1 · Start with the idea
A free-body diagram is a force bookkeeping sheet. Imagine cutting one object away from everything touching it, then replacing each interaction with a force or moment. The diagram tells you what belongs in the equations—and helps you avoid counting the same interaction twice.
2 · Why it matters in real machines
Support reactions determine what bearings, brackets, foundations and connections must carry. A force can be modest yet produce a large moment because it acts far from a support. Before sizing a component, identify how load reaches it and how load leaves it.
3 · Core concepts
Build a free-body diagram
- Choose the object or assembly boundary.
- Draw external applied loads, self-weight where relevant and support reactions.
- Show load locations, dimensions and coordinate directions.
- Replace each support with the reactions allowed by the idealized model.
- Write a sign convention before writing equations.
In a planar idealization, a pin support can provide two force components; a roller can provide one force normal to its supporting surface; a fixed support can also provide a reaction moment. Real connections need a defensible approximation. Do not draw a reaction moment at an ideal pin merely to make the algebra work.
Equilibrium and moments
ΣFx = 0 ΣFy = 0 ΣMO = 0
MO = F × perpendicular distance
These are planar rigid-body equilibrium equations. The vector form is ΣF = 0 and ΣM = 0; accelerating systems require dynamics. A force’s moment depends on the perpendicular distance from the chosen point to its line of action. A pure couple has a moment but zero resultant force.
A distributed load q in N/m over length L has a resultant qL in N. For a uniform load it acts at the midpoint; a nonuniform load acts at the centroid of its load diagram. Do not replace every distributed load at the midpoint.
Trusses, frames and friction
An ideal pin-jointed truss with loads only at the joints has members that act as two-force members: axial tension or compression. Frames generally carry bending and shear as well as axial force. Applying a transverse load between truss joints can invalidate the simple two-force assumption.
For a simple dry-contact model, static friction adjusts as needed up to a limit: |f| ≤ μsN. It equals μsN only at impending slip. Here N denotes the normal reaction, not the newton unit. Kinetic friction is often approximated by μkN during sliding. Coefficients depend on actual surfaces and conditions; do not select a safety-critical value from appearance.
Centre of gravity
For discrete masses in a uniform gravitational field, x̄ = Σ(mixi)/Σmi, with equivalent formulas in other directions. A symmetric uniform shape has its centre of gravity at its geometric centre; asymmetric density or attached hardware can move it. A geometric centroid is not automatically the centre of gravity of a mixed-material assembly.
4 · Worked example
Hoist-load beam—static classroom model only. A pin-supported beam at A and roller-supported beam at B has span L = 2.00 m [6.562 ft]. A specified stationary point force P = 4.00 kN [899.2 lbf] acts 0.750 m [2.461 ft] from A. A uniform downward load q = 0.400 kN/m [27.41 lbf/ft] represents included beam/ancillary weight. Assume these values include everything stated; do not infer a lifted mass or rated capacity.
Take upward forces as positive and counterclockwise moments as positive. The distributed load has resultant qL = 0.800 kN [179.8 lbf], acting 1.00 m [3.281 ft] from A.
ΣMA = 0:
RB(2.00 m) − (4.00 kN)(0.750 m) − (0.800 kN)(1.00 m) = 0
RB = 1.90 kN [427.1 lbf], upward
ΣFy = 0: RA + 1.90 − 4.00 − 0.800 = 0
RA = 2.90 kN [651.9 lbf], upward
ΣFx = 0: RAx = 0
Independent moment check about B: −(2.90)(2.00) + (4.00)(1.25) + (0.800)(1.00) = 0 kN·m. Force check: 2.90 + 1.90 = 4.80 kN, equal to total downward load. The reaction nearest the point load is larger, which matches intuition.
Second case: put the same point force at midspan, 1.00 m from A. Symmetry gives 2.40 kN [539.5 lbf] at each support. Moving the load changes load sharing without changing total downward force.
Reactions alone do not establish beam or hoist safety. This model excludes dynamic effects, trolley movement, impact, lateral loads, stability, deflection, connections, local effects and any applicable design/inspection requirements. Actual load cases and support conditions require qualified assessment [VERIFY]. Do not conduct a lifting test from this example.
5 · Common mistakes
- Using distance along a sloping member instead of perpendicular moment arm.
- Placing a uniform load’s resultant at an end rather than at its midpoint.
- Mixing clockwise-positive and counterclockwise-positive signs in one equation.
- Putting action and reaction forces on the same free body when one belongs to the removed object.
- Assuming static friction is always at its maximum.
- Treating negative reaction results as arithmetic errors before checking their physical meaning and the support’s ability to resist uplift.
6 · In the field
Sketch the load path from load to attachment, member, support and foundation. If the connection behaves differently from your assumed pin, roller or fixed support, reconsider the model. Public diagrams can explain these principles without showing site layouts or equipment identifiers.
Hoisting, rigging and stored-energy work require employer procedures, proper training and qualified supervision. A free-body diagram is an analysis tool, not permission to lift.
7 · Quick checks
Why is a force times distance not always the correct moment?
The distance must be perpendicular to the force’s line of action. Using an arbitrary distance produces the wrong moment.
Why is the near support reaction larger in the example?
Moment equilibrium puts more of the off-centre point load on the nearer support. The uniform load contributes equally to both.
Does static friction always equal μsN?
No. Its magnitude adjusts up to that limit. Equality is the limiting, impending-slip condition in the simple model.
8 · Free resources & references
- OpenStax: conditions for static equilibrium ↗
Free explanation of force and moment balance.
- OpenStax: equilibrium examples ↗
Free worked examples and practice.
9 · Mini-project
On graph paper, draw a 200 mm [7.87 in] line representing the 2.00 m beam at 1:10 scale. Mark the point load at 75 mm [2.95 in] from A and the distributed-load resultant at 100 mm [3.94 in]. Recalculate reactions for the point load at 0.50, 1.00 and 1.50 m [1.64, 3.28 and 4.92 ft]. Keep this as a paper/spreadsheet exercise: no suspended objects or improvised hoists.