5.1.1 The isolation philosophy

A free-body diagram is an explicit act of separation. The analyst selects a body or a system, cuts every physical connection to the surroundings, and replaces each cut by the force and couple that the removed part exerted on the retained part. Weight acts at the mass centre. Contact forces act at contact points. Internal forces between parts of the retained system cancel in pairs and never appear. What remains is a closed force picture that contains every external influence and nothing else. Correctness of all later algebra depends entirely on completeness of this picture. A missing force corrupts every equation that follows, while an invented force corrupts the solution with equal certainty. The discipline is therefore to draw first, with lines of action, senses, and labels, and to write equations only afterward.

Support idealizations follow one rule. Each reaction component replaces one kinematically restrained motion. A roller on a smooth surface restrains motion normal to the surface and supplies one normal reaction. A smooth pin restrains translation in the plane and supplies two force components, commonly resolved into horizontal and vertical parts. A fixed support restrains translation and rotation and supplies two force components plus one couple moment. A flexible cable supplies one tensile force along its length. A smooth contact supplies one normal force. These models are not arbitrary conventions. Each one expresses the constraint that the support permits certain displacements freely and resists the complementary displacements with reactions that do workless constraint forces in ideal theory.

5.1.2 Derivation of the equilibrium equations from Newton laws

Consider a rigid body of mass denoted by m, mass centre denoted by G, mass moment of inertia about G denoted by I sub G. Let the vector sum of all external forces be denoted by capital Sigma F, and let the sum of moments of all external forces about G be denoted by capital Sigma M sub G. Newton second law for translation of the mass centre and Euler law for rotation about the mass centre state

$$\sum \mathbf{F} = m\,\mathbf{a}_G, \qquad \sum \mathbf{M}_G = I_G\,\alpha\,\mathbf{k},$$

where a sub G is the acceleration of G, alpha is the angular acceleration, and k is the out-of-plane unit vector. These two vector relations govern all plane motion.

Static equilibrium is the special case of rest, continued rest, or uniform translation without rotation in an inertial frame. In the frame fixed to the ground, rest means vanishing velocity and vanishing acceleration of every material point. Hence a sub G equals zero and alpha equals zero. Substitution gives the vector equilibrium conditions

$$\sum \mathbf{F} = \mathbf{0}, \qquad \sum \mathbf{M}_G = 0.$$

Resolution of the force vector equation along two orthogonal in-plane axes x and y gives two scalar equations. The moment equation gives the third scalar equation. The familiar triple is therefore

$$\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0,$$

where O is any point in the plane, not necessarily G. The freedom to take moments about any point follows from the transfer theorem for moments. If the resultant force vanishes, the resultant moment is the same about every point, so vanishing of the moment about G implies vanishing about every O, and conversely. In practice the analyst chooses O at the intersection of two unknown reaction lines, so that both unknowns drop out of the moment equation and the third unknown stands alone. Moment arms must always be true perpendicular distances from O to the lines of action. The most persistent source of error is a moment arm measured along an inclined distance rather than its perpendicular projection.

Three scalar equations determine at most three unknowns. A body with more than three reaction components is externally indeterminate to first order and cannot be solved by equilibrium alone. Frames and machines with internal hinges are handled by dismantling the assembly into members, drawing a free-body diagram for each member with action-reaction pairs at the connections, and writing three equations per member. Determinacy is then judged on the assembled count.

The diagram above shows the canonical inclined-plane free body. Weight W acts vertically through the mass centre. Normal reaction N acts perpendicular to the plane. Friction F acts parallel to the plane, opposing the impending relative slip. The resultant contact force R is the vector sum of N and F and tilts away from the normal by the friction angle discussed in the next section.

5.2.1 The Coulomb model

Dry friction between clean metallic or rough nonlubricated surfaces is described by the Coulomb model. Let N be the normal contact force and F the tangential friction force. At rest with no tendency to slip, F adjusts itself to whatever value below the limiting value is needed for equilibrium. The limiting value at impending slip is proportional to N,

$$F_{\max} = \mu_s\,N,$$

where mu sub s is the coefficient of static friction. The inequality governing stick is therefore

$$F \le \mu_s\,N,$$

with equality only at the point of impending motion. Once gross sliding begins, the kinetic friction force is approximately

$$F_k = \mu_k\,N,$$

where mu sub k is the kinetic coefficient, generally somewhat smaller than mu sub s. Friction opposes relative motion or impending relative motion at the contact, never the absolute motion of the body. Direction must be assigned by asking which way the contact point would slip if friction vanished, then drawing F against that slip direction.

