Study Companion: S11 — Equilibrium in Two Dimensions

S10 built the free-body diagram: one isolated body, every contact replaced by the force it can exert, the unknowns counted. This lesson turns the diagram into answers. For a body whose forces lie in one plane, equilibrium is exactly three scalar equations, so a well-supported body has at most three unknown reactions, and three equations find them. The skill is not in the algebra but in choosing which three equations to write: a good choice gives one unknown per equation, and spotting special force systems (concurrent, parallel, two forces, three forces) cuts the work further. The lesson also shows when the three equations are not enough: too many supports, or supports badly placed.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 3/3 ("Equilibrium Conditions"), with Sample Problems 3/1–3/4. Marghitu, Dupac and Madsen, Statics with MATLAB, sections 4.1 ("Equilibrium Equations") and 4.4 ("Two-Force and Three-Force Members"), with Examples 4.2 and 4.7. Harper, Solving Statics Problems in MATLAB, "2D Equilibrium" (Problem 3/25 and Sample Problem 3/4). Counter-clockwise moments are positive. US-unit data are converted to SI where they first appear.

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

S10 left the switchgear shaft with its free-body diagram drawn: a radial bearing A, a thrust bearing B 200\,\mathrm{mm} further on, and the crank 60\,\mathrm{mm} beyond B, where the spring pulls with 500\,\mathrm N. Seen square to both the shaft and the spring's line, the spring pull, and both bearing forces, all lie in one plane.

The bearings take the spring's pull, so the natural guess is that they share it: 250\,\mathrm N each. Is that right? If not, which bearing works harder, and by how much? And what holds the shaft from turning? Three equations will settle it in the Synthesis, and the guess fails badly.

Nugget 01: Three Equations Hold a Body in a Plane

Claim. A body acted on by forces in one plane is in equilibrium if and only if \Sigma F_x = 0, \Sigma F_y = 0 and \Sigma M_O = 0 about any one point O. Zero force plus zero moment about one point already guarantees zero moment about every point.

The picture. Take moments about a second point Q instead of P. Each force's arm from Q is its arm from P plus the fixed step \mathbf r_{QP}.

Summing (\mathbf r_{QP} + \mathbf r_{PA_i})\times\mathbf F_i over all forces splits into two parts: \mathbf r_{QP}\times\Sigma\mathbf F, which is zero because the forces add to zero, and \Sigma\mathbf r_{PA_i}\times\mathbf F_i, which is the moment about P, also zero. The moment about Q is therefore zero, wherever Q is. So a fourth equation, a moment about another point, can only repeat what the three already say.

Check with numbers. Forces of 30\,\mathrm N up at (0, 0) and 30\,\mathrm N down at (2, 0)\,\mathrm m add to zero. About P(0, 0): M_P = 2 \times (-30) = -60\,\mathrm{N\cdot m}. About Q(5, 3)\,\mathrm m: the first force gives (-5)(30) - (-3)(0) = -150, the second (-3)(-30) - (-3)(0) = 90, total -60\,\mathrm{N\cdot m}. Same, as the picture says: with \Sigma\mathbf F = \mathbf 0 the moment does not depend on the point. Add a +60\,\mathrm{N\cdot m} couple and the body is in equilibrium about both points at once.

The idea. Meriam (3/3) writes S10's vector equations in planar scalar form, Eqs. 3/2. They are necessary (any imbalance would accelerate the body) and sufficient (once they hold, nothing is left unbalanced). A body moving at constant velocity has zero acceleration, so it is in translational equilibrium too. The three equations are independent: one can hold without the others. A block sliding faster and faster along a level floor has \Sigma F_y = 0 but \Sigma F_x \ne 0; a flywheel spinning up about its fixed mass centre has both force sums zero but \Sigma M \ne 0. Marghitu (4.1) states the conditions as \Sigma\mathbf F = \mathbf 0 and \Sigma\mathbf M_P = \mathbf 0 for every P, and proves the step in the picture: one point is enough.

The equation (physical law: Newton's second law with zero acceleration, in components; Meriam's Eqs. 3/2):

Forces in N, moments in N·m, O any point on or off the body. Valid for a rigid body (or any body treated as one) whose forces lie in, or project onto, one plane.

Worked number (Marghitu's Example 4.7). A uniform 8\,\mathrm m cantilever OD of mass 250\,\mathrm{kg} is built in at O. It carries 3000\,\mathrm N at A (x = 1\,\mathrm m, up and to the right at 45^\circ), 3500\,\mathrm N at B (x = 3\,\mathrm m, up and to the left at 30^\circ), 4000\,\mathrm N at D (x = 8\,\mathrm m, down and to the left at 30^\circ) and a 10\,\mathrm{kN\cdot m} counter-clockwise couple. Three unknowns: F_{Ox}, F_{Oy}, M_O, drawn positive.

