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Statics

Field: Newtonian Mechanics

The branch of mechanics that is concerned with the analysis of loads on physical systems in static equilibrium.

Sequence of Expressions

Let BB be a continuous, rigid body occupying a domain Ω⊂R3\Omega \subset \mathbb{R}^3. The state of equilibrium is defined by the condition that the body is subjected only to external forces and moments, and these effects balance out. Mathematically, this requires satisfying the following system of partial differential equations and integral constraints:\n\n1. **Force Equilibrium (Cauchy's Equation):** The balance of linear momentum requires that the divergence of the stress tensor σ\sigma plus the body force density b⃗\vec{b} must vanish everywhere within the domain Ω\Omega:\n∇⋅σ+b⃗=0⃗in Ω\nabla \cdot \sigma + \vec{b} = \vec{0} \quad \text{in } \Omega\n\n2. **Moment Equilibrium (Rotational Balance):** The balance of angular momentum requires that the moment generated by the stress tensor and body forces relative to any point must vanish. This is often expressed as the equilibrium of the moment tensor, which simplifies to the condition that the stress tensor must be symmetric (σij=σji\sigma_{ij} = \sigma_{ji}) and that the body forces must be conservative (if b⃗\vec{b} is derived from a potential). For a general body, the condition is:\n∑j=13(xi∂σij∂xj+bi)=0for i=1,2,3\sum_{j=1}^{3} (x_i \frac{\partial \sigma_{ij}}{\partial x_j} + b_i) = 0 \quad \text{for } i=1, 2, 3 \n\n3. **Boundary Conditions:** The equilibrium must also hold on the boundary ∂Ω\partial\Omega. The traction vector t⃗\vec{t} acting on the boundary must satisfy the force balance:\n∫∂Ωt⃗⋅n⃗ dS=0⃗\int_{\partial\Omega} \vec{t} \cdot \vec{n} \, dS = \vec{0} \n\nWhere n⃗\vec{n} is the outward unit normal vector, and σij\sigma_{ij} are the components of the Cauchy stress tensor.
Let Q\mathcal{Q} be the set of physical quantities under consideration. We define two disjoint subsets, S⊂Q\mathcal{S} \subset \mathcal{Q} and V⊂Q\mathcal{V} \subset \mathcal{Q}, such that Q=S∪V\mathcal{Q} = \mathcal{S} \cup \mathcal{V}.\n\n1. **Scalar Quantities (S\mathcal{S}):** A quantity S∈SS \in \mathcal{S} is characterized by its mapping into the underlying field of real numbers, R\mathbb{R}. For any physical context C\mathcal{C}, SS is represented by a function fS:Context→Rf_S: \text{Context} \to \mathbb{R}, such that its value is determined solely by its magnitude, S=fS(C)S = f_S(\mathcal{C}).\n\n2. **Vector Quantities (V\mathcal{V}):** A quantity V⃗∈V\vec{V} \in \mathcal{V} is an element of a real vector space Rn\mathbb{R}^n (where nn is the dimension of the physical space, e.g., n=3n=3 for Newtonian Mechanics). V⃗\vec{V} is uniquely represented by its components relative to an orthonormal basis {e^i}i=1n\{\hat{e}_i\}_{i=1}^n: \nV⃗=∑i=1nVie^i\vec{V} = \sum_{i=1}^n V_i \hat{e}_i \nwhere Vi∈RV_i \in \mathbb{R} are the scalar components. The physical nature of V⃗\vec{V} requires that its transformation under a coordinate change x→x′x \to x' is governed by the Jacobian matrix RR: Vi′=∑j=1nRijVjV'_i = \sum_{j=1}^n R_{ij} V_j. This directional dependence is formalized by the requirement that the quantity must satisfy the closure property under vector addition and scalar multiplication within the structure of Rn\mathbb{R}^n.