5.2.2 Angle of friction derived

At impending slip, the contact carries N normal to the surface and F equal to mu sub s times N along the surface. The resultant R therefore makes a definite angle with the normal. From the right triangle formed by N as adjacent side and F as opposite side,

$$\tan \phi = \frac{F}{N} = \frac{\mu_s\,N}{N} = \mu_s,$$

so that

$$\phi = \arctan \mu_s.$$

This angle phi is the angle of friction. The derivation shows that phi is not a new material constant. It is the same Coulomb information expressed as an angle, which makes graphical reasoning possible. A block on an incline of inclination theta has weight component W sin theta down the plane and W cos theta into the plane. Slip impends when the ratio of tangential to normal demand equals mu sub s, namely tan theta equals mu sub s. Hence the angle of repose, the steepest incline on which the block just holds, equals phi. This identity is a direct consequence of the tangent relation above.

5.2.3 Cone of friction

Fix the normal direction at a contact point. As the direction of impending slip rotates through a full circle in the tangent plane, the resultant R at limiting friction traces a right circular cone about the normal with half-angle phi. The interior of the cone is the stick region. If the resultant demanded by equilibrium lies strictly inside the cone, the required ratio F over N is below mu sub s and the contact sticks. If the demanded resultant lies on the cone wall, slip impends. If equilibrium would demand a resultant outside the cone, no static solution exists and slip or tip must occur. This geometric test replaces repeated inequality checks and explains why rough contacts tolerate oblique loads up to a definite obliquity but no further.

5.2.4 Screw thread mechanics and the self-lock condition derived

A square-threaded power screw is an inclined plane wrapped around a cylinder. Let d sub m be the mean thread diameter, p the pitch, and L the lead, with L equal to p for a single start and equal to the number of starts times p for a multiple start. Unwrap one turn of the helix into a right triangle whose base is the mean circumference pi times d sub m and whose height is the lead L. The helix angle alpha satisfies

$$\tan \alpha = \frac{L}{\pi\,d_m}.$$

Raising a load W against the thread is equivalent to pushing the load up the unwrapped incline of angle alpha with friction angle phi. Resolve forces on the load along and normal to the incline. Let P be the equivalent tangential effort at mean radius. Along the incline, P cos alpha must balance the downslope weight component W sin alpha plus the friction force, while the normal pressure follows from transverse equilibrium. Including the friction tilt through the standard inclined-plane analysis gives the compact force relation

$$P = W\,\tan(\alpha + \phi).$$

The torque about the screw axis is P times the mean radius d sub m over two, so the raising torque is

$$M_{\mathrm{raise}} = W\,\tan(\alpha + \phi)\,\frac{d_m}{2}.$$

Lowering the load reverses the sense of impending relative slip, so friction assists the holding effort rather than opposing the driving effort. The same derivation with the friction angle subtracted gives

$$M_{\mathrm{lower}} = W\,\tan(\phi - \alpha)\,\frac{d_m}{2}.$$

Self-locking means the load cannot drive the screw backward when the applied torque is removed. That requires the lowering torque to remain positive in the holding sense, so that a positive effort is still needed to lower the load. Positivity of tan of phi minus alpha for acute angles is equivalent to

$$\alpha \lt \phi.$$

When helix angle is smaller than friction angle, the thread holds without a brake. When helix angle exceeds friction angle, the screw overhauls and the load descends on its own. Thread efficiency for raising, defined as useful work W times L divided by input work two pi times M sub raise, simplifies to

$$\eta = \frac{\tan \alpha}{\tan(\alpha + \phi)},$$

and the self-lock inequality is equivalent to efficiency below one half. Large friction therefore guarantees holding but penalizes efficiency, which is the central design tradeoff of power screws.

Quick example — screw self-lock check: Single-start square thread with lead $L = 8$ mm, mean diameter $d_m = 40$ mm, friction $\mu_s = 0.15$.

Helix angle from $\tan\alpha = L/(\pi d_m) = 0.0637$, so $\alpha \approx 3.64^\circ$, and friction angle $\phi = \tan^{-1}(0.15) \approx 8.53^\circ$.

Since $\alpha \lt \phi$, the thread self-locks, with efficiency $\eta = \tan\alpha/\tan(\alpha + \phi) \approx 0.295 \lt 0.5$.