• \Sigma F_x = 0: F_{Ox} + 3000\cos45^\circ - 3500\cos30^\circ - 4000\cos30^\circ = 0, so F_{Ox} = -2121.3 + 3031.1 + 3464.1 = 4373.9\,\mathrm N.
• \Sigma F_y = 0: F_{Oy} + 2121.3 + 1750 - 2000 - 2452.5 = 0, so F_{Oy} = 581.2\,\mathrm N (the weight, 250 \times 9.81 = 2452.5\,\mathrm N, acts at x = 4\,\mathrm m).
• \Sigma M_O = 0: M_O + 1(2121.3) - 4(2452.5) + 3(1750) + 10\,000 - 8(2000) = 0, so M_O = 8438.7\,\mathrm{N\cdot m}.

So F_{Ox} = 4.374\,\mathrm{kN}, F_{Oy} = 0.581\,\mathrm{kN}, M_O = 8.439\,\mathrm{kN\cdot m} counter-clockwise, as Marghitu's program prints.

Real example. A balcony slab cantilevered from a building is this beam. The wall must supply the slab's vertical load and the bending moment at the root, which is why balcony reinforcement is heaviest where the slab meets the wall.

Plausible wrong answer. "If the forces add to zero, the body is in equilibrium." Two equal and opposite forces on different lines add to zero and still form a couple: the check above has \Sigma\mathbf F = \mathbf 0 and M = -60\,\mathrm{N\cdot m}, so it spins.

Check. A body has \Sigma F_x = \Sigma F_y = 0 and \Sigma M_A = 0. Is \Sigma M_B zero for every other point B? (Yes; that is the lemma.)

Narration spine. A body with three arrows. A point P and the moment readout 0. A second point Q is dragged around the screen; the vector \mathbf r_{QP} and the dotted arms follow it, and the readout stays at 0. Remove one force: the force sum shows a gap, and now the readout changes as Q moves.

Bridge. Three equations is the most a planar body ever needs. Some force systems need fewer, and recognising them saves work.

Nugget 02: Fewer Kinds of Force, Fewer Equations

Claim. If all the forces lie on one line, one equation is enough. If they meet at one point, or are all parallel, two are enough. Only a general planar system needs all three.

The picture. Meriam's four categories (Fig. 3/3).

For collinear forces, every force lies along x: there is nothing to sum along y, and every force has zero arm about any point on the line. For concurrent forces, take moments about the meeting point: every arm is zero, so the moment equation is automatically satisfied. For parallel forces along x, \Sigma F_y is automatically zero, but the forces can still turn the body, so one moment equation remains.

Check with numbers (Meriam's Sample Problem 3/1). A bridge-truss joint carries 8\,\mathrm{kN} to the right, 16\,\mathrm{kN} to the left, 3\,\mathrm{kN} down, an unknown member force T pulling at 40^\circ above the horizontal and an unknown C pushing at 20^\circ from the vertical. The forces are concurrent: two equations.

that is, 0.766T + 0.342C = 8 and 0.643T - 0.940C = 3, so T = 9.09\,\mathrm{kN} and C = 3.03\,\mathrm{kN}. Meriam then shows three more routes to the same answer. Axes x' along T and y' across it give one equation in C alone: -C\cos20^\circ - 3\cos40^\circ - 8\sin40^\circ + 16\sin40^\circ = 0, so C = 3.03\,\mathrm{kN} directly. Vector algebra, setting the \mathbf i and \mathbf j coefficients of \Sigma\mathbf F = \mathbf 0 to zero, gives the same pair of equations. And the force polygon: lay the known forces head to tail, then close it with lines in the directions of T and C; where they meet fixes both lengths.

The idea. Meriam's categories: (1) collinear forces need only the force equation along their line; (2) forces concurrent at a point O need only the two force equations, since the moment about O is zero (this includes the equilibrium of a particle); (3) parallel forces need the force equation along their direction and one moment equation about an axis normal to their plane; (4) a general planar system needs all three. Marghitu (4.1) adds that for concurrent forces with no couples the only condition is that the forces add to zero. Meriam's hint: choose axes to suit the problem, for instance one perpendicular to an unknown, so that unknown drops out.

The equation (mathematical identity: Eqs. 3/2 with the automatically satisfied ones removed):

Worked number (Meriam's Sample Problem 3/2, a 1000\,\mathrm{lb} load, 4448\,\mathrm N, on three pulleys). Each pulley turns freely, so moments about its centre give equal tensions on its two sides whatever its radius: T_1 r = T_2 r. Pulley A (parallel forces) gives 2T_1 = 4448, T_1 = T_2 = 2224\,\mathrm N; pulley B gives T_3 = T_4 = 1112\,\mathrm N; and the cable tension is T = T_3 = 1112\,\mathrm N (250\,\mathrm{lb}). Pulley C's bearing (concurrent forces) then carries F_x = 1112\cos30^\circ = 963\,\mathrm N and F_y = 556\,\mathrm N, a total of 1112\,\mathrm N.

Real example. Every joint of a pin-jointed truss is a concurrent system, two equations per joint. Part 4's method of joints is nothing more than Sample Problem 3/1 repeated at each joint.

Plausible wrong answer. "For the truss joint, take moments too, to get a third equation for a third unknown." About the joint the moment equation reads 0 = 0. A concurrent system can never fix more than two unknowns.

Check. Three forces act on a ring and all pass through its centre. How many unknowns can equilibrium find? (Two.)

