Hibbeler Engineering Mechanics Cheatshee

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1. General Principles Newton's First Law: A particle remains at rest or continues to move with constant velocity if there is no unbalanced force acting on it. Newton's Second Law: The acceleration of a particle is proportional to the net force acting on it and inversely proportional to its mass. $\sum \vec{F} = m\vec{a}$ Newton's Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear. Newton's Law of Gravitational Attraction: $F = G \frac{m_1 m_2}{r^2}$, where $G = 66.73 \times 10^{-12} \text{ m}^3/(\text{kg} \cdot \text{s}^2)$ Weight: $W = mg$, where $g = 9.81 \text{ m/s}^2$ or $32.2 \text{ ft/s}^2$ 2. Force Vectors 2.1. 2D Force Systems Cartesian Vectors: $\vec{F} = F_x \hat{i} + F_y \hat{j}$ Magnitude: $F = \sqrt{F_x^2 + F_y^2}$ Direction: $\theta = \arctan\left(\frac{F_y}{F_x}\right)$ Components: $F_x = F \cos\theta$, $F_y = F \sin\theta$ Resultant Force: $\vec{F}_R = \sum \vec{F} = (\sum F_x)\hat{i} + (\sum F_y)\hat{j}$ 2.2. 3D Force Systems Cartesian Vectors: $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ Magnitude: $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$ Direction Cosines: $\cos\alpha = \frac{F_x}{F}$, $\cos\beta = \frac{F_y}{F}$, $\cos\gamma = \frac{F_z}{F}$ $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ Unit Vector: $\vec{u}_F = \frac{\vec{F}}{F} = \cos\alpha \hat{i} + \cos\beta \hat{j} + \cos\gamma \hat{k}$ Position Vector: $\vec{r} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$ Force along a Line: $\vec{F} = F \vec{u} = F \frac{\vec{r}}{r}$ Dot Product: $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta = A_x B_x + A_y B_y + A_z B_z$ Projection of $\vec{A}$ onto $\vec{B}$: $A_B = \vec{A} \cdot \vec{u}_B$ 3. Equilibrium of a Particle Conditions for Equilibrium: $\sum \vec{F} = 0$ 2D: $\sum F_x = 0$, $\sum F_y = 0$ 3D: $\sum F_x = 0$, $\sum F_y = 0$, $\sum F_z = 0$ Free-Body Diagram (FBD): Essential for identifying all forces. 4. Force System Resultants 4.1. Moment of a Force Scalar (2D): $M_O = Fd$ (force magnitude $\times$ perpendicular distance) Vector (3D): $\vec{M}_O = \vec{r} \times \vec{F}$ $\vec{M}_O = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$ Varignon's Theorem: The moment of a force about a point is equal to the sum of the moments of its components about the same point. 4.2. Moment of a Couple Couple: Two parallel forces with the same magnitude, opposite direction, and separated by a perpendicular distance $d$. Magnitude: $M = Fd$ (independent of the point of moment calculation) Vector: $\vec{M} = \vec{r} \times \vec{F}$ (where $\vec{r}$ is from any point on $-\vec{F}$ to any point on $\vec{F}$) 4.3. Resultant of a Force System Resultant Force: $\vec{F}_R = \sum \vec{F}$ Resultant Couple Moment: $\vec{M}_{R_O} = \sum \vec{M}_O$ (sum of moments of all forces about point O) Wrench: For 3D systems, any system of forces and couples can be reduced to a resultant force $\vec{F}_R$ and a resultant couple moment $\vec{M}_R$ acting at a specific point. If $\vec{F}_R$ and $\vec{M}_R$ are parallel, it's a wrench. 5. Equilibrium of a Rigid Body 5.1. Conditions for Equilibrium $\sum \vec{F} = 0 \implies \sum F_x = 0, \sum F_y = 0, \sum F_z = 0$ $\sum \vec{M}_O = 0 \implies \sum M_x = 0, \sum M_y = 0, \sum M_z = 0$ (about any point O) 5.2. Supports and Reactions Support Type 2D Reactions 3D Reactions Cable/Rope 1 force (tension) 1 force (tension) Smooth Surface 1 normal force 1 normal force Roller 1 normal force 1 normal force Pin (hinge) 2 forces ($F_x, F_y$) 3 forces ($F_x, F_y, F_z$) Fixed (cantilever) 2 forces ($F_x, F_y$), 1 moment ($M_z$) 3 forces ($F_x, F_y, F_z$), 3 moments ($M_x, M_y, M_z$) 5.3. Two- and Three-Force Members Two-Force Member: A member subjected to forces only at two points. The forces must be equal, opposite, and collinear along the line connecting the two points. Three-Force Member: A member subjected to forces only at three points. The forces must be concurrent or parallel. 6. Trusses, Frames, and Machines 6.1. Trusses Assumptions: Members are connected by pins, loads applied at joints, members are two-force members. Method of Joints: Apply particle equilibrium ($\sum F_x=0, \sum F_y=0$) at each joint. Method of Sections: Cut through members to isolate a section, then apply rigid body equilibrium ($\sum F_x=0, \sum F_y=0, \sum M_O=0$). Zero-Force Members: If only two non-collinear members connect at a joint and no external load or reaction acts at that joint, both members are zero-force members. If three members connect at a joint, two of which are collinear, and no external load or reaction acts at that joint, the third member is a zero-force member. 