Study Companion: S10 — Isolating the System: the Free-Body Diagram

Part 2 built the tools that describe forces: components, moments, couples and resultants. Part 3 puts them to work on bodies at rest. Equilibrium itself is a two-line law: the resultant force and the resultant couple are both zero. The hard part is not the law but the list of forces it is applied to. Leave out one force that acts on the body, or include one that does not, and every number after that is wrong. This lesson is about building that list: choosing the body, cutting it free from everything it touches, and replacing each contact by the force it can actually exert. The picture that results is the free-body diagram, and Meriam calls it the most important single step in mechanics.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 3/1 ("Introduction") and 3/2 ("System Isolation and the Free-Body Diagram"), with Figs. 3/1 and 3/2 and the free-body-diagram exercises 3/A–3/C, and the three-dimensional support table of article 3/4 (Fig. 3/8). Marghitu, Dupac and Madsen, Statics with MATLAB, sections 4.2 ("Supports") and 4.3 ("Free-Body Diagrams"). Harper has no free-body-diagram problem. Both books treat the topic without numbers; the numbers here are our own. Unknown forces are drawn in an assumed sense; counter-clockwise is positive for planar moments.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

The owner's switchgear contact has a steel shaft carried in two bearings. A crank of radius 60\,\mathrm{mm} is keyed to one end, and a spring anchored 240\,\mathrm{mm} from the shaft pulls the crank pin with 500\,\mathrm N. A latch holds the crank at 60^\circ until the contact is tripped. This is the crank of S04, Nugget 05, now mounted on a real shaft.

The bearings are what wear out. How hard does each one push on the shaft? A first guess says the two bearings share the 500\,\mathrm N spring pull, 250\,\mathrm N each. Before that guess can be tested, there is a more basic question: which forces act on the shaft at all, and how many of them are unknown? Is it two unknowns, or four, or more? Until that list exists, no equation can be written. This lesson builds the list (Synthesis); S11 then solves it, and the 250\,\mathrm N guess turns out to be badly wrong.

Nugget 01: Equilibrium Is a Statement About One Isolated Body

Claim. A body is in equilibrium when the resultant of all forces acting on it is zero: \mathbf R = \Sigma\mathbf F = \mathbf 0 and \mathbf M = \Sigma\mathbf M = \mathbf 0. Those sums mean nothing until we say which body, and which forces cross its boundary.

The picture. A 2\,\mathrm{kg} lamp hangs from the ceiling on a 0.5\,\mathrm{kg} chain. The same physical object gives two different force lists, depending on where the boundary is drawn.

Draw the boundary round the lamp alone and the chain is outside it: the chain pulls up on the lamp, and that pull is on the diagram. Draw it round lamp and chain together and the same pull is now inside the system. It is internal, it appears twice with opposite signs (chain on lamp, lamp on chain), and it drops out. The ceiling's pull, which was absent before, now crosses the boundary.

Check with numbers. Lamp alone: W = 2 \times 9.81 = 19.62\,\mathrm N down, so the chain must pull 19.62\,\mathrm N up. Lamp and chain: weights 19.62 + 4.905 = 24.53\,\mathrm N down, so the ceiling must pull 24.53\,\mathrm N up. Both diagrams satisfy \Sigma F = 0; they describe different systems, not different physics.

The idea. Meriam (3/1–3/2) defines a mechanical system as a body or group of bodies conceptually isolated from all others: rigid or not, solid, fluid or a mixture. Once chosen, it is treated as a single body. The free-body diagram (FBD) shows that system alone, with every force applied to it by the bodies imagined removed, plus body forces such as weight. Forces may be contact or remote, internal or external, applied or reactive, concentrated or distributed, and a force on a rigid body is a sliding vector (transmissibility, S03). Equilibrium needs only the external ones.

The equation (physical law: Newton's second law with zero acceleration; Meriam's Eq. 3/1, necessary and sufficient):

Every symbol sums over forces crossing the chosen boundary, in N and N·m. Bodies that can be treated in a plane (S11) need the planar form; the rest are three-dimensional (S12).

Worked number. In the puzzle, isolate the shaft with its crank. The spring is outside, so its 500\,\mathrm N pull appears. Isolate shaft, crank and spring together, and the spring's pull on the crank becomes internal; instead, the force at the spring's anchor appears, and it is the same 500\,\mathrm N because a light spring in equilibrium pulls equally at both ends.

Real example. The car-body manipulator that opens Meriam's chapter turns a body shell slowly through many orientations. Its accelerations are tiny, so designers treat it as in equilibrium; what they isolate (the shell, the gripper, the whole arm) decides which joint loads they can find.

Plausible wrong answer. "Put the force the lamp exerts on the chain on the lamp's diagram." That force acts on the chain. A diagram shows forces on the system by the outside, never the reverse.