Narration spine. Four panels light in turn. In the collinear one, the y and moment equations grey out. In the concurrent one, the moment arms shrink to the meeting point and the moment equation greys out. In the parallel one, \Sigma F_y greys out. The truss joint's polygon then closes, and T and C are read off: 9.09 and 3.03 kN.

Bridge. Some bodies, not just force systems, are special: a link loaded at only two points can push or pull in only one direction.

Nugget 03: Two-Force Members Pull Straight Along the Line of Their Pins

Claim. A body acted on by only two forces is in equilibrium only if the two forces are equal, opposite and collinear: both along the line joining their points of application, whatever the body's shape.

The picture. A bent link pinned at both ends, with no load in between.

\Sigma\mathbf F = \mathbf 0 forces the two to be equal and opposite. But equal and opposite forces on different lines are a couple (S05), and a couple cannot be balanced by nothing, so \Sigma M = 0 forces them onto one line, the line AB. The bend in the link is irrelevant: only the two pins matter.

Check with numbers (Marghitu's Problem 4.5). Link 1, AB of length l, is pinned to a wall at A; link 2 is pinned to the wall at C and to link 1 at B, with the angle ABC = \theta. A force P at D, AD = 2l/3, acts at \theta to the horizontal. Link 2 carries no load between its pins, so it is a two-force member: its force on link 1 acts along CB, one unknown F. Moments about A on link 1: F\sin\theta \cdot l counter-clockwise against P\sin\theta \cdot 2l/3 clockwise, so F = 2P/3. With P = 1000\,\mathrm N, F = 666.7\,\mathrm N in compression, for any \theta and any l.

The idea. Meriam (3/3): for a two-force member to be in equilibrium the forces must be equal, opposite and collinear, and the member's shape does not matter (Fig. 3/4); its weight is taken as negligible compared with the applied forces. Marghitu (4.4) defines it as a body whose forces and moments are equivalent to two forces at different points, and gives the same argument: \mathbf F_A = -\mathbf F_B from the force sum, and the couple they would otherwise form must vanish, so they act along the line AB. Their magnitude is not found by the member itself; it needs the rest of the structure.

The equation (mathematical identity: Eqs. 3/2 applied to two forces):

So a two-force member contributes one unknown (its tension or compression), not the two of a general pin.

Worked number. In S10's slider-crank, the connecting rod BC carries forces only at its two pins: a two-force member. The slider's free-body diagram then has the rod force along the rod, one unknown, not S10's C_x and C_y. With the rod at 15^\circ to the guide and a 400\,\mathrm N load on the slider along the guide: rod force 400/\cos15^\circ = 414.1\,\mathrm N and guide force 400\tan15^\circ = 107.2\,\mathrm N.

Real example. The hydraulic cylinders on an excavator arm are pinned at both ends and carry no side loads: two-force members. Their force is along the cylinder axis, which is why it equals the oil pressure times the piston area.

Plausible wrong answer. "A curved link pushes along its own curve, so its force at each pin is tangent to the link." The force is along the straight line between the pins; for a sharply bent link that can be far from the link's own direction at the pin.

Check. A light, curved strut is pinned at (0, 0) and (3, 4)\,\mathrm m. What is the direction of its force? (Along (3, 4)/5: at 53.1^\circ to x.)

Narration spine. A bent link with two pins. Two random arrows appear at the pins; the force readout shows a gap, and they swing into equal and opposite; the link spins (couple); they swing onto the dashed line AB and it stops. The link is then reshaped into a straight bar, then an S-curve: the arrows never move.

Bridge. Two forces must be collinear. What must three forces be?

Nugget 04: Three Forces in Equilibrium Meet at a Point

Claim. If a body is in equilibrium under three forces, their lines of action meet at one point (or are all parallel). So knowing two lines fixes the third, and the three forces close a triangle.

The picture. Meriam's Sample Problem 3/4, a jib crane: a 5\,\mathrm m beam pinned at A, held by a cable at B at 25^\circ, carrying its weight and a 10\,\mathrm{kN} load.

Combine the beam's weight and the load into one 14.66\,\mathrm{kN} force. The beam now has three forces: that one, the cable tension T along the cable, and the pin force \mathbf A of unknown direction. If \mathbf A did not pass through O, where the other two lines cross, it would have a moment about O that nothing balances. So \mathbf A points from A at O, and the three forces, laid head to tail, must close.

Check with numbers. Measure x from the pin A. The weight 95 \times 5 \times 9.81 = 4.66\,\mathrm{kN} acts at 2.38\,\mathrm m (the beam's middle, 0.12\,\mathrm m right of the pin's line) and the load at 3.38\,\mathrm m. Their resultant, 14.66\,\mathrm{kN}, acts at (4.66 \times 2.38 + 10 \times 3.38)/14.66 = 3.062\,\mathrm m (S06's principle of moments). The cable runs from B (4.88\,\mathrm m along, 0.25\,\mathrm m above A) at 25^\circ; at x = 3.062\,\mathrm m it is 0.25 + (4.88 - 3.062)\tan25^\circ = 1.098\,\mathrm m above A. So O = (3.062, 1.098)\,\mathrm m and \mathbf A acts at \arctan(1.098/3.062) = 19.72^\circ. The closed triangle gives T = 19.61\,\mathrm{kN} and A = 18.88\,\mathrm{kN}.

