Sources and References

These notes synthesize material from several standard engineering mechanics texts. The primary reference is R.C. Hibbeler, Engineering Mechanics: Statics (15th edition), which the course follows closely. Supplementary derivations and alternative presentations draw from Beer, Johnston, Mazurek, and Cornwell, Vector Mechanics for Engineers: Statics, and from Meriam, Kraige, and Bolton, Engineering Mechanics: Statics. Conceptual framing, especially around free-body thinking and internal forces, is informed by the MIT 2.001 Mechanics and Materials I notes, the Stanford CEE 121 Statics course materials, and the Cambridge Engineering Part IA Mechanics notes. Classical treatments of friction and structural analysis also benefit from Crandall, Dahl, and Lardner, An Introduction to the Mechanics of Solids.

Chapter 1: Foundations of Mechanics

Statics is the branch of classical mechanics that studies bodies at rest or in uniform motion under the action of forces. Although the subject rests on only a handful of postulates, mastering it is the prerequisite for every downstream topic in engineering: dynamics, deformable-body mechanics, structural analysis, machine design, and finite-element modelling. The whole edifice is built on Newton’s three laws together with the law of gravitation, specialized to the case where the acceleration of every particle is zero.

Fundamental Concepts

Mechanics treats four primitive quantities: length, time, mass, and force. In the Newtonian framework, length, time, and mass are taken as independent base quantities and force is a derived quantity defined by Newton’s second law, \( \mathbf{F} = m\mathbf{a} \). An alternative (common in statics) treats force as a primitive and mass as derived. Both viewpoints are consistent for our purposes.

A particle is an idealization: a body whose dimensions are small enough that its geometry does not affect the problem. A rigid body is a body in which the distance between any two points is constant — deformations are ignored. Statics works almost exclusively with these two idealizations.

Newton’s laws, stated for an inertial frame, read:

1. A particle at rest remains at rest, and a particle in uniform rectilinear motion continues in that motion, unless acted on by an unbalanced force.
2. The resultant force on a particle equals the time rate of change of its linear momentum; for constant mass, \( \mathbf{F} = m\mathbf{a} \).
3. To every action there is an equal and opposite reaction: forces between two bodies are collinear, equal in magnitude, and oppositely directed.

For statics, \( \mathbf{a}=\mathbf{0} \), so the resultant of all forces acting on a particle must vanish.

Units and Numerical Practice

The SI base units used throughout are the metre, kilogram, and second, with the newton \( (1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2) \) derived. Engineers frequently report numbers using three significant figures; the underlying data rarely justify more, and carrying spurious digits obscures the dominant uncertainty. Consistent unit checking is the cheapest form of error detection available, and you should form the habit of carrying units through every calculation.

Chapter 2: Force Vectors

Vectors, Scalars, and the Parallelogram Law

A scalar is a quantity characterized entirely by a real magnitude: mass, temperature, speed, energy. A vector has both magnitude and direction and obeys the parallelogram (or equivalently, triangle) rule of addition. Force, position, velocity, and moment are vectors.

Two forces \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \) acting at the same point combine into a single resultant \( \mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 \) whose magnitude and direction are given by the diagonal of the parallelogram they span. When the geometry is awkward, the law of cosines and law of sines give the magnitude and orientation directly. For forces \( F_1 \) and \( F_2 \) separated by an interior angle \( \theta \) (measured between their tails),

\[ R = \sqrt{F_1^{2} + F_2^{2} - 2 F_1 F_2 \cos(180^{\circ} - \theta)}. \]

Cartesian Representation

Although the parallelogram law is conceptually fundamental, computation is easier in components. Adopt a right-handed Cartesian basis \( (\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}) \); any vector \( \mathbf{A} \) is written

\[ \mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}, \]

with magnitude \( \left|\mathbf{A}\right| = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}} \) and direction cosines \( \cos\alpha = A_x/|\mathbf{A}| \), etc. The direction cosines satisfy \( \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1 \).