Check. For the lamp-and-chain system, why does the chain's own weight now appear? (The chain is inside the boundary, and gravity is an external body force on it.)

Narration spine. The hanging lamp. A dashed boundary snaps round the lamp: an upward chain arrow and a weight arrow appear. The boundary grows to take in the chain: the chain arrow fades (internal), the ceiling arrow and the chain's weight appear, and the readouts change from 19.62 to 24.53.

Bridge. Removing the ceiling or the chain means replacing it by a force. But which force? A chain can only pull. What can other contacts do?

Nugget 02: A Contact That Only Pushes or Pulls Gives One Unknown

Claim. A support exerts force only in the directions in which it stops the body moving. A cable, a smooth surface, a roller and a smooth guide each stop motion in one direction only, so each gives one force of known direction and unknown magnitude: one unknown.

The picture. Remove the contacting body and ask what the isolated body would then do. The contact force opposes exactly that motion.

Cut the cable and the block falls away from the cable's anchor, so the cable pulls away from the body, along itself. Take away the smooth surface and the ball drops into where the surface was, so the surface pushes, along the normal. A smooth surface cannot grip, so it has no component along itself. A roller is the same: it can roll along the surface, so it pushes only normal to it. The collar slides freely along the guide and is stopped only across it.

Check with numbers. A cable at 30^\circ above the horizontal pulling with T = 800\,\mathrm N has components 800\cos30^\circ = 692.8\,\mathrm N and 800\sin30^\circ = 400.0\,\mathrm N. A roller on a 20^\circ incline pushing with N = 1.5\,\mathrm{kN} has a horizontal component 1.5\sin20^\circ = 0.513\,\mathrm{kN} and a vertical one 1.5\cos20^\circ = 1.410\,\mathrm{kN}. In each case the direction is fixed by the geometry, and only the magnitude (T or N) is unknown.

The idea. Meriam's Fig. 3/1 (cases 1, 2, 4, 5) lists these contacts. Because of its flexibility a cable, belt, rope or chain cannot resist bending, shear or compression, so it exerts only tension, tangent to itself where it is attached. If its weight is small compared with its tension it hangs straight; if not, it sags and its tension changes direction and size along its length (Part 5). Smooth surfaces in contact push normal to their common tangent; no real surface is perfectly smooth, but many are close enough. A roller, rocker or ball support transmits a compressive force normal to the supporting surface. Marghitu (4.2.1) describes the roller the same way: it cannot exert a force parallel to the surface on which it moves, nor a couple. Bridges sit on rollers so they can expand and contract with temperature.

The equation (material model: a frictionless contact; \mathbf n is the unit normal, pointing into the body):

with \mathbf e the unit vector along the cable, away from the body. The signs N \ge 0 and T \ge 0 are physical: a surface cannot pull, a cable cannot push.

Worked number. A crate rests in a corner against a smooth vertical wall at A. The wall's force is horizontal, away from the wall: one unknown, A. If the crate's 40\,\mathrm{kg} were held by the wall alone, there would be nothing to carry its 392.4\,\mathrm N weight, because A has no vertical component. The floor has to supply it.

Real example. A highway bridge girder rests on a fixed bearing at one end and an expansion bearing (a roller or sliding pad) at the other. On a hot day the girder grows by millimetres; the roller lets it, because it carries no force along the girder.

Plausible wrong answer. "The roller on the incline pushes straight up, like the floor." It pushes normal to the incline: 20^\circ from vertical, with a 0.513\,\mathrm{kN} horizontal part that a vertical force would miss.

Check. A ladder rests against a smooth wall. Which way does the wall push? (Horizontally, away from the wall, normal to it.)

Narration spine. Four panels. In each, the contacting body fades out and the isolated body starts to move (falls, slides, rolls); the motion freezes and a red arrow grows opposing it, along the cable or along the normal. The roller's arrow tilts with the incline, and its horizontal shadow reads 0.513 kN.

Bridge. One-unknown contacts allow sliding in some direction. A pin, a rough surface or a weld allows less, and must give more.

Nugget 03: Pins, Rough Contacts and Welds Give Two or Three

Claim. A pin free to turn stops sliding in every direction of its plane, so it gives a force of unknown size and direction: two unknowns. A rough surface adds a friction component to the normal: two. A pin not free to turn, or a built-in end, also stops rotation and adds a couple: three.

The picture. Same test as before: what motion does the contact prevent?

A pin holds the bar's end in place against sliding in any direction in the plane, so it must be able to push in any direction. We do not know which, so we draw two components, R_x and R_y. It cannot stop the bar turning about the pin, so no couple. A welded or built-in end stops sliding along and across the beam and also stops the beam turning: an axial force F, a transverse (shear) force V and a couple M, the bending moment.