The idea. Meriam (3/3): a three-force member is in equilibrium only if the three lines of action are concurrent; otherwise one force would have a moment about the intersection of the other two. The exception is three parallel forces, which meet "at infinity". The force polygon must close (Fig. 3/5), which makes concurrency a tool for graphical solutions, and a body with more than three forces can often be reduced to three by combining the known ones. Marghitu (4.4) proves the full theorem: the three forces are coplanar (the moment about the line through two of the application points shows the third lies in their plane) and either concurrent or parallel.

The equation (geometric relationship, from \Sigma M_O = 0 about the meeting point and \Sigma\mathbf F = \mathbf 0):

Worked number (Sample Problem 3/4 by algebra, as Meriam does it first). Moments about A remove A_x and A_y. Split T at B into components (Varignon): T\cos25^\circ has arm 0.25\,\mathrm m, T\sin25^\circ has arm 4.88\,\mathrm m.

2.2889T = 44.89, so T = 19.61\,\mathrm{kN}. Then \Sigma F_x = 0: A_x = 19.61\cos25^\circ = 17.77\,\mathrm{kN}; \Sigma F_y = 0: A_y = 14.66 - 19.61\sin25^\circ = 6.37\,\mathrm{kN}; A = 18.88\,\mathrm{kN} at \arctan(6.37/17.77) = 19.72^\circ, the direction concurrency gave.

Real example. A wall-mounted jib crane in a workshop is exactly this. When the trolley moves along the beam, the point O slides along the cable's line, and the pin force swings to follow it (Nugget 07).

Plausible wrong answer. "The pin force at A acts along the beam, like the cable acts along the cable." The beam is not a two-force member: its weight and the load act between its ends. The pin force is 19.72^\circ above the beam.

Check. A ladder leans on a smooth wall and stands on a rough floor; its weight acts at its middle. Where must the floor's total force point? (Through the point where the weight's vertical line meets the wall's horizontal force.)

Narration spine. The crane beam with its two vertical loads; they merge into one green arrow at 3.062 m. The cable's line is extended, and the vertical line of the merged load rises to meet it at O. A line from A to O appears and the pin force snaps onto it. The three arrows slide head to tail into a triangle, labelled 14.66, 19.61 and 18.88 kN.

Bridge. Moments about A found T in one step because the two other unknowns passed through A. Can we always write equations that isolate one unknown each?

Nugget 05: Any Three Independent Equations Will Do

Claim. The three equations need not be two force sums and one moment sum. Two moment sums and one force sum, or three moment sums, work equally well, provided the points are chosen so the equations are independent. Choosing moment centres where unknowns cross gives one unknown per equation.

The picture. Why \Sigma M_A = 0, \Sigma M_B = 0 and \Sigma F_x = 0 leave no room for a resultant (Meriam's Fig. 3/6).

Zero moment about A allows only a resultant force through A (it cannot be a couple, which has the same moment about every point). Zero moment about B as well puts that force on the line AB. If it also has no component along x, and AB is not perpendicular to x, it has nothing left: R = 0. With three moment centres A, B, C not on one line, the force would have to lie on AB and pass through C as well, which is impossible unless it is zero.

Check with numbers (Marghitu's Example 4.2). A 10\,\mathrm{kg} mass hangs at A from two cables anchored at B and C on a ceiling, a_1 = 3\,\mathrm m left and a_2 = 5\,\mathrm m right of the point O above A, with A at a_3 = 1\,\mathrm m below the ceiling. Marghitu's first method solves the two force equations together. His second takes moments about B, where T_{AB} has no arm: one equation, one unknown,

and moments about C likewise give T_{AB} = 98.1 \times 5\sqrt{10}/8 = 193.887\,\mathrm N. Both methods agree.

The idea. Meriam (3/3) gives the two alternative sets: \Sigma F_x = 0, \Sigma M_A = 0, \Sigma M_B = 0 with A and B not on a line perpendicular to x; and \Sigma M_A = \Sigma M_B = \Sigma M_C = 0 with A, B, C not collinear. Equations that are not independent carry duplicated information: three moment equations about three points on one line can at best find two unknowns, and the third equation reduces to 0 = 0. Meriam's approach to solving problems: (1) identify knowns and unknowns; (2) choose the body unambiguously and draw its complete FBD; (3) choose convenient right-handed axes and moment centres, generally a point through which as many unknowns pass as possible; (4) state the governing principle before each step; (5) match the number of independent equations to the number of unknowns; (6) solve and check, ideally against an estimate made beforehand.

The equation (mathematical identity: equivalent forms of Eqs. 3/2):

Worked number (Meriam's Sample Problem 3/3). A uniform 100\,\mathrm{kg} I-beam, 8\,\mathrm m long, rests on rollers; a vertical cable at C, 6\,\mathrm m from end A, lifts end B to 3\,\mathrm m above A. Roller A and the weight are vertical, so the cable must be too: parallel forces. Moments about A: P(6\cos\theta) - 981(4\cos\theta) = 0, so P = 654\,\mathrm N, and \cos\theta cancels: the answer does not depend on the angle. \Sigma F_y = 0: R = 981 - 654 = 327\,\mathrm N. The angle comes from geometry, \sin\theta = 3/8, \theta = 22.0^\circ.