6.2. Frames and Machines Components are generally multi-force members. Disassemble the frame/machine into its component parts. Draw FBDs for each component and for the entire structure. Apply rigid body equilibrium equations to each FBD. Remember Newton's Third Law for internal forces. 7. Internal Forces Procedure: Determine external reactions. Pass an imaginary section through the point of interest. Draw FBD of either segment. Apply equilibrium equations to find internal normal force (N), shear force (V), and bending moment (M). Sign Convention (for positive internal forces): Normal Force (N): Tension is positive (pulling away from section). Shear Force (V): Causes clockwise rotation of the element (down on right face, up on left face). Bending Moment (M): Causes compression in the top fibers and tension in the bottom fibers (smiles). Shear and Moment Diagrams: $\frac{dV}{dx} = w(x)$ (slope of shear diagram = distributed load intensity) $\frac{dM}{dx} = V(x)$ (slope of moment diagram = shear force) $\Delta V = \int w(x) dx$ (change in shear = area under load diagram) $\Delta M = \int V(x) dx$ (change in moment = area under shear diagram) 8. Friction Static Friction: $F_s \le \mu_s N$ $F_s^{max} = \mu_s N$ (occurs at impending motion) Kinetic Friction: $F_k = \mu_k N$ (occurs when motion is happening) $\mu_s > \mu_k$ Angle of Static Friction: $\phi_s = \arctan(\mu_s)$ Angle of Repose: Angle at which object on incline starts to slide. $\tan\theta = \mu_s$ Wedges: Analyze equilibrium by isolating wedges and applying friction laws. Flat Belts: $T_2 = T_1 e^{\mu_s \beta}$ (where $\beta$ is angle of contact in radians) 9. Center of Gravity and Centroid Center of Gravity: Point where the entire weight of a body can be considered to act. $\bar{x} = \frac{\sum W_i x_i}{\sum W_i}$, $\bar{y} = \frac{\sum W_i y_i}{\sum W_i}$, $\bar{z} = \frac{\sum W_i z_i}{\sum W_i}$ Centroid (Area): Geometric center of an area. $\bar{x} = \frac{\int x dA}{\int dA} = \frac{\sum \tilde{x}_i A_i}{\sum A_i}$ $\bar{y} = \frac{\int y dA}{\int dA} = \frac{\sum \tilde{y}_i A_i}{\sum A_i}$ Centroid (Volume): Geometric center of a volume. $\bar{x} = \frac{\int x dV}{\int dV} = \frac{\sum \tilde{x}_i V_i}{\sum V_i}$ $\bar{y} = \frac{\int y dV}{\int dV} = \frac{\sum \tilde{y}_i V_i}{\sum V_i}$ $\bar{z} = \frac{\int z dV}{\int dV} = \frac{\sum \tilde{z}_i V_i}{\sum V_i}$ Pappus-Guldinus Theorems: Area of Surface of Revolution: $A = \theta \bar{r} L$ (where $\theta$ is angle of revolution in radians, $\bar{r}$ is distance from axis to centroid of line, $L$ is length of line) Volume of Body of Revolution: $V = \theta \bar{r} A$ (where $\bar{r}$ is distance from axis to centroid of area, $A$ is area) 10. Moments of Inertia 10.1. Area Moment of Inertia Definition: $I_x = \int y^2 dA$, $I_y = \int x^2 dA$ Polar Moment of Inertia: $J_O = \int r^2 dA = I_x + I_y$ Parallel-Axis Theorem: $I = \bar{I} + Ad^2$ (where $\bar{I}$ is moment of inertia about centroidal axis, $A$ is area, $d$ is distance between parallel axes). Radius of Gyration: $k = \sqrt{\frac{I}{A}}$ 10.2. Mass Moment of Inertia Definition: $I_x = \int y^2 dm$, $I_y = \int x^2 dm$, $I_z = \int r^2 dm$ Parallel-Axis Theorem: $I = \bar{I} + md^2$ (where $\bar{I}$ is mass moment of inertia about centroidal axis, $m$ is mass, $d$ is distance between parallel axes). 11. Virtual Work Principle of Virtual Work for a Particle: $\delta U = \sum F_i \delta r_i = 0$ for equilibrium. Principle of Virtual Work for a Rigid Body: $\delta U = 0$ for equilibrium. $\sum F_x \delta x + \sum F_y \delta y + \sum M \delta\theta = 0$ Conservative Forces: Forces for which virtual work can be expressed as the negative of the change in potential energy ($\delta U = -\delta V$). Potential Energy: $V = V_g + V_e$ Gravitational: $V_g = Wy$ Elastic (Spring): $V_e = \frac{1}{2}ks^2$ Criterion for Equilibrium: $\frac{dV}{ds} = 0$ Stability of Equilibrium: Stable: $\frac{d^2V}{ds^2} > 0$ (minimum potential energy) Unstable: $\frac{d^2V}{ds^2} Neutral: $\frac{d^2V}{ds^2} = 0$ (constant potential energy)