Engineering Mechanics: Distributed Forces, Centers of Mass, and Centroids

Overview and Classification of Distributed Forces

• Fundamental Concept of Concentrated vs. Distributed Forces:

  • Forces are treated as concentrated along their lines of action and at their points of application when analyzing external effects on bodies as a whole.

  • Exact physical concentrated forces do not exist because every external force applied mechanically is distributed over a finite contact area.

  • When the dimension $b$ of a contact area is negligible compared to other pertinent body dimensions (such as wheel-to-wheel distance), actual distributed contact forces are replaced by a concentrated resultant force $R$.

• Physical Examples of Contact Force Distributions:

  • Automobile Tire on Pavement: The force exerted by pavement on a tire is distributed over a finite contact area, which expands if the tire is soft. Replaced by a single concentrated resultant $R$ for whole-vehicle equilibrium analysis.

  • Ball Bearings: Contact between a hardened steel ball and its race occurs over an extremely small, yet finite, contact area.

  • Truss Pin Joints: Pin forces acting on a two-force member are distributed over the contact surface between the pin and the hole, as well as across internal cut sections.

  • Structural Engineering Application: The Gateshead Millennium Bridge spanning the River Tyne in the United Kingdom rotates about a horizontal axis along its span to allow ship passage; its design required determining the cumulative effect of its weight distribution across a range of rotational orientations.

• Internal Force Distributions vs. External Body Analysis:

  • When analyzing internal stress and strain distributions near contact locations, forces cannot be modeled as concentrated; actual distributed loading must be evaluated using material properties from theories of elasticity and plasticity.

  • Forces applied over regions non-negligible in size compared to overall body dimensions require mathematical integration of force intensities across the entire contact domain.

• Three Primary Categories of Distributed Forces:

  • Line Distribution:

  • Occurs when a force is distributed along a line, such as a continuous vertical load supported by a suspended cable.

  • Loading intensity $w$ is expressed as force per unit length.

  • SI units: newtons per meter ($\text{N/m}$).

  • U.S. Customary units: pounds per foot ($\text{lb/ft}$).

  • Area Distribution:

  • Occurs when a force acts over a surface area, such as hydraulic water pressure against the inner face of a dam.

  • Intensity is termed pressure for fluid forces and stress for internal solid forces.

  • Basic SI unit: pascal ($\text{Pa}$), defined as one newton per square meter ($\text{N/m}^2$).

  • Conversion factor: $6895\,\text{Pa} = 1\,\text{lb/in.}^2$

  • Practical SI scales: kilopascal ($\text{kPa} = 10^3\,\text{Pa}$) for fluid pressures; megapascal ($\text{MPa} = 10^6\,\text{Pa}$) for structural stress.

  • Practical U.S. Customary unit: pounds per square inch ($\text{lb/in.}^2$).

  • Volume Distribution (Body Forces):

  • Forces distributed over the entire volume of a body are classified as body forces.

  • Most common body force: gravitational attraction acting on every differential mass element in a body (e.g., heavy cantilevered structures).

  • Gravitational force intensity is expressed by specific weight $\rho g$, where $\rho$ is mass density (mass per unit volume) and $g$ is acceleration due to gravity.

  • SI units for specific weight: $(\text{kg/m}^3)(\text{m/s}^2) = \text{N/m}^3$

  • U.S. Customary units for specific weight: pounds per cubic foot ($\text{lb/ft}^3$) or pounds per cubic inch ($\text{lb/in.}^3$).

Principles of Mass Center and Center of Gravity

• Center of Gravity Definition and Physical Concept:

  • Suspending a body of mass $m$ from various support points (e.g., points $A$, $B$, $C$) yields equilibrium under cord tension and the resultant weight force $W$.

  • The line of action of the resultant weight force $W$ is collinear with the suspension cord.

  • Assuming a uniform, parallel gravitational field, all lines of action intersect at a unique point $G$, defined as the center of gravity.