Check with numbers. Suppose the equations later give R_x = -300\,\mathrm N and R_y = 400\,\mathrm N. The pin's force is R = \sqrt{300^2 + 400^2} = 500\,\mathrm N, acting at \theta = 126.87^\circ from +x: up and to the left. The minus sign only said that R_x acts opposite to the arrow drawn. For a rough contact with normal force N = 600\,\mathrm N and friction F = 150\,\mathrm N, the total contact force is \sqrt{600^2 + 150^2} = 618.5\,\mathrm N at \arctan(150/600) = 14.04^\circ from the normal.

The idea. Meriam's Fig. 3/1 case 6: a freely hinged pin supports a force in any direction in the plane normal to the pin axis, shown either as R_x, R_y or as a magnitude R and direction \theta. A pin not free to turn also supports a couple M. Case 7: a built-in or fixed support carries F, V and M, which Meriam explains as the resultant of the complex stresses over the beam's cross section, reduced to a force–couple system (S05). Case 3: a rough surface can carry a tangential, frictional component F as well as the normal N of the resultant contact force R (Part 6). Marghitu (4.2.1) gives the pin \mathbf F_{01} = F_{01x}\mathbf i + F_{01y}\mathbf j and the fixed support \mathbf F_{01} = F_{01x}\mathbf i + F_{01y}\mathbf j, \mathbf M_{01} = M_{01z}\mathbf k. His subscript 01 means "by body 0 (the ground) on body 1"; we keep Meriam's letters.

The equation (geometric relationship: magnitude and direction from components):

Either form has two unknowns. A pin can be described by two components, or by a size and an angle, but never by one number.

Worked number. A diving board is built into the pool edge. A 70\,\mathrm{kg} diver stands at its tip. The FBD of the board has three unknowns at the built-in end: F (zero here, since no load acts along the board), V and M. The diver's 686.7\,\mathrm N weight is known; finding V and M is S11's job.

Real example. Door hinges are pins: a door swings freely, so a hinge carries no couple about its own axis. A shelf bracket screwed flat to a wall is a built-in end: it carries the shelf's load and stops it tipping, which is why the top screws of a loaded bracket are the ones that pull out.

Plausible wrong answer. "A pin can stop a bar rotating, so it carries a couple." Only if it is not free to turn: rusted, clamped or keyed. A pin free to turn has no couple, by the definition of free.

Check. A pin reaction comes out as R_x = 120\,\mathrm N, R_y = -50\,\mathrm N. Which way does it act? (Right and down, 130\,\mathrm N at -22.6^\circ.)

Narration spine. The pinned bar is tugged by an unseen hand in different directions; at each tug the pin's reaction swings to oppose it, showing why its direction is unknown. It splits into R_x and R_y. Then a beam end is welded: the tug arrows, and now a twist, are all opposed: F, V and a curled M.

Bridge. Contacts are not the only forces. Gravity acts without touching, and springs act with a force that depends on how far they are stretched.

Nugget 04: Weight, Springs and Remote Forces Belong on the Diagram Too

Claim. Gravity is a force on every particle of the body; its resultant is the weight W = mg, acting through the centre of gravity G. A spring exerts a force set by its deformation, F = kx for a linear spring. Both go on the FBD, unless they are shown to be negligible.

The picture. The force of a spring depends on how far it is stretched or compressed from its neutral length, not on its length.

Stretch the spring by x and it pulls the block back towards the neutral position; compress it and it pushes. For a linear spring the force grows in proportion; for a hardening spring it grows faster than proportionally, for a softening one slower.

Check with numbers. The puzzle's spring has stiffness k = 20\,\mathrm{kN/m} and is stretched x = 25\,\mathrm{mm} beyond its neutral length: F = 20\,000 \times 0.025 = 500\,\mathrm N. The shaft is steel, 20\,\mathrm{mm} in diameter and 0.30\,\mathrm m long: m = 7850 \times \pi (0.010)^2 \times 0.30 = 0.740\,\mathrm{kg}, so W = 0.740 \times 9.81 = 7.26\,\mathrm N. That is 1.45\,\% of the spring force. Dropping it is an assumption, and now it is a stated, measured one.

The idea. Meriam's Fig. 3/1 case 8: gravitational attraction is distributed through the body; its resultant is W = mg through G, directed towards the centre of the earth. G is often found by symmetry; otherwise by experiment or calculation (Part 5). Remote magnetic and electric forces have the same overall effect on a rigid body as contact forces of equal size and direction. Case 9: a linear spring gives F = kx, with k the stiffness or modulus and x the deformation from the neutral position; the force is tensile if the spring is stretched, compressive if compressed. Nonlinear springs harden or soften. Case 10: a linear torsional (clockwork) spring gives a moment proportional to its angular deflection, M = k_T\theta. Meriam warns that Fig. 3/1 is not a set of FBDs, only the elements from which they are built.