Trap: a double start doubles $L$ and can push $\alpha$ past $\phi$ into overhaul.

5.2.5 Wedge mechanics

A wedge of small included angle alpha converts a driving effort P into a large transverse force on a load W. In the frictionless ideal, horizontal and vertical equilibrium of the wedge and the load give

$$P = W\,\tan \alpha,$$

so the mechanical advantage W over P is the cotangent of alpha and grows large for thin wedges. Friction on each sliding face adds its friction angle to the effective wedge angle in the force triangle. With friction angles phi one and phi two on the two faces, the effort to drive the wedge forward involves tan of alpha plus phi one plus phi two in the exact force polygon, while the effort to hold or withdraw involves the difference of those angles. The wedge stays in place under the load alone, resisting being squeezed back out, when the included angle is smaller than the sum of the face friction angles,

$$\alpha \lt \phi_1 + \phi_2.$$

This is the wedge self-lock condition. It has the same structure as the screw condition, with the screw helix playing the role of the wedge angle and the thread flank friction playing the role of the face friction. Driving a wedge with an angle above this sum produces a reversible wedge that the load can expel, while an angle below the sum produces a locking wedge of the kind used in tool holders, clamps, and cotters.

5.2.6 Belt friction derived by the capstan differential element

Consider a flexible belt in contact with a drum over a contact angle beta. Let T one be the tight-side tension on the side where motion impends and T two the slack-side tension. Isolate a differential element subtending angle d beta at the drum centre. The element carries tension T on one end and T plus dT on the other, normal pressure dN directed radially inward, and limiting friction mu sub s times dN directed tangentially against the impending motion.

Radial equilibrium of the half-element gives dN equal to T times d beta to first order, because the two tension vectors each tilt by d beta over two from the tangent and their inward radial components sum to T times d beta with higher-order terms vanishing in the limit. Tangential equilibrium gives dT equal to mu sub s times dN at impending slip. Elimination of dN yields the differential relation

$$\frac{dT}{T} = \mu_s\,d\beta.$$

Integration from the slack side to the tight side over the full contact gives

$$\int_{T_2}^{T_1}\frac{dT}{T} = \mu_s\int_{0}^{\beta}d\beta, \qquad \ln\frac{T_1}{T_2} = \mu_s\,\beta,$$

hence the capstan formula

$$\frac{T_1}{T_2} = e^{\mu_s\,\beta}.$$

The contact angle beta in the exponent must be expressed in radians, because the step dN equals T times d beta uses arc length equal to radius times angle, which holds only in radian measure. Use of degrees inflates the exponent by a factor near fifty seven and produces an absurd dimensionless ratio larger by scores of orders of magnitude, which is the diagnostic signal of the error. The ratio grows exponentially with wrap, so modest extra wrap multiplies holding capacity dramatically, which is the principle behind rope capstans, belt drives, and band brakes.

Quick example — belt ratio: Slack tension $T_2 = 250$ N, friction $\mu_s = 0.30$, contact $\beta = \pi$ rad.

Capstan ratio gives $T_1/T_2 = e^{\mu_s\beta} = e^{0.3\pi} \approx 2.566$.

Tight tension is $T_1 = 250 \times 2.566 \approx 642$ N.

Trap: radians in the exponent — degrees would give nonsense.

5.3.1 Idealizations

A plane truss is an assembly of straight prismatic members joined by smooth pins at joints, loaded only at joints, with member weights neglected or lumped at the joints. Each member is then a two-force member. Forces act only at the two end pins, are equal, opposite, and collinear with the member axis, so each member carries pure axial force, tensile when pulling away from the joint and compressive when pushing into the joint. Bending and shear in members vanish by construction under these idealizations. Real trusses with gusset plates and distributed weight approximate this behavior when loads are applied at panel points and members are slender.

5.3.2 Determinacy count and proof sketch of m equals two j minus three

Let m be the number of members and j the number of joints including supports. Each joint in the plane supplies two independent equilibrium equations, sum of F sub x equals zero and sum of F sub y equals zero, for a total of two j equations. Unknowns consist of m member axial forces plus three global support reactions for a simply supported truss on a pin and a roller. Equating the number of independent equations to the number of unknowns for internal and external determinacy gives

$$2j = m + 3, \qquad m = 2j - 3.$$

This is the determinacy count for a simple internally determinate truss with minimal external supports. Fewer members indicate an internal mechanism, while more members indicate internal indeterminacy requiring compatibility of deformation beyond equilibrium. The count is necessary but not sufficient. A truss satisfying the count can still be instantaneously unstable if members are arranged with concurrent or parallel axes leaving a joint without stiffness in some direction, so geometry must also be inspected.