  • In exact physical reality, gravity forces converge toward Earth's center of attraction and vary with distance, meaning lines of action are slightly non-concurrent; however, for bodies small relative to Earth, the uniform parallel field assumption holds perfectly.

• Mathematical Derivation via Moment Principle:

  • The moment of the resultant gravitational force $W$ about any axis equals the sum of moments of the infinitesimal weight elements $dW$ about that same axis.

  • Total weight equation:     $W = \int dW$

  • Moment equality about coordinate axes:     $\bar{x} W = \int x\,dW$

  • Scalar coordinate equations for center of gravity (Equations 5/1a):     $\bar{x} = \frac{\int x\,dW}{W}$     $\bar{y} = \frac{\int y\,dW}{W}$     $\bar{z} = \frac{\int z\,dW}{W}$

  • Substituting $W = mg$ and $dW = g\,dm$ eliminates gravitational acceleration $g$, yielding center of mass equations (Equations 5/1b):     $\bar{x} = \frac{\int x\,dm}{m}$     $\bar{y} = \frac{\int y\,dm}{m}$     $\bar{z} = \frac{\int z\,dm}{m}$

• Vector Formulation and Density Expressions:

  • Position vector of an elemental mass $dm$: $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$

  • Position vector of mass center $G$: $\bar{\mathbf{r}} = \bar{x}\mathbf{i} + \bar{y}\mathbf{j} + \bar{z}\mathbf{k}$

  • Vector expression for mass center location (Equation 5/2):     $\bar{\mathbf{r}} = \frac{\int \mathbf{r}\,dm}{m}$

  • Substituting mass element in terms of variable density $\rho$ and volume element $dV$ ($dm = \rho\,dV$) yields coordinate equations (Equations 5/3):     $\bar{x} = \frac{\int x\rho\,dV}{\int \rho\,dV}$     $\bar{y} = \frac{\int y\rho\,dV}{\int \rho\,dV}$     $\bar{z} = \frac{\int z\rho\,dV}{\int \rho\,dV}$

• Center of Mass vs. Center of Gravity Comparison:

  • Center of mass depends strictly on mass distribution and is independent of gravitational fields.

  • Center of gravity requires a gravitational field to exist.

  • Both points coincide whenever a body is situated in a uniform and parallel gravitational field.

• Symmetry and Coordinate System Selection:

  • Axes should be chosen to simplify boundary equations (e.g., polar coordinates for circular profiles).

  • If a homogeneous body possesses a line or plane of symmetry, the center of mass always lies on that line or plane because symmetrically located mass elements produce canceling moments.

  • Examples: Homogeneous right-circular cone mass center lies on its central axis of symmetry; half right-circular cone mass center lies on its symmetry plane; half ring mass center lies on line $AB$ at the intersection of two symmetry planes.

Centroids of Lines, Areas, and Volumes

• Definition of Centroid:

  • When density $\rho$ is uniform throughout a body, $\rho$ cancels from numerator and denominator in mass center formulas, leaving purely geometric expressions.

  • The term centroid refers to the geometric center of a shape; for uniform density bodies, centroid and center of mass locations are identical.

• Centroid Categories and Integral Formulas:

  • Line Centroids (Slender Rods / Wires):

  • For length $L$, constant area $A$, constant density $\rho$: $dm = \rho A\,dL$.

  • Coordinate equations (Equations 5/4):       $\bar{x} = \frac{\int x\,dL}{L}$       $\bar{y} = \frac{\int y\,dL}{L}$       $\bar{z} = \frac{\int z\,dL}{L}$

  • The centroid $C$ of a curved line segment generally does not lie on the line itself.

  • Area Centroids (Thin Flat / Curved Surfaces):

  • For area $A$, constant thickness $t$, constant density $\rho$: $dm = \rho t\,dA$.

  • Coordinate equations (Equations 5/5):       $\bar{x} = \frac{\int x\,dA}{A}$       $\bar{y} = \frac{\int y\,dA}{A}$       $\bar{z} = \frac{\int z\,dA}{A}$

  • Numerators $\int x\,dA$, $\int y\,dA$, $\int z\,dA$ are defined as first moments of area.