The equation (physical law for weight; material model for the springs):

m in kg, g = 9.81\,\mathrm{m/s^2}, k in N/m, x in m, k_T in N·m/rad, \theta in rad.

Worked number. A torsional return spring on a door latch has k_T = 0.5\,\mathrm{N\cdot m/rad} and is wound 90^\circ (\pi/2\,\mathrm{rad}) from neutral: M = 0.5 \times \pi/2 = 0.785\,\mathrm{N\cdot m}, opposing the deflection. On the latch's FBD it appears as a couple, not a force.

Real example. The switchgear contact stores its closing energy in the spring. When the latch trips, the stored deformation is what makes the force; a spring that has crept a few millimetres shorter closes the contact more weakly, which is why such springs are tested by force at a set length.

Plausible wrong answer. "Spring force is k times the spring's length." It is k times the change in length from neutral. A 300\,\mathrm{mm} spring stretched to 325\,\mathrm{mm} gives k \times 0.025, not k \times 0.325, which would be thirteen times too big.

Check. A spring of stiffness 3\,\mathrm{kN/m} is compressed 40\,\mathrm{mm}. What force does it exert on the body at its end, and which way? (120\,\mathrm N, pushing the body away.)

Narration spine. The block on its spring is dragged out; a red arrow grows in proportion, and a point climbs the straight F = kx line. The line bends into hardening and softening curves. Cut to the shaft: a weight arrow appears at its middle, labelled 7.26\,\mathrm N, beside the 500\,\mathrm N spring arrow, and dims.

Bridge. We now have every element. The free-body diagram is the method that assembles them without missing one.

Nugget 05: Draw the Boundary, Then Walk Around It

Claim. A complete FBD comes from four steps: choose the system, draw its complete boundary, walk round that boundary replacing every contact by its force and add the body forces, then draw the axes. Where a force's sense cannot be seen, assume one; the sign of the answer corrects it.

The picture. A beam on a pin at A and a roller at B, loaded by a force F and a couple M (after Marghitu's Fig. 4.8), and its FBD.

Walking round the boundary from the left end: at A the pin is cut away and replaced by A_x and A_y; along the top, the applied F and M are kept; at B the roller is replaced by one vertical force. The beam's weight is neglected here; if it were not, it would go at the beam's centre. Nothing else touches the beam. The ground, the pin and the roller are gone, and only their forces stay.

Check with numbers. Count: A_x, A_y and B are three unknowns, and a body in a plane has three equations (Nugget 07). For a sense that can be seen, draw it correctly. Take a bell crank pinned at A with a 10\,\mathrm{kg} mass hanging 0.30\,\mathrm m to one side of the pin, held by a horizontal cable whose line passes 0.15\,\mathrm m from the pin. The weight turns the crank one way, so the cable must turn it the other; the cable pulls, so its sense is fixed, and S04's zero moment condition gives T = 98.1 \times 0.30/0.15 = 196.2\,\mathrm N.

The idea. Meriam's construction of free-body diagrams, in four steps. Step 1: decide which system to isolate; it should usually involve the unknowns wanted. Step 2: draw its complete external boundary, which isolates it from all other contacting or attracting bodies. This is often the most crucial step. Step 3: identify every force acting on the system by the removed bodies, by a systematic traverse of the whole boundary, and add body forces such as weight. Show known forces with magnitude, direction and sense; unknown ones with a symbol, and an arbitrarily assigned sense if the sense is unknown. A positive answer confirms the assumed sense; a negative one reverses it. Be consistent with the assumed senses throughout. Step 4: draw the coordinate axes on the diagram. Do not clutter it, and make force arrows distinct from other arrows. Marghitu (4.3) compresses this into two steps: draw the FBD, then apply the equilibrium equations.

Meriam's Fig. 3/2 shows four cases: a truss lifted off its supports as one rigid body (a pin at A gives A_x, A_y; a rocker at B gives one vertical force); a cantilever with its wall forces F, V, M and its weight mg; a beam resting on a smooth corner (normal to the beam) and pinned at B; and a whole mechanism isolated as one unit, where the internal forces of the cable and links never appear, so the way the cable is fastened inside cannot change the external reactions. A reaction may come out zero, which means that force does not exist for that loading.

The equation (physical law: Newton's third law, which every removed contact uses):

Marghitu writes it \mathbf F_{ij} = -\mathbf F_{ji} (Fig. 4.3: F_{10x} = -F_{01x}).

