Statics: Free-Body Diagrams and Equilibrium

Build the free-body diagram first

FE Mechanical statics questions look short because the arithmetic is simple once the model is right. Start by isolating one body, removing every support, and replacing each contact with the reactions it can actually transmit. A smooth roller gives one normal reaction. A pin (hinge) gives two force components, Ax and Ay. A fixed (cantilever) support gives two force components plus a reaction couple M. A smooth surface gives a normal force only; a rough surface adds friction along the surface.

Declare axes and a positive moment direction before touching the calculator. A standard convention is +x to the right, +y upward, and counterclockwise moments positive. The convention is not sacred, but it must be consistent from sketch to final answer. If a reaction comes out negative, the actual direction is opposite the assumed arrow — equilibrium did not fail.

The three planar equilibrium equations

For a rigid body in a plane, statics gives three independent scalar equations, exactly as printed in the FE Reference Handbook Statics section:

$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_P = 0$

Three equations solve at most three unknowns, so a 2D problem with more than three reaction unknowns is statically indeterminate and cannot be solved by statics alone. Choose the moment center P strategically: taking moments about a pin removes both of that pin's force components, often leaving a single unknown reaction.

Distributed loads and resultants

Never insert a distributed load w (N/m) into a force equation directly — first replace it with an equivalent concentrated resultant equal to the area under the load curve, acting at that area's centroid.

• Rectangular (uniform) load: R = wL, located at midspan (L/2).
• Triangular load: R = ½ w_max L, located one-third of the base from the larger (peak) end, two-thirds from the zero end.
• Trapezoidal load: split into a rectangle plus a triangle and sum the two resultants and their separate locations.

Worked example

A 6 m simply supported beam carries a uniform 4 kN/m load over its full length. Resultant R = wL = (4)(6) = 24 kN at x = 3 m. Taking moments about the left pin A: ΣM_A = 0 → R_B(6) − 24(3) = 0, so R_B = 12 kN. Then ΣF_y = 0 → R_A = 24 − 12 = 12 kN. Symmetry confirms the split. If instead the load were triangular, 0 → 9 kN/m left to right, the resultant ½(9)(6) = 27 kN acts at x = 4 m (two-thirds from the zero end), giving R_B = 27(4)/6 = 18 kN and R_A = 9 kN — the asymmetric split a triangular load produces.

Units, weight, and a repeatable pass

Units are the most common statics trap. A moment must be force × distance: N·m, lb·ft, or lb·in — never mix a length in meters with a section property in millimeters without converting. In USCS, a 200 lb load is 200 lbf (a force) in statics; weight W = mg only matters when the problem gives a mass. Do not drag the gravitational constant g_c into a pure statics balance.

Use this short routine on every equilibrium problem:

1. Isolate the body; draw all external forces, couples, dimensions, and angles.
2. Replace supports and distributed loads with the correct reaction model or resultant.
3. Pick axes and a positive moment direction.
4. Write symbolic equilibrium equations before substituting numbers.
5. Solve, then check signs, units, and physical plausibility.

A plausible answer balances both force and moment. If the vertical reactions sum to the total downward load but the moment equation is violated, the reactions are still wrong. If a reaction exceeds the total applied load, look for a long overturning lever arm before flipping signs blindly. Cables and two-force members carry load only along their axis — a 30° cable supporting a 500 N vertical pull has tension T = 500/sin30° = 1000 N, and its horizontal component 1000·cos30° = 866 N must be reacted somewhere, a detail distractor answers often omit.

Moments, couples, and force resultants

The moment of a force about a point is M = F·d, where d is the perpendicular distance from the point to the force's line of action — not the straight-line distance to the point of application. If a force is given at an angle, either resolve it into components and sum each component's moment (Varignon's theorem), or find the true perpendicular lever arm. A 200 N force applied at the end of a 0.5 m wrench at 60° to the handle gives M = 200·(0.5)·sin60° = 86.6 N·m, not 100 N·m.

A couple is two equal, opposite, non-collinear forces; it produces a pure moment M = F·d that is the same about every point and has no net force. Couples are why a fixed support must supply a reaction moment, and why you can slide a couple anywhere on a rigid body. A single force can always be replaced by an equal force at a chosen point plus a couple equal to the moment the force produced about that point — the force-couple system used to relocate loads to a convenient analysis point.

Concurrent force systems and resultants

When several forces share a common point (concurrent), there is no moment to balance and equilibrium reduces to ΣF_x = 0 and ΣF_y = 0 — the basis of a particle analysis and of every truss joint. To find a resultant, sum components: R_x = ΣF_x, R_y = ΣF_y, magnitude R = √(R_x² + R_y²), direction θ = arctan(R_y/R_x). Always resolve before combining; adding magnitudes of non-parallel forces is the most common setup error, and the FE deliberately offers that arithmetic sum as a distractor.