  • Volume Centroids (3D Solids):

  • For volume $V$, constant density $rho$:$dm = \rho\,dV.\n - Coordinate equations (Equations 5/6):\n      \bar{x} = \frac{\int x\,dV}{V}\n      \bar{y} = \frac{\int y\,dV}{V}\n      \bar{z} = \frac{\int z\,dV}{V}\n\n# Guidelines for Choosing Integration Differential Elements\n\n- **Guideline 1: Order of Element:**\n - Prefer first-order differential elements over higher-order elements to require only one integration step.\n - *Example:* Use a horizontal strip dA = x\,dy$(first-order, single integral in$y$) rather than$dx\,dy (second-order, double integral). Use circular slice $dV = \pi r^2\,dy for solid cone (single integral) rather than $dx\,dy\,dz$ (triple integral).

• Guideline 2: Continuity:

  • Choose differential elements that can be integrated continuously over the entire figure without encountering boundary equation discontinuities.

  • Vertical strips requiring multiple split integration domains due to slope changes should be avoided in favor of continuous horizontal strips.

• Guideline 3: Discarding Higher-Order Terms:

  • Higher-order differential quantities are dropped in comparison to lower-order terms.

  • Example: Area under curve vertical strip $dA = y\,dx$ drops second-order triangular element $dx\,dy$, introducing zero error in the limit.

• Guideline 4: Choice of Coordinates:

  • Align coordinate systems with geometry boundaries (rectangular coordinates for straight edges; polar coordinates for circular sectors and arcs).

• Guideline 5: Centroidal Coordinate of Element ($x_e, y_e, z_e$):

  • Moment arms in first-moment integrals must represent the exact coordinates of the centroid of the chosen differential element.

  • Rewritten integral equations with centroidal element coordinates (Equations 5/5a and 5/6a):     $\bar{x} = \frac{\int x_e\,dA}{A}$     $\bar{y} = \frac{\int y_e\,dA}{A}$     $\bar{z} = \frac{\int z_e\,dA}{A}$     $\bar{x} = \frac{\int x_e\,dV}{V}$     $\bar{y} = \frac{\int y_e\,dV}{V}$     $\bar{z} = \frac{\int z_e\,dV}{V}$

Comprehensive Analytical Derivations and Sample Problems

• Sample Problem 5/1: Centroid of a Circular Arc

  • Arc subtends total angle $2\alpha$, radius $r$, oriented symmetrically about $x$-axis ($\bar{y} = 0$).

  • Differential element arc length $dL = r\,d\theta$; element position $x = r\cos(\theta)$.

  • Total arc length $L = 2\alpha r$.

  • First moment integral:     $L \bar{x} = \int_{-\alpha}^{\alpha} (r\cos(\theta))(r\,d\theta) = 2 r^2 \sin(\alpha)$

  • Final centroid coordinate equation:     $\bar{x} = \frac{r\sin(\alpha)}{\alpha}$

  • Semicircular arc ($2\alpha = \pi$):     $\bar{x} = \frac{2r}{\pi}$

• Sample Problem 5/2: Centroid of a Triangular Area

  • Altitude $h$, base $b$ aligned along $x$-axis.

  • Horizontal differential strip element $dA = x\,dy$.

  • By similar triangles:     $\frac{x}{h-y} = \frac{b}{h} \implies x = \frac{b(h-y)}{h}$

  • Total area $A = \frac{1}{2}bh$.

  • Element centroid moment arm $y_e = y$.

  • First moment integral calculation:     $A \bar{y} = \int_0^h y \left(\frac{b(h-y)}{h}\right) dy = \frac{b h^2}{6}$

  • Final centroid location:     $\bar{y} = \frac{h}{3}$

  • Geometric property: Centroid lies at the intersection of medians, positioned at one-third altitude from any base.