Worked number. If the beam's pin reaction comes out A_x = -40\,\mathrm N with the arrow drawn to the right, the pin pushes the beam 40\,\mathrm N to the left, and the beam pushes the pin 40\,\mathrm N to the right. Same magnitude, three sentences.

Real example. A car lift's arm is isolated with its pins and hydraulic ram cut away. Inspectors check its pins against the forces on exactly this diagram; a force left off it is a pin that was never checked.

Plausible wrong answer. "If I guess a sense wrongly, the answer is wrong." The answer comes out negative, and the magnitude is right. The only error is to reverse an assumed sense halfway through the working.

Check. On the mechanism of Meriam's Fig. 3/2 example 4, why does the cable inside not appear? (It is internal: both its pulls lie inside the boundary and cancel.)

Narration spine. A pencil traces the boundary of the beam, starting at A. At each contact it pauses: the pin dissolves into two green arrows, the roller into one. The loads stay. The axes appear last. A counter in the corner reads "unknowns: 3".

Bridge. In a plane, supports stop at most two slidings and one turning. In space a body can slide three ways and turn three ways. What do supports do there?

Nugget 06: In Space a Support Can Stop Six Motions

Claim. A body in space can translate along three axes and rotate about three. Each support supplies a force component for each translation it prevents and a couple component for each rotation it prevents: from one unknown (a smooth surface) to six (a weld).

The picture. The switchgear shaft, isolated. Its axis is z; the crank lies in the x-y plane.

Ask of each support which motions it stops. The radial bearing A holds the shaft against moving sideways (x and y) but lets it slide along its axis and turn: two forces. The thrust bearing B also stops axial sliding: three forces. Neither stops the shaft spinning about z; the latch does, with the couple M_L. The bearings do not resist tilting either, because each is short compared with the 200\,\mathrm{mm} between them. Tilting is resisted by the pair of bearings pushing in opposite directions, which is a couple made of forces already drawn.

Check with numbers. Unknowns: 2 + 3 + 1 = 6, and a body in space has six equations (Nugget 07). The spring force is known: \mathbf F = -491.0\mathbf i - 94.5\mathbf j\ \mathrm N, along the line from the crank pin (30.0, 52.0)\,\mathrm{mm} to the anchor (-240, 0)\,\mathrm{mm}, a distance of 275.0\,\mathrm{mm}. A ball-and-socket joint whose force came out as (120, -50, 300)\,\mathrm N carries \sqrt{120^2 + 50^2 + 300^2} = 327.0\,\mathrm N: three components, one force.

The idea. Meriam's Fig. 3/8 extends Fig. 3/1 to space:

(Meriam's thrust bearing is drawn on a y-axis shaft.) Meriam notes that the bearing couples must in some cases be assumed zero to keep the problem statically determinate. Marghitu (4.2.2) says the same in general form: a bearing can exert forces in all three directions and couples about the two axes perpendicular to the shaft, but never a couple about the shaft's own axis; many radial bearings exert no couples, or no couples and no axial force. The ball and socket lets the body rotate freely, so it exerts a force but no couple. Meriam adds advice for 3D diagrams: draw forces in their true physical sense when it can be seen, because the diagram is then closer to the real problem; assume a sense only when it cannot.

The equation (definition: the count of unknowns a support adds):

Worked number. If both shaft bearings were thrust bearings, B_z and an A_z would both resist axial load: seven unknowns for six equations, and nothing could fix how the axial load splits between them. That is why machines put the thrust bearing at one end only: the other end must float to allow for thermal growth.

Real example (one you can see). Pull a heavy door open and look at the hinges: the top hinge's screws are pulled out of the frame and the bottom hinge is pushed in. Two hinges, each a short pin, resist the door's tipping moment as a pair of opposite forces, just as the shaft's two bearings do.

Plausible wrong answer. "Each bearing also resists tilting, so give each two couples." Then the shaft has ten unknowns and only six equations. For short bearings in a pair the couples are near zero; model them as point supports and let the pair resist tilting.

Check. A sign is held by a ball-and-socket joint at one corner and three cables. How many unknowns? (3 + 3 = 6.)

Narration spine. The shaft in perspective, with its bearings, crank and spring. The shaft is nudged along each of the six motions in turn; at each, the support that stops it lights up and grows an arrow (or a curl, for the latch). The counter climbs to six. A ghost seventh arrow at A appears and is crossed out.

Bridge. Six unknowns and six equations sounds like a match. But is counting enough, and what if we cut the machine somewhere else?

Nugget 07: Cut Anywhere, Then Count Unknowns Against Equations

Claim. Any part of a machine can be isolated, and internal forces become external the moment the cut passes through them. For each diagram, a body in a plane gives at most three independent equations and a body in space six; more unknowns than that, or supports that cannot stop some motion, means the equations alone will not do.