Addition of vectors in Cartesian form is trivial — add components — and this is why components dominate practical work. A unit vector along a line from point \( A \) to point \( B \) is \( \hat{\mathbf{u}}_{AB} = (\mathbf{r}_B - \mathbf{r}_A)/\left|\mathbf{r}_B - \mathbf{r}_A\right| \); multiplying by a scalar magnitude \( F \) produces a force directed along that line.

The Dot Product

The scalar product of two vectors is

\[ \mathbf{A} \cdot \mathbf{B} = \left|\mathbf{A}\right|\left|\mathbf{B}\right| \cos\theta = A_x B_x + A_y B_y + A_z B_z, \]

where \( \theta \) is the angle between them. It is the workhorse for two recurring tasks: finding the angle between two vectors, and projecting one vector onto the direction of another. The scalar projection of \( \mathbf{A} \) onto a unit vector \( \hat{\mathbf{u}} \) is \( A_u = \mathbf{A}\cdot\hat{\mathbf{u}} \), and the vector projection is \( \mathbf{A}_u = (\mathbf{A}\cdot\hat{\mathbf{u}})\,\hat{\mathbf{u}} \). This projection construction will reappear when we compute the moment of a force about a specified axis.

Chapter 3: Equilibrium of a Particle

A particle is in equilibrium when the vector sum of all forces on it is zero. This yields three scalar equations in three dimensions, or two in the plane. The art lies in two steps: drawing the free-body diagram (FBD) correctly, and choosing a coordinate system that makes the equations easy to solve.

Free-Body Diagrams

An FBD isolates the particle from its surroundings and shows every force acting on it — applied loads, cable tensions, spring forces, contact reactions, gravity. Any force that does not cross the imaginary boundary around the isolated particle does not appear. Drawing a clean FBD is not artistic polish; it is the entire conceptual content of the problem.

Planar and Three-Dimensional Problems

In two dimensions the equilibrium equations are \( \sum F_x = 0 \) and \( \sum F_y = 0 \). A good coordinate choice often aligns one axis with an unknown force, so that unknown appears in only one equation. In three dimensions \( \sum F_x = 0 \), \( \sum F_y = 0 \), \( \sum F_z = 0 \). You almost always write each force in Cartesian components using the position-vector construction of Chapter 2, then sum components.

Chapter 4: Force System Resultants

The Moment of a Force

When a force acts on a rigid body along a line that does not pass through a chosen reference point \( O \), it tends to rotate the body about \( O \). This rotational tendency is the moment of the force.

The cross product \( \mathbf{r}\times\mathbf{F} \) can be expanded via the determinant

\[ \mathbf{r}\times\mathbf{F} = \left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ r_x & r_y & r_z \\ F_x & F_y & F_z\end{array}\right|. \]

In two dimensions the moment collapses to the scalar \( M = x F_y - y F_x \), positive counterclockwise by convention.

Varignon’s theorem (the principle of moments) states that the moment of a force equals the sum of moments of its components. This is nothing more than the distributive law of the cross product over addition, but it is enormously useful: you can replace an inconvenient force by convenient components and compute moments from each separately.

Moment About an Axis

Sometimes we want only the component of \( \mathbf{M}_O \) along a particular line — for example, the twisting effect a force produces about a shaft. If \( \hat{\mathbf{u}}_a \) is a unit vector along the axis passing through \( O \),

\[ M_a = \hat{\mathbf{u}}_a \cdot (\mathbf{r}\times\mathbf{F}), \]

written as a scalar triple product. This extracts the projection of \( \mathbf{M}_O \) onto the axis and discards components perpendicular to it.

Couples

A couple is a pair of equal, opposite, non-collinear forces. Its resultant force is zero, but it produces a moment that is independent of the reference point — a couple is a free vector. The couple moment has magnitude \( F\,d \) where \( d \) is the perpendicular distance between the two lines of action. Couples are the pure rotational actuators of statics.