• Sample Problem 5/3: Centroid of Circular Sector Area

  • Sector radius $r$, total included angle $2\alpha$, symmetrical about $x$-axis ($\bar{y} = 0$).

  • Solution I (Concentric Partial Rings):

  • Differential ring radius $r_0$, thickness $dr_0$, area $dA = 2 r_0 \alpha\,dr_0$

  • Arc element centroid coordinate $x_e = \frac{r_0 \sin(\alpha)}{\alpha}$

  • Total area $A = \alpha r^2$

  • Integration:       $A \bar{x} = \int_0^r \left(\frac{r_0 \sin(\alpha)}{\alpha}\right) (2 r_0 \alpha\,dr_0) = \frac{2}{3} r^3 \sin(\alpha)$       $\bar{x} = \frac{2 r \sin(\alpha)}{3 \alpha}$

  • Solution II (Triangular Elements):

  • Differential triangle area $dA = \frac{1}{2} r^2\,d\theta$

  • Triangle element centroid coordinate $x_e = \frac{2}{3} r \cos(\theta)$

  • Integration over $\theta \in [-\alpha, \alpha]$ produces identical result:       $\bar{x} = \frac{2 r \sin(\alpha)}{3 \alpha}$

  • Semicircular area ($2\alpha = \pi$):     $\bar{x} = \frac{4r}{3\pi}$

• Sample Problem 5/4: Centroid of Area Under Curve $x = k y^3$ from $x=0$ to $x=a$

  • Curve passes through $(a, b)$, yielding $k = \frac{a}{b^3}$, or $y = b \left(\frac{x}{a}\right)^{1/3}$.

  • Solution I (Vertical Element $dA = y\,dx$):

  • Total area calculation:       $A = \int_0^a y\,dx = \int_0^a b \left(\frac{x}{a}\right)^{1/3} dx = \frac{3}{4} ab$

  • Centroid coordinate $\bar{x}$ ($x_e = x$):       $A \bar{x} = \int_0^a x y\,dx = \frac{3}{7} a^2 b \implies \bar{x} = \frac{4}{7} a$

  • Centroid coordinate $\bar{y}$ ($y_e = \frac{y}{2}$):       $A \bar{y} = \int_0^a \left(\frac{y}{2}\right) y\,dx = \frac{1}{2} \int_0^a b^2 \left(\frac{x}{a}\right)^{2/3} dx = \frac{3}{10} a b^2 \implies \bar{y} = \frac{2}{5} b$

  • Solution II (Horizontal Element $dA = (a - x)\,dy$):

  • Element centroid moment arm $x_e = x + \frac{a-x}{2} = \frac{a+x}{2}$; $y_e = y$.

  • Integration yields identical coordinates: $\bar{x} = \frac{4}{7} a$ and $\bar{y} = \frac{2}{5} b$.

• Sample Problem 5/5: Hemispherical Volume Centroid

  • Radius $r$, base on $x-z$ plane ($\bar{x} = 0, \bar{z} = 0$). Surface equation: $y^2 + z^2 = r^2$.

  • Total hemisphere volume V = \frac{2}{3} \pi r^3$.\n - *Solution I (Circular Slice Slices Parallel to x-z plane):*\n - Slice thickness dy$, radius$z = \sqrt{r^2 - y^2}$, volume element$dV = \pi z^2\,dy = \pi (r^2 - y^2)\,dy.\n - Element centroid coordinate y_e = y$.

  • Moment integral:       $V \bar{y} = \int_0^r y \pi (r^2 - y^2)\,dy = \frac{\pi r^4}{4}$

  • Result:       $\bar{y} = \frac{\frac{\pi r^4}{4}}{\frac{2}{3} \pi r^3} = \frac{3}{8} r$

  • Solution II (Cylindrical Shell Elements):

  • Length $y = \sqrt{r^2 - z^2}$, radius $z$, thickness $dz$.

  • Volume element $dV = 2\pi z y\,dz = 2\pi z \sqrt{r^2 - z^2}\,dz$.