The picture. A slider-crank (after Marghitu's Fig. 4.10): crank AB pinned to the frame at A, connecting rod BC, slider at C pushed by a force F.

Isolated as a whole, the pins at B and C are inside: their forces cancel in pairs and vanish. Only the frame's pin at A (A_x, A_y) and the guide's normal force N act, with the applied F: three unknowns. Isolate the slider alone and the cut passes through the pin C: the rod's force on the slider, C_x and C_y, is now external. With N, three unknowns again. Each diagram is a different question, and each gets its own equations.

Check with numbers. In a plane: the beam of Nugget 05 on a pin and a roller, 2 + 1 = 3 unknowns, three equations: determinate. On two pins, 2 + 2 = 4 unknowns: degree of redundancy 4 - 3 = 1. A cantilever with a fixed end and a roller at the far end: 3 + 1 = 4, degree 1. In space: a fixed end plus a ball and socket, 6 + 3 = 9 unknowns, degree 9 - 6 = 3. And a beam on three rollers on a level floor, 1 + 1 + 1 = 3 unknowns, has three vertical forces: a load with a horizontal component, such as 1\,\mathrm{kN} at 30^\circ to the floor with its 0.866\,\mathrm{kN} along the beam, cannot be resisted. The count matches; the beam still slides.

The idea. Marghitu (4.3): a free-body diagram can isolate a whole system or any part of it; \mathbf F_{ij} is the force by link i on link j, and \mathbf F_{ij} = -\mathbf F_{ji}. A pin joint between two links has two unknowns (both components); a slider joint has two, the magnitude of the force normal to the slide and the position x at which it acts. A planar FBD gives three independent scalar equations, \Sigma F_x = 0, \Sigma F_y = 0, \Sigma M_P = 0; moments about further points give more equations, but they are not independent of the first three. A body in space gives six: \Sigma F_x = \Sigma F_y = \Sigma F_z = 0, \Sigma M_x = \Sigma M_y = \Sigma M_z = 0. So at most three (planar) or six (space) unknown forces or couples per diagram. A body has redundant supports when it has more than the minimum needed for equilibrium; designers use them for strength and safety. The degree of redundancy is the number of reactions minus the number of independent equations. A body has improper supports if it will not stay in equilibrium under the loads: it moves when loaded. Meriam treats the same ideas as adequacy of constraints in 3/3 and 3/4 (S11, S12).

The equation (definition):

It is valid only when the supports are proper. Counting cannot see that three parallel rollers miss a direction.

Worked number. The switchgear shaft: six unknowns, six equations, degree 0, and the supports are proper: A and B stop every translation and every tilt, the latch stops the spin. With the thrust bearing replaced by a radial one, there would be five unknowns, and nothing would stop axial sliding: improper, even though five is less than six.

Real example. Bridges with three or more spans over continuous girders are deliberately redundant: lose one bearing and the girder can still stand. The price is that the reactions cannot be found by statics alone; they depend on how the girder bends (a mechanics-of-materials question).

Plausible wrong answer. "Take moments about a fourth point to get a fourth equation." It is a combination of the first three and adds nothing. A planar diagram never yields more than three independent equations.

Check. A beam has a built-in end and a pin at the other end. Degree of redundancy? (3 + 2 - 3 = 2.)

Narration spine. The slider-crank moves, then freezes. A boundary is drawn round all of it: pins B and C glow and their paired arrows cancel. The boundary shrinks to the slider: the pair at C splits, and one half stays as C_x, C_y. A tally of unknowns against equations runs beside each diagram; under the three rollers it reads "3 = 3" while the beam slides away.

Bridge. The diagram is drawn and counted. Back to the switchgear shaft.

Synthesis

The switchgear shaft's free-body diagram is in Nugget 06. Its boundary runs round the shaft and crank. Every contact has been replaced: the radial bearing A by A_x, A_y; the thrust bearing B by B_x, B_y, B_z; the latch by the couple M_L about the shaft axis; the spring by a known 500\,\mathrm N along the line from the crank pin to the anchor. The weight, 7.26\,\mathrm N, is left off by a stated choice. There are six unknowns and six equations, and the supports stop every motion. So the bearing forces can be found, and the "250\,\mathrm N each" guess can be tested, which is S11's job. Every statics problem is solved on a free-body diagram, and the diagram is right only if each force on it acts on the chosen body, comes from a contact crossed by its boundary or from gravity, and points the way that contact can push.

• Central idea: isolate one system, and replace every contact across its boundary by the force (and couple) it can exert; then count.
• Key relationship: \mathbf R = \mathbf 0, \mathbf M = \mathbf 0 for the forces on the FBD; at most three unknowns per planar diagram and six per spatial one.
• Most important assumption: each support is idealised: smooth, frictionless pins, short bearings with no couples, light springs. Real contacts carry a little of everything; the model keeps what matters.
• Open question: with six equations and six unknowns, how do we solve them quickly? And which way does each bearing push? (S11 and S12.)

