A free body diagram (FBD) is a sketch of one body cut free from everything around it, showing every external force and couple moment that acts on it. It is the first step of every statics problem: if the FBD is right, the equilibrium equations almost write themselves.
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What a free body diagram shows
Draw the body alone, with its supports and neighbours removed. In their place, draw the forces those supports and neighbours exert on the body. Add the body’s own weight at its center of gravity, any applied loads, the key dimensions and angles, and a coordinate system.
Only external forces go on the diagram. Forces that one part of the body exerts on another part cancel in pairs (Newton’s third law), so they never appear unless you cut the body to expose them.
How to draw one, step by step
The same five steps work for a particle, a beam, a bracket or a whole machine.
Pick the body. Decide exactly what is inside the boundary: one member, a joint, or an assembly.
Sketch its outline isolated from the supports, roughly to scale.
Draw the known forces: weight W = mg at the center of gravity, applied loads, and couple moments, each with its magnitude and direction.
Replace every support with its reaction forces and moments. If a direction is unknown, assume one; a negative answer just means it acts the other way.
Label dimensions and angles, choose x–y axes, and count unknowns against the available equations.
Support reactions you will meet most
Each support stops some motions. A reaction appears for every motion it stops: a force for a blocked translation, a couple moment for a blocked rotation. Once the FBD is drawn, the next step is solving the equilibrium equations.
Cable or rope: one tension force, along the cable, pulling away from the body.
Roller, rocker or smooth surface: one force normal to the surface.
Smooth pin or hinge: two force components (or one force at an unknown angle).
Fixed support: two force components and one couple moment.
Two-force member (pinned at both ends, no load in between): one force along the line joining the pins.
Rough surface: a normal force plus a friction force tangent to the surface.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Weight
W = m·g
m = mass (kg), g = 9.81 m/s²
Hibbeler, Engineering Mechanics: Statics
Force components
Fx = F·cos θ, Fy = F·sin θ
θ measured from the +x axis
Beer & Johnston, Vector Mechanics for Engineers: Statics
Particle equilibrium
ΣFx = 0, ΣFy = 0
all forces on the FBD
Hibbeler, Engineering Mechanics: Statics
Moment of a force
M_O = F·d
d = perpendicular distance from point O to the force’s line of action
Meriam & Kraige, Engineering Mechanics: Statics
Worked examples
Example: A housing hung from two cables
Given
A 40 kg pump housing hangs from a ring at point O.
Cable OA runs up and to the left at 30° above the horizontal.
Cable OB runs up and to the right at 45° above the horizontal.
Take g = 9.81 m/s². Cables and ring are weightless.
Find
The tension in each cable, T_A and T_B.
Solution
Isolate the ring. Three forces act on it: T_A along OA, T_B along OB, and the housing’s weight straight down.
Weight of the housing:W = m·g = 40 kg × 9.81 m/s² = 392.4 N
Sum forces in x (right positive):−T_A·cos 30° + T_B·cos 45° = 0 → T_A = T_B·(0.7071 / 0.8660) = 0.8165·T_B
Sum forces in y (up positive):T_A·sin 30° + T_B·sin 45° − 392.4 N = 0
Substitute T_A:0.5(0.8165·T_B) + 0.7071·T_B = 1.1154·T_B = 392.4 N → T_B = 351.8 N
Back-substitute:T_A = 0.8165 × 351.8 N = 287.3 N
Check: both answers are positive, so both cables are in tension, as assumed. A cable can only pull, so a negative tension would mean the FBD was wrong.
Answer: T_A ≈ 287 N and T_B ≈ 352 N. The steeper cable carries more of the load.
Example: A uniform bar leaning against a smooth wall
Given
A uniform 12 kg bar, 2.5 m long, rests with end A on a rough floor and end B against a smooth vertical wall.
The bar makes 60° with the floor.
Take g = 9.81 m/s².
Find
The wall reaction N_B, the floor reactions N_A and F_A, and the smallest friction coefficient that prevents slipping.
Solution
FBD of the bar: weight W at the midpoint; at B a single horizontal force N_B pushing away from the smooth wall; at A a vertical normal force N_A and a horizontal friction force F_A pointing toward the wall, because the foot tends to slide outward.
Weight:W = 12 kg × 9.81 m/s² = 117.7 N
Geometry from A: B is 2.5 cos 60° = 1.25 m across and 2.5 sin 60° = 2.165 m up. W acts 0.625 m across from A.
Sum moments about A, counterclockwise positive. This removes N_A and F_A, which pass through A:N_B(2.165 m) − 117.7 N(0.625 m) = 0 → N_B = 73.56 N·m / 2.165 m = 33.98 N
Sum forces in x:F_A − N_B = 0 → F_A = 33.98 N
Sum forces in y:N_A − 117.7 N = 0 → N_A = 117.7 N
Friction needed to hold the bar:μ_min = F_A / N_A = 33.98 / 117.7 = 0.289
Answer: N_B ≈ 34.0 N, N_A ≈ 117.7 N, F_A ≈ 34.0 N, and the floor must provide μ ≥ 0.289 or the bar slides.
Common mistakes
Leaving a support on the diagram. The body must be completely isolated. Every contact with the outside world becomes a force or a moment.
Forgetting the weight, or putting it in the wrong place. Weight acts at the center of gravity. For a uniform bar that is the midpoint, not an end.
Giving a pin a moment, or a fixed support no moment. A pin lets the body rotate, so it carries no moment. A fixed support does not, so it carries one.
Drawing internal forces. Forces between parts of the same body cancel. Show them only after you cut the body to expose them.
Mixing up the force on the body with the force from the body. Draw what the support does to the body. The equal and opposite force acts on the support, which is a different FBD.
In industry
Frequently asked questions
Do I include the weight of every member?
Include it unless the problem says the member is weightless or its weight is negligible next to the applied loads. If you include it, place it at the member’s center of gravity.
What if I guess a reaction direction wrong?
Nothing breaks. Solve with your assumed direction; a negative result means the force acts the opposite way with the same magnitude. The exception is a cable or a smooth contact, which can only pull or only push, so a negative value there means the setup is wrong.
Is an FBD the same for statics and dynamics?
The drawing is the same. In statics the forces sum to zero. In dynamics they sum to mass times acceleration, so many courses also draw a kinetic diagram next to the FBD.
How many unknowns can one FBD solve?
A 2D rigid body gives three independent equations, so it can solve three unknowns. A particle gives two in 2D. More unknowns means the problem is statically indeterminate, or you need another FBD.