Real example. Weighing a car one axle at a time uses exactly this. The front scale reading and a moment about the rear contact locate the mass centre; no simultaneous equations needed.

Plausible wrong answer. "Three moment equations about three points always give three unknowns." Not if the points are collinear: about A, B and a C on line AB, the third equation is a combination of the first two.

Check. For a beam on a pin at A and a roller at B, which single equation gives the roller force directly? (\Sigma M_A = 0: both pin components pass through A.)

Narration spine. A body with a ghost resultant arrow. \Sigma M_A = 0 pins the arrow's line through A; \Sigma M_B = 0 swings it onto AB; \Sigma F_x = 0 shrinks it to nothing. Then the cable-hung mass: a moment centre is dropped on B, one cable's arm vanishes, and T_{AC} = 187.58\,\mathrm N appears alone.

Bridge. So far every body had just enough supports. What if it has too many, or the wrong ones?

Nugget 06: Counting Is Not Enough: Constraints Must Be Adequate

Claim. Three equations find three unknown reactions only if the supports are just enough and properly placed. More supports than needed make the body statically indeterminate; three supports badly placed leave it free to move.

The picture. Four ways to hold a plate with links (Meriam's Fig. 3/7). Each link is a two-force member, one unknown.

In (a), two links fix the corner A and the third, away from A, stops the plate turning about A. In (b) the third link's line runs through A: its force has no moment about A, so a load with a moment about A cannot be balanced. In (c) all three links are vertical: nothing resists a horizontal push. In (d) a fourth link holds a plate that was already fixed: four unknowns, three equations.

Check with numbers. A truss of span 4\,\mathrm m is pinned at both A and B and carries a horizontal 6\,\mathrm{kN} load 2\,\mathrm m above AB. Four unknowns. \Sigma M_A = 0 gives B_y \times 4 = 6 \times 2, so B_y = 3\,\mathrm{kN}; \Sigma F_y = 0 gives A_y = -3\,\mathrm{kN} (down). \Sigma F_x = 0 gives only A_x + B_x = -6\,\mathrm{kN}. How the 6\,\mathrm{kN} splits between the two pins is not decided by statics: it depends on how the truss deforms, and on whether it was forced to fit between pins slightly the wrong distance apart.

The idea. Meriam (3/3) defines a constraint as a restriction of movement: a roller, ball or rocker constrains normal to its surface only; a collar on a guide normal to the guide; a pin in both directions but not against rotation unless it is not free to turn; a fixed support against rotation as well. If the rocker under the truss of S10's Fig. 3/2 became a second pin, only the sum A_x + B_x could be found. If the pin at B in Fig. 3/2 example 3 were not free to turn, it would add a couple: four unknowns. A body with more supports than are needed is statically indeterminate; the removable supports are redundant, and their number is the degree of statical indeterminacy (Marghitu's degree of redundancy), the number of unknown external forces minus the number of independent equations. A body with just enough is statically determinate. Count before solving, or effort is wasted. The unknowns may be forces, couples, distances or angles. Adequacy: three constraints do not guarantee equilibrium. With them placed as in (a) the body is completely fixed and the constraints adequate (proper); as in (b) or (c) it is incompletely fixed under partial (improper) constraints; as in (d) it has excessive fixity with a redundant constraint.

The equation (definition):

Zero means determinate only if the constraints are also adequate.

Worked number. A beam is pinned at A and also pinned at B, 5\,\mathrm m away, and loaded vertically. Unknowns A_x, A_y, B_x, B_y: degree 1. For vertical loads, \Sigma F_x = 0 gives A_x = -B_x but not their value, which is set by any temperature change: a 5\,\mathrm m steel beam warmed 20\,\mathrm K would grow 12 \times 10^{-6} \times 20 \times 5000 = 1.2\,\mathrm{mm} if free, and the pins prevent it.

Real example. Long pipelines and rail bridges are deliberately given one sliding or roller support per span; with pins at both ends, a hot day would load the supports with forces no statics calculation could predict.

Plausible wrong answer. "Three unknowns and three equations, so it is solvable." Case (c) has three unknowns and three equations and still slides sideways under any horizontal load. Check that every motion is stopped.

Check. A plate is held by three links whose lines all meet at one point. Adequate? (No: nothing resists rotation about that point.)

Narration spine. Four plates in a row. Each is nudged: sideways, up, and with a twist. Plate (a) holds against all three. Plate (b) rotates about A when twisted; plate (c) slides when pushed sideways. Plate (d) holds, but its four link forces flicker between many values that all balance: the indeterminacy.

Bridge. With adequate supports, three equations give the reactions for one position of the load. Engineers want them for every position.

Nugget 07: Reactions Are Functions of Where the Load Sits

Claim. Write the equilibrium equations with a load's position as a variable and every reaction becomes a function of that position. Its largest value, which is what a support must be designed for, may sit at an end of the range or at a stationary point inside it.

