Statics
Statics is the study of rigid bodies in equilibrium, where every force and moment balances to zero. The exam rewards a disciplined free-body diagram followed by the equilibrium equations, applied to particles, beams, trusses, frames, and problems involving friction. Master the bookkeeping of forces and moments and the rest is arithmetic.
Force vectors and components
A force is a vector with magnitude and direction. Resolve it into rectangular components with F_x = F·cos θ and F_y = F·sin θ, measuring θ from the horizontal. A 100 N force at 30° above horizontal has a horizontal component 100·cos 30° ≈ 86.6 N and vertical component 100·sin 30° = 50 N. Working in components is what makes equilibrium solvable, because forces add component-by-component rather than by their magnitudes. Choosing convenient x–y axes (sometimes aligned with an incline) simplifies the algebra.
Resultants of concurrent forces
Concurrent forces (all passing through one point) combine by vector addition. Two perpendicular forces of magnitude a and b give a resultant √(a² + b²): 30 N and 40 N at right angles produce √(30² + 40²) = 50 N. For two forces of equal magnitude F separated by angle θ, the resultant is 2·F·cos(θ/2). The general method is to sum all x-components and all y-components, then take R = √(ΣF_x² + ΣF_y²) with direction arctan(ΣF_y/ΣF_x).
Moments and couples
A moment (torque) measures the tendency of a force to rotate a body about a point: M = F·d, where d is the perpendicular distance from the point to the force's line of action. A 100 N force perpendicular to a 0.5 m wrench produces 50 N·m. A couple is two equal, opposite, parallel forces separated by distance d; its moment M = F·d is the same about every point, and it produces pure rotation with no net force — two 50 N forces 0.4 m apart give a 20 N·m couple. The classic error is using slant distance rather than the perpendicular distance to the line of action.
Equilibrium and free-body diagrams
A body in equilibrium satisfies ΣF_x = 0, ΣF_y = 0, and ΣM = 0 about any convenient point — three independent equations in two dimensions. The free-body diagram, showing every applied load and support reaction, is the essential first step; omitting a reaction or a weight is the most common failure. A smart choice of moment center (one that passes through unknown forces) eliminates them from the moment equation and often solves a support reaction in a single step.
Trusses, frames, and supports
Idealized trusses carry only axial force (tension or compression) in two-force members pinned at their ends. The method of joints applies ΣF_x = ΣF_y = 0 at each pin; the method of sections cuts through the members of interest and applies all three equilibrium equations to one side. Zero-force members can often be spotted by inspection. Support types set the reaction count: a roller gives one reaction, a pin gives two, and a fixed support gives three. Matching unknown reactions to available equations tells you whether a structure is statically determinate.
Friction and area properties
Impending sliding is resisted by static friction up to F_max = μ·N, proportional to the normal force N and independent of contact area; once moving, kinetic friction (a lower μ) applies. A 200 N normal force with μ = 0.3 resists up to 60 N before sliding. Area properties round out statics: the centroid locates an area's geometric center, and the area moment of inertia I (with the parallel-axis theorem I = I_c + A·d²) feeds directly into strength-of-materials formulas for bending stress and column buckling.
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