Equivalent Force-Couple Systems

Any system of forces and couples acting on a rigid body can be replaced by a single force \( \mathbf{R} = \sum \mathbf{F}_i \) acting at a chosen point \( O \), together with a couple moment \( \mathbf{M}_O^{R} = \sum \mathbf{M}_{i/O} \) equal to the sum of moments of the original forces about \( O \) plus any couples already present. Two systems are equivalent if both \( \mathbf{R} \) and \( \mathbf{M}_O^{R} \) match for some (hence every) point \( O \).

In two dimensions a non-zero force system can often be reduced further to a single resultant force with no couple, by sliding the force to a new line of action whose perpendicular distance from \( O \) is \( d = M_O^{R}/R \). In three dimensions the general reduction is a wrench: a force aligned with a moment.

Distributed Loads

A load spread along a line, such as a beam under its own weight, is described by an intensity \( w(x) \) in newtons per metre. It is equivalent to a single concentrated force whose magnitude equals the area under \( w(x) \) and whose line of action passes through the centroid of that area:

\[ F_R = \int_{L} w(x)\,dx, \qquad \bar{x} = \frac{1}{F_R}\int_{L} x\,w(x)\,dx. \]

For common shapes the results memorize: a uniform load of intensity \( w_0 \) over length \( L \) reduces to \( w_0 L \) at the midpoint; a linearly varying (triangular) load from \( 0 \) to \( w_0 \) reduces to \( \tfrac{1}{2} w_0 L \) acting at \( \tfrac{2}{3} L \) from the zero end.

Chapter 5: Equilibrium of Rigid Bodies

For a rigid body in equilibrium, both the resultant force and resultant moment (about any point) must vanish. In planar problems this gives three scalar equations, in three-dimensional problems six. The central task is again drawing the correct FBD: identifying supports, replacing them with the reactions they can transmit, and accounting for every applied load and couple.

Support Reactions

Typical planar supports and the reactions they impose:

Support type	Reactions transmitted
Roller, rocker, smooth surface	One normal force
Pin or hinge	Two components of force (no couple)
Fixed (built-in)	Two force components and a couple
Smooth collar on rod	One normal force
Short link (two-force member)	Force along the link

Three-dimensional supports generalize these: a ball-and-socket joint transmits three force components, a journal bearing two forces and two couples (typically), and a thrust bearing adds one more force. The key rule is that a support can transmit only those actions it is kinematically capable of preventing.

Two-Force and Three-Force Members

The proof is immediate: force balance makes them equal and opposite; moment balance about either point forces their common line of action to pass through the other point. Truss members, idealized as pin-connected rods with loads applied only at the joints, are the canonical example. Identifying a two-force member up front often slashes the number of unknowns in a problem.

A three-force member in equilibrium requires that the three forces be either concurrent (lines of action meet at one point) or all parallel. This follows from moment balance taken about the intersection of two of them.

Statical Determinacy

A planar rigid body has three degrees of freedom; three independent reaction components can hold it in equilibrium. A system with more unknown reactions than independent equations is statically indeterminate — rigid-body equilibrium alone cannot solve it, and additional information about deformation is required. Conversely, a system with too few reactions is a mechanism and cannot hold static loads. A subtler failure is improper constraint: reactions that, although numerically sufficient, are arranged so that one equilibrium equation is automatically satisfied and another cannot be, for example three parallel reactions or three reactions whose lines of action meet at a common point.

Chapter 6: Structural Analysis

Plane Trusses

A truss is an assembly of slender members connected at their ends by frictionless pins and loaded only at those joints. Under these idealizations each member is a two-force member carrying pure tension or pure compression along its axis. The approximation is accurate for bolted or welded joints whose rigidity is small compared to member slenderness.

A truss is statically determinate and stable if \( m + r = 2j \), where \( m \) is the number of members, \( r \) the number of reaction components, and \( j \) the number of joints. Satisfying this count is necessary but not sufficient; geometric arrangement also matters.