The picture. Harper's version of the jib crane: the 10\,\mathrm{kN} load at a distance c to the left of B, from c = 0 (at the cable end) to c = 5\,\mathrm m.

The cable tension falls in a straight line as the load moves towards the pin, because T comes from one moment equation that is linear in c. The pin force falls, levels off and turns up slightly: it is the length of a vector whose two components change at different rates, so it has a true minimum inside the range.

Check with numbers. Nugget 04's moment equation with the load at 4.88 - c from A: 2.2889\,T = 10(4.88 - c) + 4.66 \times 2.38, so T = 26.17 - 4.369c\ \mathrm{kN}. At c = 1.5\,\mathrm m, Sample Problem 3/4's position, T = 19.61\,\mathrm{kN}, as before. Then A_x = T\cos25^\circ, A_y = 14.66 - T\sin25^\circ and A = \sqrt{A_x^2 + A_y^2}. Harper finds the extremes: T_{\max} = 26.15\,\mathrm{kN} at c = 0, T_{\min} = 4.31\,\mathrm{kN} at c = 5\,\mathrm m, A_{\max} = 23.97\,\mathrm{kN} at c = 0, and A_{\min} = 13.28\,\mathrm{kN} at c = 4.57\,\mathrm m, from dA/dc = 0. (Harper rounds his coefficients to 26.15 and 4.367; carried unrounded, the values are 26.17, 4.32, 23.99 and 13.29\,\mathrm{kN}, still at c = 4.57\,\mathrm m.)

The idea. Harper's point is about method. MATLAB's max and min on a grid of c return the largest or smallest computed value, which may be at an end of the range and need not be where the slope is zero; for T there is no stationary point at all. Only A_{\min} is a true stationary point, and he finds it two ways: [y, i] = min(A) on a 0.01\,\mathrm m grid (c(i) = 4.57), and by solving dA/dc = 0 symbolically. His second problem (Meriam's Problem 3/25) turns the question around: a pickup truck weighs 3600\,\mathrm{lb} (16.01\,\mathrm{kN}) with its centre of gravity 45\,\mathrm{in} (1.143\,\mathrm m) behind the front axle and 67\,\mathrm{in} (1.702\,\mathrm m) ahead of the rear. What load W_L, placed x behind the rear axle, makes the front and rear wheel forces equal? Equal wheel forces N give \Sigma F_y = 0: 2N = 16.01 + W_L; and \Sigma M_B = 0 about the rear contact: 16.01 \times 1.702 - N \times 2.845 - W_L x = 0. Eliminating N:

Harper's W_L = 39\,600/(56 + x) lb with x in inches. At x = 0 it is 3146\,\mathrm N (707\,\mathrm{lb}); at x = 1.27\,\mathrm m (50\,\mathrm{in}) only 1662\,\mathrm N.

The equation (mathematical identity: Eqs. 3/2 solved with a parameter left in):

Worked number. For the crane, where must the load be for T = 15\,\mathrm{kN}? 26.17 - 4.369c = 15, so c = 2.556\,\mathrm m from B. For the truck, a load 0.40\,\mathrm m behind the rear axle must weigh 4474/1.8224 = 2455\,\mathrm N (250.3\,\mathrm{kg}).

Real example (one you can see). A tower crane's load chart, posted in its cab, lists the allowed load against the trolley's distance out along the jib. It is this calculation: the further out the load, the larger the moment the tower and counter-jib must balance.

Plausible wrong answer. "The pin force is smallest where the cable tension is smallest." T is smallest at c = 5\,\mathrm m; A is smallest at c = 4.57\,\mathrm m. Different functions, different minima.

Check. Why can't min on a grid be trusted to give a stationary point? (It returns the smallest computed value, which may lie at the end of the range, where the slope is not zero.)

Narration spine. The crane with a sliding load. As the load glides from B to A, two live readouts, T and A, trace their curves on a graph beside it. T draws a straight line; A dips and bottoms out at c = 4.57\,\mathrm m, where a tangent flattens. The concurrency point O slides along the cable as the load moves.

Bridge. Back to the switchgear shaft, which has a parallel force system and a couple.

Synthesis

Seen square to the spring, the shaft carries three parallel forces: the spring's 500\,\mathrm N at 260\,\mathrm{mm} from A, and the bearing forces at A and at B (200\,\mathrm{mm}). Parallel forces need two equations (Nugget 02); take moments about A first, so that A drops out (Nugget 05).

\Sigma M_A = 0: B \times 0.200 = 500 \times 0.260, so B = 650\,\mathrm N. \Sigma F = 0 along the spring: A = 650 - 500 = 150\,\mathrm N, acting the same way as the spring. The guess of 250\,\mathrm N each was wrong twice over: the bearing beside the crank carries 650\,\mathrm N, more than the spring itself, and the far bearing is pulled the other way. The shaft is a lever with its fulcrum at B: the 60\,\mathrm{mm} overhang levers against the 200\,\mathrm{mm} span, so A must hold the far end down. In the end view, the spring's line passes 45.4\,\mathrm{mm} from the shaft axis, so \Sigma M_O = 0 needs a latch couple of 500 \times 0.0454 = 22.7\,\mathrm{N\cdot m} against the spring. (Solved as a full 3D problem in S12, the same numbers appear, with the thrust bearing's axial force zero.) A free-body diagram turns any static question about a body in a plane into at most three equations; choosing them well turns them into three one-line answers. Moving the crank 20\,\mathrm{mm} closer to B would cut B to 600\,\mathrm N: the overhang, not the spring, sets the bearing load.