  • Element centroid coordinate $y_e = \frac{y}{2}$.

  • Integrates to identical result: $\bar{y} = \frac{3}{8} r$.

  • Solution III (Angle Parameter Integration):

  • Differential angle $\theta$ from $0$ to $\frac{\pi}{2}$.

  • Slice radius $r \sin(\theta)$, thickness $dy = (r\,d\theta) \sin(\theta)$, shell length $y = r \cos(\theta)$.

Composite Bodies, Composite Figures, and Approximation Methods

• Composite Body Principle of Moments:

  • Bodies constructed from standard geometric parts have mass centers determined by summation rather than integration.

  • Scalar composite mass center equations (Equations 5/7):     $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$     $\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$     $\bar{z} = \frac{\sum m_i z_i}{\sum m_i}$

  • Equivalent relations apply to composite lines ($L$), areas ($A$), and volumes ($V$).

  • Cutouts, holes, and cavities are represented as components with negative mass, area, or volume.

• Approximation Methods for Irregular Boundaries:

  • Irregular Area Approximation: Area is divided into strips of width $\Delta x$ and height $h$. Strip area $\Delta A = h \Delta x$.     $\bar{x} = \frac{\sum x_e \Delta A}{\sum \Delta A}$     $\bar{y} = \frac{\sum y_e \Delta A}{\sum \Delta A}$