5.3.3 Zero-force rules proved

The zero-force rules follow directly from joint equilibrium and need no separate postulate.

First rule. Consider a joint connecting exactly two noncollinear members, with no external load and no support reaction. Resolve equilibrium along a direction perpendicular to one member. The component of the second member along that direction must vanish, and since the members are noncollinear that component is nonzero times the second member force, forcing the second force to zero. Repetition with the roles exchanged forces the first member to zero. Both members are therefore zero-force members. The proof uses only that the sine of the nonzero included angle cannot vanish.

Second rule. Consider a joint connecting exactly three members, two of which are collinear, with no external load and no support reaction. Resolve equilibrium perpendicular to the collinear pair. Only the noncollinear third member has a component in that direction, proportional to the sine of its inclination. Vanishing of the sum forces the third member force to zero. The two collinear members then balance each other through the remaining axial equation and carry equal and opposite joint forces, hence equal axial forces in the member sense. If an external load acts at the joint, neither rule applies, because the load supplies the missing component.

Quick example — zero-force truss joint: Joint with members AB along the x-axis and AC at $60^\circ$, no external load.

Transverse equilibrium gives $\sum F_y = F_{AC}\sin 60^\circ = 0$, so $F_{AC} = 0$.

Axial equilibrium then gives $\sum F_x = F_{AB} + F_{AC}\cos 60^\circ = 0$, so $F_{AB} = 0$.

Trap: any external load at the joint voids the rule, even a small one.

5.3.4 Method of joints derived

Method of joints applies the two joint equilibrium equations progressively through the truss. Starting from a joint with at most two unknown member forces, commonly a support joint with known reactions, the analyst solves sum of F sub x equals zero and sum of F sub y equals zero for the unknowns. Solved forces are transferred to adjacent joints with reversed sense by Newton third law, tension pulling away from every joint it connects and compression pushing into every joint. The procedure advances until all member forces are known. The method is systematic and best suited when most or all member forces are required. Its derivation is nothing beyond repeated application of particle equilibrium, with the two-force character of members supplying the line-of-action information.

5.3.5 Method of sections derived

Method of sections exposes interior forces by cutting the truss. Imagine a section line passing through at most three members of unknown force and separating the truss into two free bodies. Either segment is in equilibrium under the external loads on that segment plus the three cut member forces acting at the cut locations along the known member axes. Three equilibrium equations are available for the segment. By taking moments about the intersection point of two cut axes, the moments of those two unknown forces vanish identically, leaving one equation in the third force alone. Axial and transverse resolution supplies the remaining forces when needed. The power of the method lies in this choice of moment centre. It is best suited when only one or two named member forces are required, and it furnishes an independent cross-check on a joints solution because the two routes use different free bodies and different equations while describing the same structure.

The diagram above shows a triangulated truss on a pin at the left and a roller at the right, loaded at panel points. The dashed line indicates a section cut through selected members. Either side of the cut is a valid free body with cut forces shown along member axes, and moment about the intersection of two cut axes isolates the third force directly.

5.4.1 Projectile motion derived by integration

Consider a particle launched with speed u at elevation theta above the horizontal in uniform gravity g with no aerodynamic drag. Choose x horizontal and y vertical upward with origin at launch. Accelerations are

$$a_x = 0, \qquad a_y = -g.$$

Integration with initial velocities u cos theta and u sin theta gives

$$v_x = u\cos\theta, \qquad v_y = u\sin\theta - g\,t,$$

and a second integration with zero initial position gives

$$x = u\cos\theta\,t, \qquad y = u\sin\theta\,t - \frac{1}{2}g\,t^2.$$

Setting y equal to zero for level ground and discarding the launch instant gives time of flight

$$T = \frac{2u\sin\theta}{g}.$$

Substitution of T into x gives range

$$R = u\cos\theta\,T = \frac{u^2\sin 2\theta}{g},$$

using the double-angle identity. Maximum height follows from vanishing of v sub y, which occurs at half the flight time, giving

$$H = \frac{(u\sin\theta)^2}{2g}.$$

Elimination of t between x and y gives the trajectory parabola

$$y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}.$$

Range is proportional to sin of twice the launch angle and therefore peaks when twice the angle equals a right angle, namely at forty five degrees, with complementary launch angles sharing the same level range in vacuo. Drag, wind, and elevated targets modify these results, but the vacuum parabola remains the reference from which corrections depart.

