8.1 Engineering Mechanics — Statics & Dynamics

8.1 Engineering Mechanics — Statics & Dynamics

Engineering Mechanics provides the fundamental physical principles governing forces, moments, equilibrium, and motion in structures and machinery. For the Registered Electrical Engineer (REE) Licensure Examination under Engineering Sciences and Allied Subjects (ESAS), candidates must master both rigid-body statics and particle/rigid-body dynamics to solve mechanical design, structural loading, and equipment mounting problems.

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1. Force Systems, Moments, and 2D Static Equilibrium

A force is a vector quantity defined by magnitude, direction, and point of application. In two-dimensional force analysis, concurrent and non-concurrent force systems are evaluated using scalar projections along rectangular coordinate axes.

Force Vectors and Moment Calculation

• Force Components: $F_x = F \cos\theta, \quad F_y = F \sin\theta, \quad F = \sqrt{F_x^2 + F_y^2}$
• Moment of a Force: The moment $M_O$ of a force about a point $O$ measures its rotational tendency: $M_O = F \cdot d$ where $d$ is the perpendicular moment arm from point $O$ to the line of action of force $F$.
• Varignon's Theorem: The moment of a force about any point is equal to the sum of the moments of its rectangular components about that same point: $M_O = F_y x - F_x y$

Conditions for 2D Rigid-Body Equilibrium

For a rigid body in static equilibrium under a coplanar force system, the resultant force and resultant moment must vanish:

$\sum F_x = 0$

$\sum F_y = 0$

$\sum M_z = 0$

Free-Body Diagrams (FBD) & Support Reactions

Constructing an accurate Free-Body Diagram (FBD) is the crucial first step in solving static equilibrium problems:

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2. Coulomb Dry Friction & Impending Motion

Friction is the resistive tangential force developed at the contact interface of two bodies opposing relative motion.

Laws of Coulomb Friction

1. Static Friction ($F_s$): Exists when relative motion is absent. The friction force adjusts dynamically to balance applied shear loads up to a maximum threshold: $F_s \le F_{s,\text{max}} = \mu_s N$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal contact force.
2. Impending Motion: At the threshold of slipping, $F_s = F_{s,\text{max}} = \mu_s N$. The angle of static friction $\phi_s$ is defined by: $\tan\phi_s = \mu_s$
3. Kinetic Friction ($F_k$): Once slipping occurs, friction drops to a constant kinetic value: $F_k = \mu_k N \quad (\text{where } \mu_k < \mu_s)$

Belt Friction (Capstan Formula)

For a flexible belt or rope wrapped around a stationary drum or pulley with total wrap angle $\beta$ (in radians):

$T_2 = T_1 e^{\mu \beta}$

where $T_2$ is the larger tension (impending motion toward $T_2$) and $T_1$ is the smaller tension.

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3. Centroids and Area Moments of Inertia

Centroids of Plane Areas

The centroid $(\bar{x}, \bar{y})$ represents the geometric center of a planar area $A$:

$\bar{x} = \frac{\int x\ dA}{A} = \frac{\sum A_i x_i}{\sum A_i}, \quad \bar{y} = \frac{\int y\ dA}{A} = \frac{\sum A_i y_i}{\sum A_i}$

Area Moments of Inertia

The second moment of area measures structural resistance to bending about a specific axis:

$I_x = \int y^2\ dA, \quad I_y = \int x^2\ dA$

• Polar Moment of Inertia ($J_O$): Resistance to torsional twisting about an axis perpendicular to the plane: $J_O = \int r^2\ dA = I_x + I_y$
• Radius of Gyration ($k$): The distance from an axis at which the total area could be concentrated without altering its moment of inertia: $k_x = \sqrt{\frac{I_x}{A}}, \quad k_y = \sqrt{\frac{I_y}{A}}$

Parallel-Axis Theorem (Steiner's Theorem)

The moment of inertia about any parallel axis $x$ located a distance $d$ from the centroidal axis $x_c$ is:

$I_x = I_{xc} + A d^2$

Geometric Formulas for Common Cross-Sections

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4. Particle Kinematics & Kinetics

Kinematics analyzes motion without regard to forces, while kinetics relates forces to motion.

Kinematics Equations (Constant Acceleration)

For linear rectilinear motion with constant acceleration $a$:

$v = v_0 + a t$

$s = s_0 + v_0 t + \frac{1}{2} a t^2$

$v^2 = v_0^2 + 2 a (s - s_0)$

Curvilinear Motion (Normal-Tangential Components)

When a particle moves along a curved path of radius of curvature $\rho$:

$a_t = \frac{dv}{dt} \quad (\text{tangential acceleration altering speed})$

$a_n = \frac{v^2}{\rho} \quad (\text{normal acceleration directing path curvature})$

$a_{\text{total}} = \sqrt{a_t^2 + a_n^2}$

Newton's Second Law of Motion

$\sum \vec{F} = m \vec{a} \implies \sum F_x = m a_x, \quad \sum F_y = m a_y$

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5. Work-Energy, Impulse-Momentum & D'Alembert's Principle

Work-Energy Principle

The work done by all forces acting on a particle equals the change in kinetic energy:

$U_{1-2} = \Delta T = T_2 - T_1$

$U_{1-2} = \int_{s_1}^{s_2} F \cos\theta\ ds, \quad T = \frac{1}{2} m v^2$

• Conservation of Mechanical Energy: In conservative force fields (gravitational, elastic spring): $T_1 + V_1 = T_2 + V_2$ where gravitational potential energy $V_g = m g h$ and elastic spring energy $V_e = \frac{1}{2} k x^2$.