  • Irregular Volume Approximation: Cross-sectional areas $A$ normal to $x$ are plotted versus $x$. Strip volume element \Delta V = A \Delta x$.\n    \bar{x} = \frac{\sum x \Delta V}{\sum \Delta V} = \frac{\sum x A \Delta x}{\sum A \Delta x}\n\n- **Sample Problem 5/6: Composite Area Centroid Calculation**\n - *Part 1 (Rectangle 12\,\text{in.} \times 10\,\text{in.}):*\n - Area A_1 = 120\,\text{in.}^2\n - Centroid coordinates: x_1 = 6\,\text{in.}$,$y_1 = 5\,\text{in.}\n - Moments: A_1 x_1 = 720\,\text{in.}^3$,$A_1 y_1 = 600\,\text{in.}^3\n - *Part 2 (Right Triangle, Base 6\,\text{in.}$, Height$10\,\text{in.}$, attached at$x=12\,\text{in.}$,$y=0):*\n - Area A_2 = 30\,\text{in.}^2\n - Centroid coordinates: x_2 = 12 + \frac{1}{3}(6) = 14\,\text{in.}$,$y_2 = \frac{1}{3}(10) = \frac{10}{3}\,\text{in.} \n - Moments: A_2 x_2 = 420\,\text{in.}^3$,$A_2 y_2 = 100\,\text{in.}^3\n - *Part 3 (Circular Hole Cutout):*\n - Area A_3 = -14.14\,\text{in.}^2\n - Centroid coordinates: x_3 = 6\,\text{in.}$,$y_3 = 1.273\,\text{in.}\n - Moments: A_3 x_3 = -84.8\,\text{in.}^3$,$A_3 y_3 = -18\,\text{in.}^3\n - *Part 4 (Triangular Hole Cutout):*\n - Area A_4 = -8\,\text{in.}^2\n - Centroid coordinates: x_4 = 12\,\text{in.}$,$y_4 = 4\,\text{in.}\n - Moments: A_4 x_4 = -96\,\text{in.}^3$,$A_4 y_4 = -32\,\text{in.}^3\n - *Summation Totals:*\n - Total Area \sum A = 120 + 30 - 14.14 - 8 = 127.9\,\text{in.}^2\n - Total First Moment in x$:$\sum A x = 720 + 420 - 84.8 - 96 = 959\,\text{in.}^3\n - Total First Moment in y$:$\sum A y = 600 + 100 - 18 - 32 = 650\,\text{in.}^3\n - *Final Centroid Coordinates:*\n    \bar{x} = \frac{959}{127.9} = 7.50\,\text{in.}\n    \bar{y} = \frac{650}{127.9} = 5.08\,\text{in.}\n\n- **Sample Problem 5/7: Volume Centroid Approximation of an Irregular Body**\n - Body total length L = 1\,\text{m}$, divided into five equal intervals$\Delta x = 0.2\,\text{m}:\n - Interval 1 (0 - 0.2\,\text{m}$): Average Area$A_{av} = 3\,\text{m}^2$, Volume$V_1 = 0.6\,\text{m}^3$,$x_1 = 0.1\,\text{m}$,$V_1 x_1 = 0.060\,\text{m}^4\n - Interval 2 (0.2 - 0.4\,\text{m}$): Average Area$A_{av} = 4.5\,\text{m}^2$, Volume$V_2 = 0.90\,\text{m}^3$,$x_2 = 0.3\,\text{m}$,$V_2 x_2 = 0.270\,\text{m}^4\n - Interval 3 (0.4 - 0.6\,\text{m}$): Average Area$A_{av} = 5.2\,\text{m}^2$, Volume$V_3 = 1.04\,\text{m}^3$,$x_3 = 0.5\,\text{m}$,$V_3 x_3 = 0.520\,\text{m}^4\n - Interval 4 (0.6 - 0.8\,\text{m}$): Average Area$A_{av} = 5.2\,\text{m}^2$, Volume$V_4 = 1.04\,\text{m}^3$,$x_4 = 0.7\,\text{m}$,$V_4 x_4 = 0.728\,\text{m}^4\n - Interval 5 (0.8 - 1.0\,\text{m}$): Average Area$A_{av} = 4.5\,\text{m}^2$, Volume$V_5 = 0.90\,\text{m}^3$,$x_5 = 0.9\,\text{m}$,$V_5 x_5 = 0.810\,\text{m}^4\n - *Summation Totals:*\n - Total Volume \sum V = 4.48\,\text{m}^3\n - Total Moment \sum V x = 2.388\,\text{m}^4\n - *Final Centroid Coordinate:*\n    \bar{x} = \frac{2.388}{4.48} = 0.533\,\text{m}\n\n- **Sample Problem 5/8: Mass Center of Bracket-and-Shaft Combination**\n - Material properties:\n - Sheet metal vertical face mass density = 25\,\text{kg/m}^2\n - Sheet metal horizontal base mass density = 40\,\text{kg/m}^2\n - Steel shaft mass density = 7.83\,\text{Mg/m}^3 = 7830\,\text{kg/m}^3\n - Symmetry simplification: \bar{x} = 0\n - *Part 1 (Semicircular Sheet Metal Top Plate, r = 50\,\text{mm}):*\n - Mass m_1 = \frac{1}{2}\pi(0.05)^2 \times 25 = 0.098\,\text{kg}\n - Centroid: y_1 = 0\,\text{mm}$,$z_1 = \frac{4(50)}{3\pi} = 21.2\,\text{mm}\n - Moments: m_1 y_1 = 0\,\text{kg}\cdot\text{mm}$,$m_1 z_1 = 2.08\,\text{kg}\cdot\text{mm}\n - *Part 2 (Vertical Rectangular Sheet Metal Plate, 100\,\text{mm} \times 150\,\text{mm}):*\n - Mass m_2 = (0.10)(0.15) \times 25 = 0.562\,\text{kg}\n - Centroid: y_2 = 0\,\text{mm}$,$z_2 = -75.0\,\text{mm}\n - Moments: m_2 y_2 = 0\,\text{kg}\cdot\text{mm}$,$m_2 z_2 = -42.19\,\text{kg}\cdot\text{mm}\n - *Part 3 (Triangular Sheet Metal Cutout, Base 100\,\text{mm}$, Height$75\,\text{mm}, negative mass):*\n - Mass m_3 = -\frac{1}{2}(0.075)(0.10) \times 25 = -0.094\,\text{kg}\n - Centroid: y_3 = 0\,\text{mm}$,$z_3 = -[150 - 25 - \frac{1}{3}(75)] = -100.0\,\text{mm}\n - Moments: m_3 y_3 = 0\,\text{kg}\cdot\text{mm}$,$m_3 z_3 = 9.38\,\text{kg}\cdot\text{mm}\n - *Part 4 (Horizontal Base Sheet Metal Plate, 100\,\text{mm} \times 150\,\text{mm}):*\n - Mass m_4 = (0.10)(0.15) \times 40 = 0.600\,\text{kg}\n - Centroid: y_4 = 50.0\,\text{mm}$,$z_4 = -150.0\,\text{mm}\n - Moments: m_4 y_4 = 30.0\,\text{kg}\cdot\text{mm}$,$m_4 z_4 = -90.00\,\text{kg}\cdot\text{mm}\n - *Part 5 (Solid Steel Shaft, r = 25\,\text{mm}$, Length$100\,\text{mm}):*\n - Volume V = \pi(0.025)^2(0.10) = 1.963 \times 10^{-4}\,\text{m}^3\n - Mass m_5 = (1.963 \times 10^{-4}) \times 7830 = 1.476\,\text{kg}\n - Centroid: y_5 = 75.0\,\text{mm}$,$z_5 = 0\,\text{mm}\n - Moments: m_5 y_5 = 110.7\,\text{kg}\cdot\text{mm}$,$m_5 z_5 = 0\,\text{kg}\cdot\text{mm}\n - *Summation Totals:*\n - Total Mass \sum m = 0.098 + 0.562 - 0.094 + 0.600 + 1.476 = 2.642\,\text{kg}\n - Total First Moment in y$:$\sum m y = 0 + 0 + 0 + 30.0 + 110.7 = 140.7\,\text{kg}\cdot\text{mm}\n - Total First Moment in z$:$\sum m z = 2.08 - 42.19 + 9.38 - 90.00 + 0 = -120.73\,\text{kg}\cdot\text{mm}\n - *Final Mass Center Coordinates:*\n    \bar{y} = \frac{140.7}{2.642} = 53.3\,\text{mm}\n    \bar{z} = \frac{-120.73}{2.642} = -45.7\,\text{mm}$</p></li></ul></li></ul><h4>Overview and Classification of Distributed Forces - <strong>Fundamental Concept of Concentrated vs. Distributed Forces:</strong> - Forces are treated as concentrated along their lines of action and at their points of application when analyzing external effects on bodies as a whole. - Exact physical concentrated forces do not exist because every external force applied mechanically is distributed over a finite contact area. - When the dimension$b$of a contact area is negligible compared to other pertinent body dimensions (such as wheel-to-wheel distance), actual distributed contact forces are replaced by a concentrated resultant force$R$. - <strong>Physical Examples of Contact Force Distributions:</strong> - <em>Automobile Tire on Pavement:</em> The force exerted by pavement on a tire is distributed over a finite contact area, which expands if the tire is soft. Replaced by a single concentrated resultant$R$for whole-vehicle equilibrium analysis. - <em>Ball Bearings:</em> Contact between a hardened steel ball and its race occurs over an extremely small, yet finite, contact area. - <em>Truss Pin Joints:</em> Pin forces acting on a two-force member are distributed over the contact surface between the pin and the hole, as well as across internal cut sections. - <em>Structural Engineering Application:</em> The Gateshead Millennium Bridge spanning the River Tyne in the United Kingdom rotates about a horizontal axis along its span to allow ship passage; its design required determining the cumulative effect of its weight distribution across a range of rotational orientations. - <strong>Internal Force Distributions vs. External Body Analysis:</strong> - When analyzing internal stress and strain distributions near contact locations, forces cannot be modeled as concentrated; actual distributed loading must be evaluated using material properties from theories of elasticity and plasticity. - Forces applied over regions non-negligible in size compared to overall body dimensions require mathematical integration of force intensities across the entire contact domain. - <strong>Three Primary Categories of Distributed Forces:</strong> - <strong>Line Distribution:</strong> - Occurs when a force is distributed along a line, such as a continuous vertical load supported by a suspended cable. - Loading intensity$w$is expressed as force per unit length. - SI units: newtons per meter ($ ext{N/m}$). - U.S. Customary units: pounds per foot ($ ext{lb/ft}$). - <strong>Area Distribution:</strong> - Occurs when a force acts over a surface area, such as hydraulic water pressure against the inner face of a dam. - Intensity is termed <em>pressure</em> for fluid forces and <em>stress</em> for internal solid forces. - Basic SI unit: pascal ($ ext{Pa}$), defined as one newton per square meter ($ ext{N/m}^2$). - Conversion factor:$6895 \, ext{Pa} = 1 \, ext{lb/in}^2$- Practical SI scales: kilopascal ($ ext{kPa} = 10^3 \, ext{Pa}$) for fluid pressures; megapascal ($ ext{MPa} = 10^6 \, ext{Pa}$) for structural stress. - Practical U.S. Customary unit: pounds per square inch ($ ext{lb/in}^2$). - <strong>Volume Distribution (Body Forces):</strong> - Forces distributed over the entire volume of a body are classified as body forces. - Most common body force: gravitational attraction acting on every differential mass element in a body (e.g., heavy cantilevered structures). - Gravitational force intensity is expressed by specific weight$