The diagram above shows the parabolic trajectory with launch velocity resolved into horizontal and vertical components. Horizontal velocity remains constant in the absence of drag, vertical velocity decreases linearly under gravity, vanishes at the apex height H, and the particle lands at range R on level ground.

Quick example — projectile range: Launch speed $u = 20$ m/s at $\theta = 30^\circ$, gravity $g = 9.81$ m/s squared.

Flight time is $T = 2u\sin\theta/g = 20/9.81 \approx 2.04$ s, and range is $R = u^2\sin 2\theta/g \approx 35.3$ m.

Peak height is $H = (u\sin\theta)^2/(2g) \approx 5.10$ m.

Trap: forty-five degrees is best only for level ground in vacuo — drag and raised targets move the optimum.

5.4.2 Circular motion and the normal tangential decomposition

A particle moving along a curved path of local radius of curvature rho is described most naturally by unit vectors tangent and normal to the path. Let s be arc length, v equal to d s over d t the speed, e sub t the forward tangent, and e sub n the inward normal toward the centre of curvature. Differentiation of the position vector and use of the turning rate of the tangent gives velocity purely tangential,

$$\mathbf{v} = v\,\mathbf{e}_t,$$

and acceleration with two components,

$$\mathbf{a} = \dot{v}\,\mathbf{e}_t + \frac{v^2}{\rho}\,\mathbf{e}_n.$$

The tangential part a sub t equals d v over d t and measures rate of change of speed. The normal part a sub n equals v squared over rho, directed toward the centre of curvature, and measures rate of change of direction. Uniform circular motion has vanishing tangential part and a pure centripetal normal part. Nonuniform motion combines both. The decomposition explains why a fast vehicle on a tight curve demands large lateral friction even at constant speed, and why the demand grows with the square of speed and inversely with radius.

5.4.3 Instant-centre theory for rigid-body plane motion

A rigid body in plane motion has a velocity field fully determined by the velocity of one reference point plus rotation about that point. At each instant there exists a point in the extended plane of the body, the instant centre, whose velocity is zero. Denote angular velocity by omega. Every other point P moves instantaneously in a circle about the instant centre IC with magnitude

$$v_P = \omega\,r_{P/IC},$$

directed perpendicular to the line joining P to the instant centre, with sense set by omega. The instant centre is located as the intersection of the perpendiculars to two known velocity directions. For rolling without slipping, the contact point has zero velocity and is therefore the instant centre, so the centre velocity equals omega times the rolling radius and the top of the wheel moves at twice the centre speed. The instant centre is an instantaneous property. It migrates as motion proceeds and its acceleration is generally nonzero, so it cannot be used directly as a moment centre for kinetics without accounting for its acceleration.

5.4.4 Coriolis acceleration derivation sketch

Consider a particle sliding with velocity v sub rel along a slotted link that itself rotates with angular velocity omega. Let the position of the particle relative to a fixed origin be the sum of the link origin position and the relative position resolved in the rotating basis. Two time differentiations produce five acceleration terms. Two belong to the transport motion of the coincident link point, one is the relative acceleration along the slot, and the remaining cross term is the Coriolis acceleration

$$\mathbf{a}_c = 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rel}},$$

of magnitude two omega v sub rel in plane motion, directed ninety degrees from the sliding direction, rotated in the sense of omega. Physically it arises from two coupled effects of equal size. The sliding motion carries the particle to larger radius where the transport speed is higher, and the rotation turns the relative velocity vector itself, each contributing omega times v sub rel. The term vanishes when either the rotation or the sliding vanishes, and it dominates the design of sliding pairs on rotating links such as crank-driven shapers and swash mechanisms.