• Central idea: planar equilibrium is three independent equations; pick the ones that isolate the unknowns.
• Key relationship: \Sigma F_x = 0, \Sigma F_y = 0, \Sigma M_O = 0; two-force members act along their pins' line; three forces are concurrent or parallel.
• Most important assumption: a rigid body with adequate, non-redundant supports; otherwise statics alone cannot give the reactions.
• Open question: the shaft's bearings are really in three dimensions. What do six equations do, and when do they reduce to fewer? (S12.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's and Harper's programs; the Python is a translation, and both print the same numbers.

Python (numpy, scipy). S11_equilibrium_2d.py solves Example 4.7 as a linear system and checks the pulleys, the two-force rod and Sample Problems 3/3 and 3/4. It runs Harper's jib-crane sweep, finding A's minimum twice, by argmin on the grid and by minimize_scalar, and evaluates Harper's truck load.

Part 1: Marghitu Example 4.7, unknowns F_Ox, F_Oy, M_O
  [F_Ox, F_Oy, M_O] = [4373.9  581.2 8438.7]
Part 2: pulleys, a two-force rod and two beams
  pulleys: T1 = 2224 N, T = 1112 N, bearing C (963, 556) N
  rod at 15 deg: rod 414.1 N, guide 107.2 N
  SP 3/3: P = 654 N, R = 327 N, theta = 22.0 deg
  SP 3/4: T = 19.61 kN, A = 18.88 kN
Part 3: Harper's jib crane, load c from B, 0 to 5 m
  T = 26.17 - 4.369 c kN
  T max 26.17 at c = 0.00, min 4.32 at 5.00
  A max 23.99 at c = 0.00, grid min 13.29 at 4.57
  bounded minimiser: A = 13.29 kN at c = 4.571 m
  T = 15 kN needs c = 2.556 m
Part 4: Harper's truck, equal front and rear wheel forces
  load 0.00 m behind the rear axle: W_L = 3146 N
  load 0.40 m behind the rear axle: W_L = 2455 N
  load 1.27 m behind the rear axle: W_L = 1662 N

The two minima agree, but only because A has a true stationary point inside the range. For T, max and min simply return the ends.

MATLAB (Octave-compatible). S11_equilibrium_2d.m does the same with the backslash operator, max, min and fminbnd.

Ex 4.7: F_Ox = 4373.9 N, F_Oy = 581.2 N, M_O = 8438.7 N m
SP 3/4 (c = 1.5 m): T = 19.61 kN, A = 18.88 kN
T max 26.17 at c = 0.00, min 4.32 at 5.00
A max 23.99 at c = 0.00, grid min 13.29 at 4.57
fminbnd: A = 13.29 kN at c = 4.571 m
x = 0.00 m: W_L = 3146 N
x = 0.40 m: W_L = 2455 N
x = 1.27 m: W_L = 1662 N

Traps

• Stopping at \Sigma\mathbf F = \mathbf 0. Two equal, opposite, offset forces add to zero and still turn the body; the moment equation is independent. (From Nugget 01.)
• Looking for a fourth equation. Moments about extra points repeat the first three; a planar body never gives more than three. (From Nugget 01.)
• Writing a moment equation for concurrent forces. About the meeting point it reads 0 = 0; two unknowns is the limit. (From Nugget 02.)
• Giving a two-force member two unknown components. Its force is along the line of its pins: one unknown. (From Nugget 03.)
• Treating a loaded beam as a two-force member. The jib's pin force is 19.72^\circ off the beam because loads act between its ends. (From Nugget 04.)
• Choosing moment centres on one line. Three collinear centres give two independent equations, not three. (From Nugget 05.)
• Trusting "three unknowns, three equations". Parallel or concurrent supports leave a motion free; four supports leave a split undetermined. (From Nugget 06.)
• Reading a grid minimum as a stationary point. min returns an end value for T; only A has dA/dc = 0 inside the range. (From Nugget 07.)
• Splitting a load equally between two supports. The shaft's bearings carry 650\,\mathrm N and 150\,\mathrm N, in opposite senses, not 250\,\mathrm N each. (From the Synthesis.)