    ho g$, where$
    ho$is mass density (mass per unit volume) and$g$is acceleration due to gravity. - SI units for specific weight:$( ext{kg/m}^3)( ext{m/s}^2) = ext{N/m}^3$- U.S. Customary units for specific weight: pounds per cubic foot ($ ext{lb/ft}^3$) or pounds per cubic inch ($ ext{lb/in}^3$).</p><h4>Principles of Mass Center and Center of Gravity - <strong>Center of Gravity Definition and Physical Concept:</strong> - Suspending a body of mass$m$from various support points (e.g., points$A$,$B$,$C$) yields equilibrium under cord tension and the resultant weight force$W$. - The line of action of the resultant weight force$W$is collinear with the suspension cord. - Assuming a uniform, parallel gravitational field, all lines of action intersect at a unique point$G$, defined as the <em>center of gravity</em>. - In exact physical reality, gravity forces converge toward Earth's center of attraction and vary with distance, meaning lines of action are slightly non-concurrent; however, for bodies small relative to Earth, the uniform parallel field assumption holds perfectly. - <strong>Mathematical Derivation via Moment Principle:</strong> - The moment of the resultant gravitational force$W$about any axis equals the sum of moments of the infinitesimal weight elements$dW about that same axis. - Total weight equation: W = extstyle rac{1}{2} imes ext{Area} imes ext{Height} - Moment equality about coordinate axes: ar{x} W = extstyle rac{1}{2} imes ext{Area} imes ext{Height} - Scalar coordinate equations for center of gravity (Equations 5/1a): ar{x} = rac{ extstyle rac{1}{2} imes ext{Area}}{W} ar{y} = rac{ extstyle rac{1}{2} imes ext{Area}}{W} ar{z} = rac{ extstyle rac{1}{2} imes ext{Area}}{W}$- Substituting$W = mg$and$dW = g imes dm$eliminates gravitational acceleration$g$, yielding center of mass equations (Equations 5/1b):$ar{x} = rac{ extstyle rac{1}{2} imes ext{Area} imes dm}{ extstyle rac{Area}} ar{y} = rac{ extstyle rac{1}{2} imes ext{Area} imes dm}{ extstyle rac{Area}} ar{z} = rac{ extstyle rac{1}{2} imes ext{Area} imes dm}{ extstyle rac{Area}}$$

    Tips 1. Visual aids can enhance understanding of force distributions and mass centers—consider incorporating diagrams where possible. 2. Use color coding to differentiate between various types of distributed forces (line, area, volume)—this can facilitate quicker comprehension during study sessions. 3. When exploring concepts of centroid, drawing real-life applications like bridges or beams can provide context and clarity.