5.5.1 Work-energy theorem proved

Start from Newton second law for a particle of mass m along its path. Let F sub t be the tangential resultant force, s arc length, v speed. Then F sub t equals m times d v over d t. Multiply by d s and use the chain relation d s equal to v times d t to obtain

$$F_t\,ds = m\,\frac{dv}{dt}\,v\,dt = m\,v\,dv.$$

Integration from position one to position two along the path gives

$$\int_{1}^{2}F_t\,ds = \frac{1}{2}m\,v_2^2 - \frac{1}{2}m\,v_1^2.$$

The left side is the work U of the resultant force along the path, and the right side is the change in kinetic energy T equal to one half m v squared. Hence

$$T_1 + \sum U_{1\to2} = T_2.$$

For a rigid body the same argument applied to every mass element adds a rotational term one half I omega squared about the mass centre or about a fixed axis, with work done by external forces and couples. Gravity work over a vertical drop h is W times h with sign set by descent versus ascent. Linear spring work from stretch s one to s two is one half k times the difference of squares s one squared minus s two squared. Normal and frictionless contact forces do no work under no-slip rolling and sliding-free constraints because their points of application have no displacement along the force. The theorem routes problems with known forces, distances, and speeds directly to the unknown speed or position without solving for time history.

5.5.2 Impulse-momentum and the meaning of restitution

Integration of Newton second law over time rather than distance gives the impulse-momentum form. For translation,

$$m\,\mathbf{v}_1 + \sum\int_{t_1}^{t_2}\mathbf{F}\,dt = m\,\mathbf{v}_2,$$

where each time integral is the linear impulse of the corresponding force. For plane rotation about the mass centre or a fixed axis, an analogous angular relation holds with moment impulses and angular momenta I times omega. This form is the natural tool when time intervals or impact forces are the data or the demand.

Direct central impact of two bodies A and B along their common line of impact is governed by conservation of total linear momentum along that line together with the restitution relation. Let approach speeds and separation speeds be measured along the line of impact. The coefficient of restitution e is defined by

$$e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}} = \frac{\text{separation speed}}{\text{approach speed}}.$$

The numerator is the rate at which the bodies move apart after contact, the denominator the rate at which they closed before contact. A value of unity describes a perfectly elastic impact in which kinetic energy of the pair along the line of impact is preserved. A value of zero describes a perfectly plastic impact in which the bodies move together after contact with maximum kinetic energy loss consistent with momentum conservation. Intermediate values describe elastoplastic contacts with partial energy loss to vibration, sound, and permanent deformation. Momentum of the pair along the line of impact is conserved for all e when external impulses during the brief contact are negligible.

5.5.3 Principle of virtual work with intuition

A virtual displacement is an imagined infinitesimal displacement consistent with the constraints at the frozen instant, denoted by delta r. The virtual work of a force F is F dot delta r. The principle of virtual work states that a rigid body in equilibrium does zero net virtual work under every admissible virtual displacement,

$$\delta U = 0.$$

The intuition is that equilibrium leaves no first-order direction in which the forces can feed energy into motion. A translation test recovers force equilibrium, because a virtual translation dotted with the resultant must vanish for arbitrary translation direction. A rotation test recovers moment equilibrium, because a virtual rotation dotted with the resultant moment must vanish for arbitrary rotation sense. The principle therefore contains the three equilibrium equations in compact form and extends naturally to mechanisms and deformable systems where direct free-body counting becomes cumbersome. It converts a statics problem into a kinematics-of-displacements problem, asking how points move with each other under the constraints rather than which forces balance which.

Equilibrium follows from Newton laws by setting mass-centre acceleration and angular acceleration to zero, giving two force equations and one moment equation in the plane, with moments valid about any point and supports modeled by restrained-motion reactions. Friction is governed by the Coulomb inequality, recast as the friction angle phi equal to arctan of mu sub s, with the cone of half-angle phi separating stick from impending slip, the screw self-lock condition that helix angle remain below phi with efficiency below one half, the wedge lock condition that included angle remain below the sum of face friction angles, and the capstan tension ratio exponential in mu times beta with beta strictly in radians. Trusses obey the determinacy count m equal to two j minus three for simple internally determinate form, with both zero-force rules proved from joint equilibrium, joints solved by progressive particle equilibrium, and sections solved by cutting at most three members and placing moments at axis intersections. Plane kinematics supplies the integrated projectile relations for flight time, range, height, and the parabolic path with maximum level range at forty five degrees, the tangential plus centripetal acceleration split, the instant-centre velocity field with rolling contact as its canonical case, and the Coriolis term twice omega cross v sub rel for sliding on rotation. Kinetics closes with the work-energy theorem proved by path integration, the impulse-momentum theorem proved by time integration, restitution as the ratio of separation speed to approach speed with momentum conserved along the line of impact, and virtual work equal to zero as the energy portrait of equilibrium. Together these results form a single failure grammar of staying put, slipping, and moving, in which design always reduces to computing the ruling number and holding the operating point inside its safe region with margin.