Method of Joints

Isolate each joint and apply particle equilibrium. Begin at a joint with no more than two unknown member forces, solve for them from \( \sum F_x = 0 \) and \( \sum F_y = 0 \), then march through the truss joint by joint. A useful convention: assume every member is in tension (arrows pull away from the joint on its FBD). A negative result reveals the member is actually in compression. Members whose forces are immediately found to be zero are called zero-force members; they are not structurally useless — they prevent buckling and maintain geometry — but they carry no load under the current loading.

Method of Sections

When only a few member forces are wanted, slicing the truss with an imaginary cut and treating one side as a rigid body is faster than marching through every joint. Choose a section crossing no more than three members with unknown forces, draw the FBD of the severed piece, and solve the three equilibrium equations. Picking moment centres cleverly — at the intersection of two unknown members — often isolates the third unknown in a single equation.

Frames and Machines

A frame is a stationary structure containing at least one multi-force member (a member loaded at more than two points). A machine is a similar assembly that moves or transmits forces to do work. Members here are no longer in pure tension or compression, so the method of joints does not apply. Instead, one dismembers the structure and draws FBDs of individual components, invoking Newton’s third law at each connection: the force that member \( A \) exerts on member \( B \) is equal and opposite to the force \( B \) exerts on \( A \). The resulting system of equilibrium equations is solved simultaneously.

Chapter 7: Internal Forces in Beams

Slicing a beam with an imaginary cut and examining the FBD of one side reveals the internal forces transmitted across the section: in a planar problem these are a normal force \( N \), a shear \( V \), and a bending moment \( M \). They are exactly those resultants that keep the isolated piece in equilibrium.

Sign Conventions

A consistent sign convention is essential because shear and moment values are compared at different sections. A common choice, matching the Hibbeler text: at a cut, \( N \) is positive in tension, \( V \) is positive when it rotates the isolated segment clockwise, and \( M \) is positive when it tends to bend the segment concave upward (“sagging positive”). Under this convention gravity loads on simply supported beams produce positive bending moments throughout most of the span.

Shear and Moment Diagrams

Plotting \( V(x) \) and \( M(x) \) along the beam gives an immediate picture of where maximum shears and moments occur — the information a designer needs to size the beam. Two approaches: piecewise equations, where one writes \( V(x) \) and \( M(x) \) as formulas on intervals between loads, and graphical integration, which builds up diagrams directly from the differential relations

\[ \frac{dV}{dx} = -w(x), \qquad \frac{dM}{dx} = V(x). \]

These say that the slope of the shear diagram equals the negative of distributed load intensity, and the slope of the moment diagram equals the shear. Integrated: the change in shear between two sections equals the negative of the area under the load diagram between them, and the change in moment equals the area under the shear diagram. At a concentrated force, \( V \) jumps by that force; at a concentrated couple, \( M \) jumps by that couple. Maximum moment occurs where \( V = 0 \) (or at a discontinuity in \( V \)).

Chapter 8: Friction

When two dry surfaces are in contact, tangential forces can be transmitted up to a limit set by the normal force and the material pair. The Coulomb (dry friction) model is simple and remarkably effective.

Laws of Dry Friction

The friction force \( F \) at a contact is tangential to the contact surface and opposes relative motion (or impending motion). In the static regime, \( F \) takes whatever value is needed to prevent sliding, up to a maximum \( F_s = \mu_s N \), where \( N \) is the normal force and \( \mu_s \) is the coefficient of static friction. Once sliding begins, kinetic friction \( F_k = \mu_k N \) applies, with \( \mu_k \) typically smaller than \( \mu_s \). Both coefficients depend on surface pairing, cleanliness, and roughness, but not (to first approximation) on contact area or sliding speed.

The ratio \( \mu_s = \tan\phi_s \) defines the angle of friction \( \phi_s \); the resultant of \( N \) and \( F \) at impending slip lies along a cone of half-angle \( \phi_s \) about the surface normal. For a block on an inclined plane just on the verge of sliding down, the angle of the plane equals \( \phi_s \).

