Free-body diagrams and equilibrium of rigid bodies

Why it matters

Every calculation in mechanics, from the axle loads on a truck to the bolt forces on an engine mount, starts with a free-body diagram (FBD). If the FBD is wrong (a missing reaction, a moment arm measured to the wrong point, a force drawn in the wrong direction), every number that follows is wrong no matter how good the algebra is. Getting equilibrium right is also the first step of every strength-of-materials problem: you cannot find a bending moment or a stress until you know the reactions.

Key ideas

Rigid body. A body whose deformation is small enough to ignore when you write the equilibrium equations. Real parts deform, but for statics we use the undeformed geometry.

Free-body diagram. Isolate the body completely from its surroundings, then replace every contact with the force (and moment) that the surroundings can exert there. An FBD shows:

• applied loads (point loads, distributed loads, couples);
• weight, acting at the centre of gravity;
• support or contact reactions, one unknown for every direction in which the support stops motion;
• internal forces only when you cut through the body (they then become external to the piece you keep).

What each support provides (plane problems).

• Roller, smooth surface or link: 1 force, perpendicular to the surface or along the link.
• Pin (hinge): 2 force components, R_x and R_y.
• Fixed (built-in) support: 2 force components plus 1 moment.
• Cable or rope: 1 tensile force along the cable, pulling away from the body.
• Rough contact: a normal force N and a friction force F tangential to the surface.

If you do not know the sense of a reaction, assume one. A negative answer simply means the actual sense is opposite.

Equilibrium. A rigid body is in equilibrium when the resultant force and the resultant moment about any point are both zero. Then it is at rest or moving with constant velocity and not changing its rotation. In a plane (coplanar) force system this gives exactly three independent scalar equations: ΣF_x = 0, ΣF_y = 0, ΣM_A = 0. Equivalent sets also work: one force equation plus two moment equations (about A and B, provided AB is not perpendicular to the force direction used), or three moment equations about three points that are not collinear. In space there are six equations: three forces and three moments.

Special force systems.

• Concurrent forces (all through one point): only 2 equations in a plane; moments are automatically zero.
• Parallel forces: 2 equations (one force sum and one moment sum).
• Two-force member: a body loaded only at two points with no couples. The two forces must be equal, opposite and along the line joining the points. Pin-ended truss bars and connecting links are two-force members.
• Three-force member: if three non-parallel forces keep a body in equilibrium, their lines of action must meet at a single point. This gives a quick graphical solution (for example, a ladder against a smooth wall).
• Lami's theorem: for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two.

Statical determinacy. If the number of independent unknown reactions equals the number of independent equilibrium equations (3 in a plane) and the supports prevent all rigid-body motion, the body is statically determinate. More unknowns means statically indeterminate; you then need deformation (compatibility) conditions, which you will meet in compound bars, propped cantilevers and fixed beams. Fewer unknowns, or reactions that are all parallel or all concurrent, means the body is a mechanism and is not properly constrained.

Distributed loads. Replace a distributed load by its resultant for the purpose of finding reactions only: a uniformly distributed load w over length a becomes a force w·a at the middle of a; a triangular load with peak w becomes w·a/2 at a/3 from the large end. Do not use this replacement when finding internal shear and moment inside the loaded span; there you cut first and then replace the part that is on the free body.

Formulas

ΣF_x = 0, ΣF_y = 0, ΣM_A = 0

• F_x, F_y: force components along chosen x and y axes (N); M_A: moment about any point A (N·m). Applies to any coplanar force system on a body in static equilibrium.

M_A = F·d

• F: force magnitude (N); d: perpendicular distance from A to the line of action of F (m). Counter-clockwise positive is the usual convention; be consistent.

M_A = x·F_y − y·F_x

• x, y: coordinates of the point of application relative to A (m). Use this when the perpendicular distance is awkward to find.

F₁ / sin α = F₂ / sin β = F₃ / sin γ (Lami's theorem)

• α, β, γ: angles opposite to F₁, F₂, F₃, i.e. the angle between the other two forces. Only for exactly three concurrent coplanar forces in equilibrium.