Linear Impulse and Momentum

The linear impulse of a force over time equals the change in linear momentum:

$\vec{J} = \int_{t_1}^{t_2} \vec{F}\ dt = m \vec{v}_2 - m \vec{v}_1 = \Delta \vec{p}$

• Coefficient of Restitution ($e$): In central line-of-impact collisions between two bodies: $e = \frac{v_{2,y} - v_{2,x}}{v_{1,x} - v_{1,y}} = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}$
  • $e = 1$: Perfectly elastic collision (no energy loss).
  • $e = 0$: Perfectly plastic collision (bodies stick together).

D'Alembert's Principle

D'Alembert's principle transforms a dynamic system into an equivalent static equilibrium system by introducing a fictitious inertia force $\vec{F}_i = -m\vec{a}$ opposite to acceleration:

$\sum \vec{F} + (-m \vec{a}) = 0$

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Solved Board Exam Examples

Example 1: 2D Equilibrium & Impending Friction on an Incline

Problem: A block of mass $m = 50\ \text{kg}$ rests on a plane inclined at $\theta = 30^\circ$ to the horizontal. The coefficient of static friction between block and plane is $\mu_s = 0.25$. A horizontal force $P$ is applied to push the block up the incline. Find the minimum force $P$ required to cause impending motion up the incline. Take $g = 9.81\ \text{m/s}^2$.

Solution:

1. Calculate weight $W$: $W = m g = 50 \times 9.81 = 490.5\ \text{N}$
2. Set up coordinate axes parallel ($x$-axis up incline) and perpendicular ($y$-axis out of incline) to the surface.
3. Resolve horizontal force $P$ and vertical weight $W$ into components:
   • $P_x = P \cos 30^\circ = 0.866025 P$
   • $P_y = -P \sin 30^\circ = -0.5 P$
   • $W_x = -W \sin 30^\circ = -490.5 \times 0.5 = -245.25\ \text{N}$
   • $W_y = -W \cos 30^\circ = -490.5 \times 0.866025 = -424.787\ \text{N}$
4. Apply equilibrium perpendicular to incline ($\sum F_y = 0$): $N - W \cos 30^\circ - P \sin 30^\circ = 0 \implies N = 424.787 + 0.5 P$
5. Express static friction at impending motion ($F_s = \mu_s N$ acting down incline): $F_s = 0.25 (424.787 + 0.5 P) = 106.197 + 0.125 P$
6. Apply equilibrium parallel to incline ($\sum F_x = 0$): $P \cos 30^\circ - W \sin 30^\circ - F_s = 0$ $0.866025 P - 245.25 - (106.197 + 0.125 P) = 0$ $(0.866025 - 0.125) P = 245.25 + 106.197$ $0.741025 P = 351.447 \implies P = \frac{351.447}{0.741025} \approx 474.27\ \text{N}$

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Example 2: Composite Section Centroid & Moment of Inertia

Problem: A symmetrical structural steel T-section has a top flange of width $b_f = 200\ \text{mm}$, flange thickness $t_f = 20\ \text{mm}$, web height $h_w = 180\ \text{mm}$, and web thickness $t_w = 20\ \text{mm}$. Calculate the distance of the centroid $\bar{y}$ from the bottom of the web and the centroidal moment of inertia $I_{\bar{x}}$.

Solution:

1. Divide the section into two rectangular segments:
   • Segment 1 (Flange): $A_1 = 200 \times 20 = 4000\ \text{mm}^2$; centroid $y_1 = 180 + 10 = 190\ \text{mm}$.
   • Segment 2 (Web): $A_2 = 180 \times 20 = 3600\ \text{mm}^2$; centroid $y_2 = 90\ \text{mm}$.
2. Compute overall centroid $\bar{y}$ from the bottom edge: $\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{(4000 \times 190) + (3600 \times 90)}{4000 + 3600} = \frac{760,000 + 324,000}{7600} = \frac{1,084,000}{7600} = 142.63\ \text{mm}$
3. Calculate centroidal moments of inertia of each segment about its own centroidal axis: $I_{c1} = \frac{b_f t_f^3}{12} = \frac{200 \times 20^3}{12} = 133,333\ \text{mm}^4$ $I_{c2} = \frac{t_w h_w^3}{12} = \frac{20 \times 180^3}{12} = 9,720,000\ \text{mm}^4$
4. Calculate vertical distances $d_i = |y_i - \bar{y}|$ to overall centroidal axis: $d_1 = |190 - 142.63| = 47.37\ \text{mm}$ $d_2 = |90 - 142.63| = 52.63\ \text{mm}$
5. Apply Parallel-Axis Theorem ($I_{\bar{x}} = \sum (I_{ci} + A_i d_i^2)$): $I_{x1} = 133,333 + 4000 \times (47.37)^2 = 133,333 + 8,975,668 = 9,109,001\ \text{mm}^4$ $I_{x2} = 9,720,000 + 3600 \times (52.63)^2 = 9,720,000 + 9,971,701 = 19,691,701\ \text{mm}^4$ $I_{\bar{x}} = 9,109,001 + 19,691,701 = 28.80 \times 10^6\ \text{mm}^4 = 28.80 \times 10^{-6}\ \text{m}^4$