bar must satisfy both translational and rotational equilibrium:

• $F_1 + F_2 - W = 0$

• $F_2 L - W \frac{L}{2} = 0$ (torque about one end)

• Solution: $F_2 = \frac{W}{2}$, $F_1 = W - F_2 = \frac{W}{2}$

Lever

A uniform rod used as a lever must satisfy equilibrium conditions:

• Vertical forces: $F = L + W + R = 800\,N$

• Torque about support point P: $Lx + W(x - d/2) - R(d - x) = 0$

• Solution: $x = \frac{L + W/2}{L + W + R} d = 0.69d$

Bar on Hinge and Rope

To find the tension $T$ and hinge force components $N$ and $V$:

• Vertical equilibrium: $V + T \sin \theta - w - W = 0$

• Horizontal equilibrium: $N - T \cos \theta = 0$

• Torque about hinge: $w \frac{L}{2} + W \frac{L}{4} - T \sin \theta \frac{3L}{4} = 0$

• Solution: $T = \frac{1}{\sin \theta} (\frac{w}{2} + W) = 2280\,N$

Leaning Ladder Problems

These examples analyze the equilibrium and friction conditions for a ladder leaning against a wall, with or without a person climbing.

• Static friction at floor: $\mu_s$

• No friction at wall.

• Equilibrium equations:

  • $N_1 + f_s + W_m + W + N_2 = 0$

  • $\sum (\vec{r}_P) = 0$

• Maximal friction: $s_{max} = \mu_s \left(1 + \frac{W_m}{W}\right) L \tan \theta - \left(\frac{W_m}{W}\right) \frac{L}{2}$

• Minimal angle for stability: $\theta_{min} = \arctan \left(\frac{1}{\mu_s}\right)$

Base of Support and Stability

Base of Support

An object at rest is stable if its center of mass is above its base of support—the area bounded by points of contact with the ground. If the CM moves outside this area, the object will tip over.

Stress, Strain, and Elastic Deformations

Definitions

• Stress: Force per unit area. $\text{Stress} = \frac{F}{A}$

• Strain: Fractional deformation. $\text{Strain} = \frac{\Delta l}{l_0}$

• Hooke's Law: For elastic deformations, stress is proportional to strain.

• Elastic Modulus: Ratio of stress to strain. $\text{Elastic Modulus} = \frac{\text{Stress}}{\text{Strain}}$

Types of Stress and Moduli

• Tensile Stress and Strain: $\text{Tensile Stress} = \frac{F_t}{A}$, $\text{Tensile Strain} = \frac{\Delta l}{l_0}$

• Compressive Stress and Strain: Defined similarly to tensile, but for compression.

• Young's Modulus (Y): $Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}$

• Bulk Stress and Strain: $p = \frac{F}{A}$, $\text{Bulk Strain} = \frac{\Delta V}{V_0}$

• Bulk Modulus (B): $B = -\frac{\Delta p}{\Delta V / V_0}$

• Shear Stress and Strain: $\text{Shear Stress} = \frac{F_t}{A}$, $\text{Shear Strain} = \frac{x}{h}$

• Shear Modulus (S): $S = \frac{F_t / A}{x / h}$

Elasticity and Plasticity

Materials obey Hooke's law up to their elastic limit. Beyond this, they may undergo plastic deformation and eventually fracture. The breaking stress is the maximum stress a material can withstand before failure.

Table: Approximate Elastic Moduli

Material	Young's Modulus, Y (Pa)	Bulk Modulus, B (Pa)	Shear Modulus, S (Pa)
Aluminum	7.0 × 1010	7.5 × 1010	2.5 × 1010
Brass	9.0 × 1010	6.0 × 1010	3.5 × 1010
Copper	11 × 1010	14 × 1010	4.4 × 1010
Crown glass	6.0 × 1010	5.0 × 1010	2.5 × 1010
Iron	21 × 1010	16 × 1010	7.7 × 1010
Lead	1.6 × 1010	4.1 × 1010	0.6 × 1010
Nickel	21 × 1010	17 × 1010	7.8 × 1010
Steel	20 × 1010	16 × 1010	7.5 × 1010

Table: Approximate Breaking Stresses

Material	Breaking Stress (Pa or N/m2)
Aluminum	2.2 × 108
Brass	4.7 × 108
Glass	10 × 108
Iron	3.0 × 108
Phosphor bronze	5.6 × 108
Steel	5 – 20 × 108

Example: Stretching Elevator Cable

A steel cable of length 3.0 m and cross-sectional area 0.20 cm2 supports a 550 kg elevator. The cable stretches 0.40 cm under load. Calculate stress, strain, and Young's modulus:

• Stress: $\frac{F}{A} = \frac{(550 \times 9.8)\,\text{N}}{2.0 \times 10^{-5}\,\text{m}^2} = 2.7 \times 10^7\,\text{Pa}$

• Strain: $\frac{\Delta l}{l_0} = \frac{0.0040\,\text{m}}{3.0\,\text{m}} = 0.00133$

• Young's Modulus: $Y = \frac{\text{stress}}{\text{strain}} = 2.0 \times 10^{10}\,\text{Pa}$

Summary

• Equilibrium requires both zero net force and zero net torque.

• The center of mass and center of gravity are crucial for stability.

• Stress and strain describe how materials deform under force; elastic moduli quantify their resistance.

• Practical examples include levers, ladders, and beams, with calculations for forces and torques.

• Material properties such as Young's modulus and breaking stress are essential for engineering applications.