Modeling a Part With a Free-Body Diagram
Atlas stands at a cluttered engineering workbench, holding a single bolt in one hand and sketching bold arrows radiating outward from a simple box outline on a whiteboard, explaining which forces act on the isolated part.
- Explain why isolating a single body is the essential first step in force analysis.
- Identify every external force acting on a body, including contact forces and gravity.
- Draw a correct free-body diagram with force vectors labeled by type, direction, and point of application.
- Apply all three conditions of static equilibrium to determine whether a body is in balance.
- Predict how adding or removing a support changes the force balance on an isolated part.
Key terms
- Free-body diagram (FBD)
- A sketch that isolates one body and shows every external force acting on it as a labeled vector.
- Normal force
- A contact force that acts perpendicular to the touching surface, pushing the body away from that surface.
- Moment (torque)
- The turning effect of a force about a point, equal to force multiplied by perpendicular distance.
- Static equilibrium
- The condition where a body has zero net force and zero net moment, so it neither translates nor rotates.
- Reaction force
- The force a support, pin, or connection exerts back on the isolated body, replacing the removed contact.
Isolating the Body
The single hardest discipline in FBD work is mentally cutting the body free from everything touching it. Every wall, pin, cable, and neighboring part is erased and replaced by the force it transmits across the cut. A pinned support resists motion in two directions and produces both horizontal and vertical reaction components; a roller resists only perpendicular to its rolling surface and produces one component. Naming the support type tells you exactly how many reaction arrows to draw.
Writing the Equilibrium Equations
Choose coordinate axes before resolving forces, because a smart axis choice eliminates terms. Inclined forces are split into components: a tension T at angle θ above horizontal contributes T·cos θ to ΣFx and T·sin θ to ΣFy. The moment equation ΣM = 0 can be taken about any point, so engineers pick the point where the most unknown forces act, since those forces have zero moment arm there and drop out of the equation, leaving one unknown to solve directly.
Counting Equations Versus Unknowns
A planar rigid body gives exactly three independent equilibrium equations: ΣFx = 0, ΣFy = 0, and ΣM = 0. If your FBD has three or fewer unknown reactions, the problem is statically determinate and fully solvable with statics alone. If it has more unknowns than equations, it is statically indeterminate and requires additional information about material deformation. Counting arrows against the three equations is your first feasibility check.
Worked examples
A 200 N sign hangs from the end of a horizontal beam supported by a pin at the wall and a cable at 30° above horizontal attached 2 m from the pin; the sign hangs at the 2 m point. Find the cable tension.
- Isolate the beam; external forces are the pin reaction at the wall, the cable tension T at 30°, and the 200 N weight pulling down at the 2 m point.
- Take moments about the pin to eliminate the unknown pin reaction (it acts at the pivot, so its moment arm is zero).
- The downward 200 N acts at 2 m, giving a clockwise moment of 200 × 2 = 400 N·m.
- The cable's vertical component T·sin 30° = 0.5T acts upward at 2 m, giving a counterclockwise moment of 0.5T × 2 = T N·m.
- Set ΣM = 0 about the pin: T − 400 = 0, so T = 400 N.
Answer: The cable tension is 400 N.
Activity
Drag each force arrow onto the correct location on the isolated bracket, then label its direction and type.
Practice
Draw the free-body diagram of a ladder leaning against a frictionless wall and resting on a rough floor, labeling every force.
A 500 N box sits on a 20° frictionless ramp; calculate the normal force the ramp surface exerts on the box.
Common mistakes to avoid
- The pulling force of a rope and the reaction cancel inside the FBD.Each body gets its own FBD; the rope tension and the support reaction act on different bodies and never cancel within a single diagram.
- Moments only matter when the object is spinning.ΣM = 0 must be checked even for a perfectly stationary body, because balanced forces alone do not guarantee the body will not rotate.
Check your understanding
A student draws a free-body diagram of a shelf bracket but keeps the wall visible in the sketch and does not draw a reaction force arrow from the wall. What is wrong with this FBD?
A traffic light of weight W hangs from two cables, each angled 30° above horizontal. On the free-body diagram of the light, the equilibrium condition ΣFy = 0 gives:
A steel beam rests on two concrete columns and carries a load. To confirm it is not rotating, which equation must also be satisfied beyond ΣFx = 0 and ΣFy = 0?
Recap
A free-body diagram isolates one body, replaces every connection with the force it exerts, and labels each arrow by type, direction, and point of application. Static equilibrium then demands ΣFx = 0, ΣFy = 0, and ΣM = 0 all hold simultaneously for a non-accelerating, non-rotating rigid body.
Reflect
Which support reactions are easiest for you to forget, and how could counting reaction components before writing equations prevent that mistake?