Let B⊂R3\mathcal{B} \subset \mathbb{R}^3 be a continuous body with a spatially varying mass density ρ(r⃗)\rho(\vec{r}). The total mass MM of the body is defined by the volume integral:\n\nM=∭Bρ(r⃗)dVM = \iiint_{\mathcal{B}} \rho(\vec{r}) dV\n\nThe position vector of the Center of Gravity, r⃗cg=(xcg,ycg,zcg)\vec{r}_{cg} = (x_{cg}, y_{cg}, z_{cg}), is defined as the weighted average of the position over the volume, where the weighting function is the density ρ(r⃗)\rho(\vec{r}). Specifically, the coordinates are given by:\n\nxcg=1M∭Bxρ(r⃗)dVx_{cg} = \frac{1}{M} \iiint_{\mathcal{B}} x \rho(\vec{r}) dV\nycg=1M∭Byρ(r⃗)dVy_{cg} = \frac{1}{M} \iiint_{\mathcal{B}} y \rho(\vec{r}) dV\nzcg=1M∭Bzρ(r⃗)dVz_{cg} = \frac{1}{M} \iiint_{\mathcal{B}} z \rho(\vec{r}) dV\n\nIn vector form, the Center of Gravity is:\n\nr⃗cg=1M∭Br⃗ρ(r⃗)dV\vec{r}_{cg} = \frac{1}{M} \iiint_{\mathcal{B}} \vec{r} \rho(\vec{r}) dV
Consider a system SS of particles with mass mm and position vector r⃗(t)\vec{r}(t) in an inertial reference frame I\mathcal{I}. The law is fundamentally stated as the rate of change of linear momentum p⃗\vec{p} being equal to the net external force F⃗net\vec{F}_{net}. Mathematically, this is expressed as:\n\nF⃗net=dp⃗dt=ddt(mv⃗)\vec{F}_{net} = \frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v})\n\nFor a system described by a continuous body occupying a volume V(t)V(t), the governing equation is derived from the Cauchy momentum equation, which relates the divergence of the stress tensor σij\sigma_{ij} to the external body force density f⃗\vec{f}: \n\n∂∂xjσij+fi=ρd2xidt2\frac{\partial}{\partial x_j} \sigma_{ij} + f_i = \rho \frac{d^2 x_i}{dt^2} \n\nWhere ρ\rho is the mass density, σij\sigma_{ij} is the Cauchy stress tensor, and fif_i represents the external body forces (e.g., gravity). In the simplified case of a single particle with constant mass mm and no external body forces, this reduces to the classical form:\n\nF⃗net=ma⃗=md2r⃗dt2\vec{F}_{net} = m\vec{a} = m \frac{d^2 \vec{r}}{dt^2}
Let F⃗∈R3\vec{F} \in \mathbb{R}^3 be a force vector acting on a point in a Cartesian coordinate system defined by the orthonormal basis B=(i^,j^,k^)B = (\hat{i}, \hat{j}, \hat{k}). The process of resolving F⃗\vec{F} involves decomposing it into its orthogonal components relative to this basis. This decomposition is formally expressed as the unique linear combination:\n\nF⃗=Fxi^+Fyj^+Fzk^\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\n\nwhere Fx,Fy,Fz∈RF_x, F_y, F_z \in \mathbb{R} are the scalar components of the force along the x,y,zx, y, z axes, respectively. These components are determined by the projection of F⃗\vec{F} onto each basis vector, utilizing the dot product operator (⋅)(\cdot): \n\nFx=F⃗⋅i^=∑k=13Fkδk,xF_x = \vec{F} \cdot \hat{i} = \sum_{k=1}^{3} F_k \delta_{k, x} \n\nIn general, for an arbitrary orthonormal basis B′=(u^1,u^2,u^3)B' = (\hat{u}_1, \hat{u}_2, \hat{u}_3), the components (F1′,F2′,F3′)(F'_1, F'_2, F'_3) are given by:\n\nFi′=F⃗⋅u^i, for i∈{1,2,3}F'_i = \vec{F} \cdot \hat{u}_i, \text{ for } i \in \{1, 2, 3\}