ΣF = 0 and ΣM = 0 in three dimensions give 6 scalar equations.

Worked examples

Example 1 (standard): static axle loads of a car. Given: car mass m = 1400 kg, wheelbase L = 2.6 m, centre of gravity 1.1 m behind the front axle, g = 9.81 m/s². Find the static loads on the front and rear axles.

1. Weight: W = m·g = 1400 × 9.81 = 13 734 N.
2. FBD: W downward at the CG; R_f upward at the front axle; R_r upward at the rear axle.
3. Moments about the front contact point: R_r·L − W·a = 0, so R_r = 13 734 × 1.1 / 2.6 = 5810.5 N.
4. Vertical forces: R_f = W − R_r = 13 734 − 5810.5 = 7923.5 N.
5. Check with moments about the rear axle: R_f = 13 734 × 1.5 / 2.6 = 7923.5 N. Matches. Answer: R_f ≈ 7.92 kN, R_r ≈ 5.81 kN (57.7 % on the front axle).

Example 2 (GATE level): ladder with a person on it. Given: a uniform ladder AB, length 5 m, weight 200 N, foot A on a rough floor, top B against a smooth vertical wall, A is 3 m from the wall (so B is 4 m high). A 600 N person stands 3.75 m up the ladder from A. Find the wall reaction, the friction force needed at A and the minimum coefficient of friction.

1. Geometry: cos θ = 3/5 = 0.6 with the floor. A point s metres along the ladder is 0.6·s horizontally from A. Ladder weight acts at s = 2.5 m, horizontal arm 1.5 m. Person at s = 3.75 m, horizontal arm 2.25 m.
2. FBD: at A, N_A upward and F_A horizontal toward the wall; at B, N_B horizontal away from the wall (smooth, so no vertical force); weights 200 N and 600 N downward.
3. Moments about A: N_B × 4 = 200 × 1.5 + 600 × 2.25 = 300 + 1350 = 1650 N·m, so N_B = 412.5 N.
4. ΣF_x = 0: F_A = N_B = 412.5 N.
5. ΣF_y = 0: N_A = 200 + 600 = 800 N.
6. Slip is just prevented when F_A ≤ μ·N_A, so μ_min = 412.5 / 800 = 0.516. Answer: N_B = 412.5 N, F_A = 412.5 N, μ_min ≈ 0.52. Check with the three-force idea for the ladder alone (no person): N_B = 200 × 1.5 / 4 = 75 N, so the person raises the friction demand more than five times. That is why ladders slip when you climb near the top.

Common mistakes

• Drawing a vertical reaction at a smooth wall. A smooth surface pushes only perpendicular to itself.
• Leaving out the fixing moment at a built-in support (cantilever), or adding a moment at a pin.
• Using the distance along the member instead of the perpendicular distance to the line of action.
• Replacing a distributed load by its resultant and then using that point load to find bending moments inside the loaded length.
• Mixing sign conventions between equations, or treating mass (kg) as force (N); multiply by g.
• Writing a fourth "independent" equation for a plane problem. Only three are independent; the rest are checks.
• Including internal forces between parts of the same free body; they cancel in pairs.

For GATE ME

Expect support reactions of beams and frames, ladder and block problems combining equilibrium with friction, two-force and three-force member reasoning, Lami's theorem for strings and pulleys, and conceptual questions on the number of reactions and statical determinacy. Practise drawing the FBD first, choosing the moment point that eliminates the most unknowns, and always checking with one extra equation.

Quick check

1. How many unknown reactions does a fixed support provide in a plane problem?
2. A pin-ended link carries no load between its ends. What is the direction of the force in it?
3. A uniform 4 m beam weighing 400 N is simply supported at its ends and carries 1 kN at 1 m from the left end. Find the left reaction.
4. Three non-parallel forces keep a body in equilibrium. What must be true of their lines of action?
5. How many independent equilibrium equations exist for a body under a general coplanar force system?

Answers: 1. Three (two forces and a moment). 2. Along the line joining the two pins. 3. R_left = 200 + 1000 × 3/4 = 950 N. 4. They must be concurrent (meet at one point). 5. Three.