Wedges, Belts, and Screws

Wedges, often used to lift heavy loads by small forces, are analyzed by drawing FBDs of each block and applying equilibrium with friction at each contact surface in the direction opposing impending motion. Belt friction on a flat or V-pulley obeys the capstan equation, \( T_2/T_1 = e^{\mu\beta} \), where \( \beta \) is the angle of contact in radians — a small coefficient can hold a very large load because of the exponential. Square-threaded screws analyze as wedges wrapped around a cylinder, and self-locking screws are those for which the lead angle is less than the friction angle. These applications extend the same Coulomb model with the geometry of the device.

Chapter 9: Center of Gravity and Centroid

The centre of gravity of a body is the point through which the resultant gravitational force acts; the centroid is its purely geometric analogue, independent of mass density. For a body of uniform density in a uniform gravitational field, the two coincide. Coordinates are defined by first moments:

\[ \bar{x} = \frac{\int x\,dW}{\int dW}, \qquad \bar{y} = \frac{\int y\,dW}{\int dW}, \qquad \bar{z} = \frac{\int z\,dW}{\int dW}, \]

or for an area \( A \),

\[ \bar{x} = \frac{\int x\,dA}{A}, \qquad \bar{y} = \frac{\int y\,dA}{A}. \]

For bodies that can be subdivided into shapes with known centroids (rectangles, triangles, circles, semicircles), the composite-body method gives

\[ \bar{x} = \frac{\sum \bar{x}_i A_i}{\sum A_i}, \]

with analogous expressions for \( \bar{y}, \bar{z} \). Holes are handled by including them with negative area. Symmetry is a powerful shortcut: if a body has an axis or plane of symmetry, the centroid lies on it.

Chapter 10: Moments of Inertia

The second moment of area (area moment of inertia) about an axis in the plane of the area is

\[ I_x = \int y^{2}\,dA, \qquad I_y = \int x^{2}\,dA, \]

with units of length\( ^4 \). This quantity appears in the flexure formula \( \sigma = My/I \) that will feature heavily in later mechanics-of-materials courses, where bending stiffness scales with \( I \) and bending stress with \( M/I \). A larger \( I \) about the bending axis means a stiffer beam; placing area far from the neutral axis (as in an I-beam) is therefore structurally efficient.

Parallel-Axis Theorem

This is the reason composite-area calculations work: tabulate each sub-area’s centroidal moment of inertia, add \( A d^{2} \) for the offset of each sub-centroid from the reference axis, and sum. The radius of gyration \( k = \sqrt{I/A} \) packages this information into a length, representing the distance at which the entire area could be concentrated to produce the same \( I \). The product of inertia \( I_{xy} = \int xy\,dA \) vanishes whenever either axis is an axis of symmetry, and is needed for the transformation of inertia tensors and for finding principal axes.

Mass Moment of Inertia

For rotational dynamics (appearing in later courses on rigid-body motion) the mass moment of inertia about an axis is

\[ I = \int r^{2}\,dm, \]

where \( r \) is perpendicular distance from the axis to the mass element \( dm \). A parallel-axis theorem holds here as well: \( I = \bar{I} + m d^{2} \). Standard results — for example, \( \tfrac{1}{2} m R^{2} \) for a solid cylinder about its axis, \( \tfrac{2}{5} m R^{2} \) for a solid sphere through its centre — will recur throughout dynamics. The statics half of the story is the integration machinery and the composite-body method; the dynamics half will use these values to relate torque to angular acceleration.

Closing Remarks

The unifying thread across every chapter is the free-body diagram together with the two vector equilibrium equations \( \sum\mathbf{F} = \mathbf{0} \) and \( \sum\mathbf{M} = \mathbf{0} \). Everything else — truss analysis, beam diagrams, friction problems, centroids — is an application of this framework combined with appropriate idealizations about how different components transmit force. Develop fluency with the FBD first. Insist on drawing one for every sub-problem, even when it seems unnecessary. The habit repays itself many times over in Dynamics, Mechanics of Deformable Solids, and every engineering analysis course that follows.