Computed views

The same check, written twice so you can work in whichever language you prefer. Both print the same verdicts.

Counting unknowns is not quite enough, and Nugget 07's improper supports show why. A computer does the full test. Write each unknown support force as a unit force at its point. Its column in the equilibrium matrix holds what it adds to \Sigma F_x, \Sigma F_y and \Sigma M (six sums in space). The rank of that matrix then says:

• rank equal to the number of equations and to the number of unknowns: proper and determinate;
• rank short of the number of equations: improper, because some motion is unresisted;
• more unknowns than the rank: indeterminate.

Python (numpy). S10_supports.py builds the matrix for five plane supports and for three versions of the switchgear shaft, and reads off matrix_rank.

Plane bodies, a 5 m beam
  pin A + roller B: 3 unknowns, rank 3: proper and determinate
  pin A + pin B: 4 unknowns, rank 3: indeterminate to degree 1
  three vertical rollers: 3 unknowns, rank 2: improper: some motion is not resisted
  three links through one point: 3 unknowns, rank 2: improper: some motion is not resisted
  built-in end: 3 unknowns, rank 3: proper and determinate
The switchgear shaft along y, bearings at y = 0 and 0.4 m
  thrust + radial + latch: 6 unknowns, rank 6: proper and determinate
  radial + radial + latch: 5 unknowns, rank 5: improper: some motion is not resisted
  thrust + thrust + latch: 7 unknowns, rank 6: indeterminate to degree 1
Spring and weights
  torsion spring 0.5 N m/rad wound 90 deg: M = 0.785 N m
  crate 40 kg: 392.4 N; diver 70 kg: 686.7 N

Three supports through one point have three unknowns, yet the rank is 2. Only the matrix sees that they cannot stop a rotation about that point.

MATLAB (Octave-compatible). S10_supports.m does the same with rank.

pin A + roller B         3 unknowns, rank 3
pin A + pin B            4 unknowns, rank 3
three vertical rollers   3 unknowns, rank 2
three concurrent links   3 unknowns, rank 2
built-in end             3 unknowns, rank 3
thrust + radial + latch: rank 6 of 6
radial + radial + latch: rank 5 of 6

Traps

• Drawing forces the body exerts, not forces on it. The lamp pulls down on the chain; that force belongs on the chain's diagram, not the lamp's. (From Nugget 01.)
• Keeping internal forces on a combined diagram. Inside one boundary they cancel in pairs; drawing one of them adds a force that is not there. (From Nugget 01.)
• Letting a cable push or a smooth surface pull or grip. A cable gives tension along itself, a smooth surface a push along the normal; neither has any other component. (From Nugget 02.)
• Drawing a roller's force vertical on an incline. It is normal to the incline: on 20^\circ, 1.5\,\mathrm{kN} has a 0.513\,\mathrm{kN} horizontal part. (From Nugget 02.)
• Giving a free pin a couple, or a weld none. A pin free to turn gives two forces; a built-in end gives F, V and M. (From Nugget 03.)
• Spring force from length instead of deformation. F = kx with x from the neutral length. (From Nugget 04.)
• Leaving weight off without saying so. Compute it and compare: 7.26\,\mathrm N against 500\,\mathrm N may be dropped, a beam's own weight often may not. (From Nugget 04.)
• Changing an assumed sense mid-solution. Assume once, keep it; a minus sign at the end reverses it. (From Nugget 05.)
• Giving every bearing couples. Two short bearings resist tilt as a force pair; giving each two couples makes the shaft indeterminate. (From Nugget 06.)
• Trusting the count alone. Three rollers give three unknowns and still let the body slide; check every motion is stopped. (From Nugget 07.)