Let R3\mathbb{R}^3 be the Euclidean space equipped with a standard orthonormal basis {i^,j^,k^}\{\hat{i}, \hat{j}, \hat{k}\} and the dot product A⃗⋅B⃗\vec{A} \cdot \vec{B} and cross product A⃗×B⃗\vec{A} \times \vec{B}. Consider a force F⃗\vec{F} applied at a point PP relative to an origin O⃗\vec{O}. The position vector r⃗\vec{r} is defined as r⃗=P⃗−O⃗\vec{r} = \vec{P} - \vec{O}. The moment of force M⃗\vec{M} about the origin O⃗\vec{O} is defined by the vector cross product: M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}. In Cartesian coordinates, if r⃗=⟨rx,ry,rz⟩\vec{r} = \langle r_x, r_y, r_z \rangle and F⃗=⟨Fx,Fy,Fz⟩\vec{F} = \langle F_x, F_y, F_z \rangle, the moment vector M⃗=⟨Mx,My,Mz⟩\vec{M} = \langle M_x, M_y, M_z \rangle is given by the determinant expansion: M⃗=∣i^j^k^rxryrzFxFyFz∣\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} M⃗=(ryFz−rzFy)i^+(rzFx−rxFz)j^+(rxFy−ryFx)k^\vec{M} = (r_y F_z - r_z F_y) \hat{i} + (r_z F_x - r_x F_z) \hat{j} + (r_x F_y - r_y F_x) \hat{k}. The magnitude of the moment is M=∣r⃗∣∣F⃗∣sin⁡(θ)M = |\vec{r}| |\vec{F}| \sin(\theta), where θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}. Alternatively, M=Mx2+My2+Mz2M = \sqrt{M_x^2 + M_y^2 + M_z^2}. This quantity represents the rotational tendency (torque) of F⃗\vec{F} about $\vec{O}.
Consider a particle of mass mm moving in a continuous time parameter tt within a Euclidean space R3\mathbb{R}^3. Let r(t)\mathbf{r}(t) be the position vector, v(t)=drdt\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} be the velocity, and Fnet(t)\mathbf{F}_{net}(t) be the net external force acting on the particle.\n\nThe Law of Inertia is formally stated as the equivalence between the vanishing of the net force and the constancy of the momentum (or velocity):\n\nFnet(t)=0  ⟺  ddt(mv(t))=0\mathbf{F}_{net}(t) = \mathbf{0} \iff \frac{d}{dt} \left( m \mathbf{v}(t) \right) = \mathbf{0} \n\nThis implies that if Fnet(t)=0\mathbf{F}_{net}(t) = \mathbf{0} for all t∈[t0,t1]t \in [t_0, t_1], then the momentum p(t)=mv(t)\mathbf{p}(t) = m \mathbf{v}(t) is constant, such that:\n\np(t)=p(t0)=mv0\mathbf{p}(t) = \mathbf{p}(t_0) = m \mathbf{v}_0 \n\nConsequently, the acceleration a(t)=dvdt\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} must be identically zero, ensuring that the particle maintains a constant velocity v(t)=v0\mathbf{v}(t) = \mathbf{v}_0 (i.e., r(t)=r0+v0(t−t0)\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 (t - t_0)).
Consider a system composed of two distinct interacting bodies, AA and BB, with respective masses mAm_A and mBm_B. Let r⃗A\vec{r}_A and r⃗B\vec{r}_B be the position vectors of the centers of mass of AA and BB. The interaction between AA and BB is characterized by a pair of forces: F⃗A on B\vec{F}_{A \text{ on } B} (the force exerted by AA on BB) and F⃗B on A\vec{F}_{B \text{ on } A} (the force exerted by BB on AA). Newton's Third Law of Motion asserts that these forces are equal in magnitude and opposite in direction, meaning they form an action-reaction pair:\n\nF⃗B on A=−F⃗A on B\vec{F}_{B \text{ on } A} = -\vec{F}_{A \text{ on } B} \n\nFurthermore, if the force F⃗A on B\vec{F}_{A \text{ on } B} is derived from a potential energy function V(r⃗AB)V(\vec{r}_{AB}), then the force pair must satisfy the condition that the total potential energy is symmetric with respect to the exchange of bodies, and the force vectors must be derivable from a generalized potential Φ\Phi: \n\nF⃗A on B=−∇A∂Φ∂r⃗ABandF⃗B on A=−∇B∂Φ∂r⃗BA\vec{F}_{A \text{ on } B} = -\nabla_A \frac{\partial \Phi}{\partial \vec{r}_{AB}} \quad \text{and} \quad \vec{F}_{B \text{ on } A} = -\nabla_B \frac{\partial \Phi}{\partial \vec{r}_{BA}} \n\nWhere r⃗AB=r⃗A−r⃗B\vec{r}_{AB} = \vec{r}_A - \vec{r}_B and r⃗BA=r⃗B−r⃗A\vec{r}_{BA} = \vec{r}_B - \vec{r}_A. The core mathematical statement remains the vector equality: F⃗B on A+F⃗A on B=0⃗\vec{F}_{B \text{ on } A} + \vec{F}_{A \text{ on } B} = \vec{0}.
