Free-body diagrams and equilibrium of rigid bodies

Why it matters

Every load calculation in a mechatronic system starts with a free-body diagram (FBD): the holding torque a robot joint motor must supply, the force on an actuator pin, the reactions on a gantry rail. If the FBD is wrong (a missing reaction, a moment arm measured to the wrong point, a force drawn the wrong way), every number after it is wrong however careful the algebra. You also cannot find a bending moment, a stress or a deflection until you know the reactions, so this topic is the entry point to all of strength of materials.

Key ideas

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

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

• applied loads: point loads, distributed loads and couples (for example a motor torque);
• the weight, acting at the centre of gravity;
• support or contact reactions, one unknown for each direction in which the support prevents motion;
• internal forces only where you cut through the body; they then act on the piece you keep as external forces.

What each support provides in a plane problem.

• Roller, smooth surface or pin-ended link: 1 force, perpendicular to the surface or along the link.
• Pin (hinge): 2 force components, R_x and R_y, and no moment.
• Fixed (built-in) support, or a joint held by a braked motor: 2 force components plus 1 moment.
• Cable, belt or rope: 1 tensile force along the cable, pulling away from the body.
• Rough contact: a normal force N plus a friction force tangential to the surface.

If you do not know the sense of a reaction, assume one. A negative answer means the real sense is opposite; do not redraw the diagram.

Equilibrium. A rigid body is in equilibrium when the resultant force and the resultant moment about any point are both zero. It is then at rest or moving with constant velocity without angular acceleration. For a coplanar force system there are exactly three independent scalar equations: ΣF_x = 0, ΣF_y = 0, ΣM_A = 0. Equivalent sets also work: one force equation and two moment equations (about A and B, provided the line AB is not perpendicular to the force direction used), or three moment equations about three non-collinear points. In space there are six: three force and three moment equations.

Special force systems.

• Concurrent forces (all through one point): only 2 useful equations in a plane; moments vanish automatically.
• Parallel forces: 2 equations, one force sum and one moment sum.
• Two-force member: a body loaded at only two points with no couple and no load in between. The two forces are equal, opposite and along the line joining the two points. Pin-ended links, truss bars and hydraulic or electric linear actuators pinned at both ends are two-force members.
• Three-force member: if three non-parallel forces keep a body in equilibrium, their lines of action meet at one point. This gives a quick graphical solution and a check.
• Lami's theorem: for three concurrent coplanar 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 equations (3 in a plane) and the supports prevent all rigid-body motion, the body is statically determinate. More unknowns means statically indeterminate: you need deformation (compatibility) conditions, which appear later 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 (improperly constrained).

Distributed loads. For finding reactions only, replace a distributed load by its resultant: a uniform load w over length a becomes 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 the replacement when finding internal shear force and bending moment inside the loaded length; cut first, then replace only the part on your free body.

Link to dynamics. If the sums are not zero, the body accelerates: ΣF = m·a_G and ΣM_G = I_G·α. D'Alembert's principle adds the inertia force −m·a_G and couple −I_G·α to the FBD so the same equilibrium equations can be used for a moving robot link.

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). Take counter-clockwise as positive (or clockwise) and stay consistent.

M_A = x·F_y − y·F_x

• x, y: coordinates of the point of application measured from 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: the angle between the other two forces, opposite F₁, F₂, F₃ respectively. Only for exactly three concurrent coplanar forces in equilibrium.

W = m·g

• m: mass (kg); g = 9.81 m/s²; W: weight (N), acting at the centre of gravity.

Worked examples

Example 1 (standard): holding torque at a robot shoulder joint. Given: a horizontal robot link 0.8 m long, uniform mass 4 kg, carrying a 5 kg payload at its tip, held stationary by the joint motor at the root, g = 9.81 m/s². Find the joint reaction force and the holding torque.

1. FBD: the joint acts like a fixed support, giving R_x, R_y and a moment M. Loads: link weight W₁ at 0.4 m, payload weight W₂ at 0.8 m.
2. Weights: W = m·g, W₁ = 4 × 9.81 = 39.24 N, W₂ = 5 × 9.81 = 49.05 N.
3. ΣF_x = 0: R_x = 0.
4. ΣF_y = 0: R_y = W₁ + W₂ = 39.24 + 49.05 = 88.29 N.
5. ΣM_joint = 0: M = W₁ × 0.4 + W₂ × 0.8 = 15.70 + 39.24 = 54.94 N·m. Answer: R_y ≈ 88.3 N upward, holding torque ≈ 54.9 N·m. The payload is only 56 % of the weight but produces 71 % of the torque, because it acts at twice the arm.

Example 2 (GATE level): boom held by a linear actuator. Given: a light horizontal boom AB, 2 m long, pinned to a wall at A, carries 1.2 kN downward at B. A linear actuator is pinned to the boom at C (1.2 m from A) and to the wall at D, 0.9 m vertically below A. Find the actuator force and the pin reaction at A.

1. The actuator is a two-force member, so its force acts along DC. DC has horizontal run 1.2 m and rise 0.9 m, length √(1.2² + 0.9²) = 1.5 m, so its direction cosines are 0.8 (horizontal) and 0.6 (vertical).
2. Assume the actuator pushes (compression) on the boom: components 0.8F away from the wall and 0.6F upward at C.
3. ΣM_A = 0: 0.6F × 1.2 = 1.2 × 2, so 0.72F = 2.4 and F = 3.333 kN. Positive, so it is in compression as assumed.
4. ΣF_x = 0: A_x + 0.8 × 3.333 = 0, so A_x = −2.667 kN (towards the wall).
5. ΣF_y = 0: A_y + 0.6 × 3.333 − 1.2 = 0, so A_y = 1.2 − 2.0 = −0.8 kN (downward).
6. Resultant: R_A = √(2.667² + 0.8²) = 2.784 kN.
7. Check with the three-force rule: the actuator line meets the vertical through B at y = −0.9 + 0.75 × 2 = 0.6 m. The pin force must point along A to that point, slope 0.6/2 = 0.3, and indeed 0.8/2.667 = 0.3. Answer: actuator force ≈ 3.33 kN (compression), R_A ≈ 2.78 kN. Note the actuator force is almost three times the load because its moment arm about A is small.

Common mistakes

• Drawing a vertical reaction at a smooth wall or roller. A smooth surface pushes only perpendicular to itself.
• Leaving out the fixing moment at a built-in support or a held motor joint, or adding a moment at a pin.
• Using the distance along a member instead of the perpendicular distance to the line of action.
• Using a distributed load's resultant to find bending moments inside the loaded length.
• Treating mass (kg) as force; 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, links and actuators treated as two-force members, ladder and block problems combining equilibrium with friction, 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 removes the most unknowns, and 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. In what direction does the force in it act?
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 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 meet at one point. 5. Three.