Self-check

1. A 3\,\mathrm{kg} lamp hangs on a 0.8\,\mathrm{kg} chain. Draw the FBD of the lamp alone and of lamp and chain together, and give every force.
2. A bell crank is pinned at A. A 25\,\mathrm{kg} mass hangs from one arm whose line is 0.24\,\mathrm m from the pin; a horizontal cable on the other arm has its line 0.12\,\mathrm m from the pin. Draw the FBD of the crank. Which way does the cable act, and how large is its pull? Which forces at A are still unknown?
3. A uniform 40\,\mathrm{kg} crate leans against a smooth vertical wall at its top corner A and stands on a rough floor at its bottom corner B. List the forces on its FBD and the number of unknowns.
4. A student draws the FBD of a 50\,\mathrm{kg} lawn roller being pushed up a 15^\circ slope by a handle force P with only P, the weight and a normal force. What is missing? Give the weight's components along and normal to the slope.
5. A prybar rests on a rough floor and lifts a block A whose underside is smooth. Describe the force the block exerts on the bar and the force the floor exerts on it. How many unknowns at the floor?
6. A bent rod is welded to a wall at A and loaded by two forces and a couple. A student shows only a vertical force at A. Correct the diagram.
7. A compression spring of stiffness 3\,\mathrm{kN/m} is 40\,\mathrm{mm} shorter than its neutral length. A torsional spring of k_T = 2\,\mathrm{N\cdot m/rad} is wound 30^\circ. What does each exert?
8. A pin reaction is found to be R_x = -240\,\mathrm N, R_y = 320\,\mathrm N with both drawn positive. Give the force's magnitude and direction. What does the pin feel?
9. A rectangular sign is held by a ball-and-socket joint at one corner and three cables. A second design adds a fourth cable. Count the unknowns and the degree of redundancy of each.
10. A beam is built in at one end and rests on a roller at the other. What is its degree of redundancy?
11. A beam rests on three rollers on a level floor and carries 1\,\mathrm{kN} at 30^\circ to the floor. Why can it not be in equilibrium, although it has three unknowns?
12. For Marghitu's slider-crank, how many unknowns are on the FBD of the crank AB alone, with the driving couple M unknown?

Fully worked solutions

1. Lamp alone: weight 3 \times 9.81 = 29.43\,\mathrm N down at its centre of gravity, chain tension up. The only two forces are collinear and vertical, so T = 29.43\,\mathrm N. Lamp and chain: weights 29.43\,\mathrm N and 0.8 \times 9.81 = 7.85\,\mathrm N down, ceiling force up; the chain–lamp force is internal. Ceiling force 3.8 \times 9.81 = 37.28\,\mathrm N.

2. Isolate the crank: weight of the mass 25 \times 9.81 = 245.25\,\mathrm N down on its arm (via the hanger); cable tension T along the cable, away from the crank; pin A_x, A_y. The weight turns the crank one way about A, so the cable must pull the other way, and a cable can only pull: its sense is fixed. Zero moment about A: T \times 0.12 = 245.25 \times 0.24, T = 490.5\,\mathrm N. A_x and A_y remain (S11 finds them: the pin must balance T horizontally and the weight vertically).

3. Weight 40 \times 9.81 = 392.4\,\mathrm N down at the centre; at A the smooth wall pushes horizontally away from the wall (1 unknown); at B the rough floor pushes up (N_B) and has a friction force F_B along the floor (2 unknowns). Three unknowns. The friction must point towards the wall, because without it the foot would slide out.

4. The slope is rough enough for the roller to roll, so it also exerts a tangential (friction) force along the slope at the contact, and the normal force must be drawn normal to the slope, not vertical. Weight 50 \times 9.81 = 490.5\,\mathrm N; along the slope 490.5\sin15^\circ = 127.0\,\mathrm N, normal 490.5\cos15^\circ = 473.8\,\mathrm N.

5. The block's smooth underside pushes on the bar normal to that surface, that is vertically down: one unknown. At the rough floor the bar feels an upward normal force and a horizontal friction force: two unknowns. With the hand force P known, the bar has three unknowns.

6. A weld is a built-in support: replace the single vertical force by A_x, A_y and a couple M_A. The two forces and the couple stay as given. Three unknowns.

7. Spring: F = 3000 \times 0.040 = 120\,\mathrm N, compressive: it pushes the bodies at its ends apart. Torsional spring: \theta = 30^\circ = 0.5236\,\mathrm{rad}, M = 2 \times 0.5236 = 1.047\,\mathrm{N\cdot m}, a couple opposing the winding.

8. R = \sqrt{240^2 + 320^2} = 400\,\mathrm N at \operatorname{atan2}(320, -240) = 126.87^\circ from +x: left and up on the body. By Newton's third law the pin feels 400\,\mathrm N right and down.

9. Ball and socket 3, each cable 1 (direction known): 3 + 3 = 6 unknowns, six equations, degree 0. With a fourth cable, 7 unknowns: degree 1, statically indeterminate (if the supports are otherwise proper).

10. Built-in end 3, roller 1: 4 - 3 = 1.

11. Each roller pushes vertically, so no support has a horizontal component; the load's horizontal part, 1\cos30^\circ = 0.866\,\mathrm{kN}, is unopposed and \Sigma F_x = 0 fails. The supports are improper: the count 3 = 3 hides the missing direction.

12. The crank is pinned to the frame at A (A_x, A_y) and to the rod at B (B_x, B_y), with the driving couple M: five unknowns for three equations. The crank cannot be solved alone; it needs the rod and slider diagrams too (Part 4's method for frames and machines).