Let BB be a rigid body modeled in an inertial Cartesian coordinate system S\mathcal{S}. Let r⃗i\vec{r}_i be the position vector of the point of application of the ii-th external force F⃗i\vec{F}_i. The system is in equilibrium if and only if the following two vector equations are simultaneously satisfied:\n\n1. **Translational Equilibrium (Force Balance):** The net external force R⃗\vec{R} acting on BB must vanish:\nR⃗=∑i=1NF⃗i=0⃗\vec{R} = \sum_{i=1}^{N} \vec{F}_i = \vec{0} \n\n2. **Rotational Equilibrium (Moment Balance):** The net external moment N⃗\vec{N} about any arbitrary origin OO must vanish:\nN⃗=∑i=1NM⃗i=∑i=1N(r⃗i×F⃗i)=0⃗\vec{N} = \sum_{i=1}^{N} \vec{M}_i = \sum_{i=1}^{N} (\vec{r}_i \times \vec{F}_i) = \vec{0} \n\nThese conditions imply that the linear momentum P⃗\vec{P} and the angular momentum L⃗\vec{L} of the body are constant in time, specifically dP⃗dt=R⃗=0⃗\frac{d\vec{P}}{dt} = \vec{R} = \vec{0} and dL⃗dt=N⃗=0⃗\frac{d\vec{L}}{dt} = \vec{N} = \vec{0}.
Let SS be a rigid body occupying a region Ω⊂R3\Omega \subset \mathbb{R}^3. The system is subjected to a finite set of external forces {F⃗1,F⃗2,…,F⃗N}\{\vec{F}_1, \vec{F}_2, \dots, \vec{F}_N\} and corresponding points of application {r⃗1,r⃗2,…,r⃗N}∈Ω\{\vec{r}_1, \vec{r}_2, \dots, \vec{r}_N\} \in \Omega. A Free Body Diagram (FBD) is the formal representation of the force vector field F⃗ext:Ω→R3\vec{F}_{ext}: \Omega \to \mathbb{R}^3 defined by the superposition of these external forces. Mathematically, the diagram asserts the existence of a set of force vectors {F⃗i}\{\vec{F}_i\} such that the system is in static equilibrium, which requires the following vector summation conditions to hold:\n\n∑i=1NF⃗i=0⃗\sum_{i=1}^{N} \vec{F}_i = \vec{0} \n\n∑i=1Nτ⃗i=∑i=1N(r⃗i−r⃗0)×F⃗i=0⃗\sum_{i=1}^{N} \vec{\tau}_i = \sum_{i=1}^{N} (\vec{r}_i - \vec{r}_0) \times \vec{F}_i = \vec{0} \n\nwhere r⃗0\vec{r}_0 is an arbitrary origin point. The FBD is the graphical visualization of the force density f⃗(r⃗)=∑i=1NF⃗iδ(r⃗−r⃗i)\vec{f}(\vec{r}) = \sum_{i=1}^{N} \vec{F}_i \delta(\vec{r} - \vec{r}_i) acting on the body, confirming that the net resultant force and net resultant torque are zero.