Self-check

1. A smooth 40\,\mathrm{kg} sphere rests on a 30^\circ incline A and against a smooth vertical wall B. Find the contact forces.
2. A 1200\,\mathrm{kg} car has its mass centre 1.3\,\mathrm m behind the front axle; the wheelbase is 2.4\,\mathrm m. Find the normal force under each tyre on level ground. State the assumptions.
3. A 1\,\mathrm m beam is pinned at O and on a roller at C. It carries 100\,\mathrm N at 0.5\,\mathrm m from O and 500\,\mathrm N at 0.8\,\mathrm m from O, both downward. Find the reactions.
4. A 20\,\mathrm{kg} mass hangs from two cables anchored 2\,\mathrm m left and 4\,\mathrm m right of the point directly above it, the mass being 1.5\,\mathrm m below the anchors. Find both tensions, each from a single moment equation, and check with the force sums.
5. Marghitu's Problem 4.5 with l = 2\,\mathrm{ft}, \theta = 45^\circ and P = 500\,\mathrm{lb}: find the force in link 2, in SI.
6. A uniform 4\,\mathrm m, 30\,\mathrm{kg} bar is pinned at its lower end A and held at 60^\circ to the horizontal by a horizontal cable at its upper end. Use the three-force principle to find the direction of the pin force, then find the cable tension and the pin force.
7. For a body in a plane, which of these sets are valid equilibrium equations? (a) \Sigma M_A = 0, \Sigma M_B = 0, \Sigma F_x = 0, with AB along y; (b) the same with AB along x; (c) \Sigma M_A, \Sigma M_B, \Sigma M_C with C the midpoint of AB.
8. Give the degree of statical indeterminacy of: (a) a truss pinned at both ends; (b) a beam on a roller at A and a pin not free to turn at B; (c) a beam on a pin and a roller.
9. A plate is held by three links whose lines meet at one point P. A load is applied whose line does not pass through P. What happens, and why?
10. For Harper's jib crane, where must the 10\,\mathrm{kN} load be for the cable tension to be 15\,\mathrm{kN}?
11. For Harper's truck, what load placed 0.40\,\mathrm m behind the rear axle equalises the front and rear wheel forces?
12. The switchgear crank is moved to 80\,\mathrm{mm} beyond bearing B, the spring still pulling 500\,\mathrm N. Find both bearing forces.

Fully worked solutions

1. Forces: weight 40 \times 9.81 = 392.4\,\mathrm N down; wall force N_B horizontal; incline force N_A normal to the incline, at 30^\circ to the vertical. All pass through the centre: concurrent, two equations. \Sigma F_y = 0: N_A\cos30^\circ = 392.4, so N_A = 453.1\,\mathrm N. \Sigma F_x = 0: N_B = N_A\sin30^\circ = 392.4\tan30^\circ = 226.6\,\mathrm N.

2. Assume the car at rest on level ground, left and right sides symmetric, no braking forces. W = 1200 \times 9.81 = 11\,772\,\mathrm N. Moments about the front contact: R \times 2.4 = 11\,772 \times 1.3, R = 6376.5\,\mathrm N on the rear axle, so 3188\,\mathrm N per rear tyre. Front axle 11\,772 - 6376.5 = 5395.5\,\mathrm N: 2698\,\mathrm N per front tyre.

3. \Sigma M_O = 0: C \times 1 = 100 \times 0.5 + 500 \times 0.8 = 450, C = 450\,\mathrm N up. \Sigma F_y = 0: O_y = 600 - 450 = 150\,\mathrm N up; O_x = 0 (no horizontal load).

4. With a_1 = 2, a_2 = 4, a_3 = 1.5\,\mathrm m, mg = 196.2\,\mathrm N. Moments about the left anchor B, where T_{AB} has no arm: T_{AC} = 196.2 \times 2\sqrt{4^2 + 1.5^2}/(1.5 \times 6) = 186.3\,\mathrm N. Moments about C: T_{AB} = 196.2 \times 4\sqrt{2^2 + 1.5^2}/(1.5 \times 6) = 218.0\,\mathrm N. Check: horizontal -218.0 \times 2/2.5 + 186.3 \times 4/4.272 = -174.4 + 174.4 = 0; vertical 218.0 \times 1.5/2.5 + 186.3 \times 1.5/4.272 = 130.8 + 65.4 = 196.2\,\mathrm N.

5. Link 2 is a two-force member; moments about A on link 1 give F = 2P/3 for any \theta and l: 333.3\,\mathrm{lb}, that is 333.3 \times 4.448 = 1483\,\mathrm N, in compression.

6. Three forces: weight 30 \times 9.81 = 294.3\,\mathrm N down at the middle, 1\,\mathrm m horizontally from A; cable tension horizontal at height 4\sin60^\circ = 3.464\,\mathrm m. Their lines meet at (1, 3.464)\,\mathrm m from A, so the pin force points at \arctan(3.464/1) = 73.9^\circ above the horizontal. Moments about A: T \times 3.464 = 294.3 \times 1, T = 84.96\,\mathrm N. The pin force balances both: A = \sqrt{294.3^2 + 84.96^2} = 306.3\,\mathrm N, and \arctan(294.3/84.96) = 73.9^\circ agrees.

7. (a) Not valid: AB along y is perpendicular to x, so a resultant along AB has no x-component and satisfies all three equations without being zero. (b) Valid: AB along x is not perpendicular to x. (c) Not valid: the three centres are collinear, so the third equation repeats the first two.

8. (a) 2 + 2 - 3 = 1. (b) 1 + 3 - 3 = 1 (pin not free to turn: two forces and a couple). (c) 2 + 1 - 3 = 0: determinate.

9. The plate turns about P. All three link forces pass through P, so their moment about P is zero, while the load's is not: \Sigma M_P = 0 cannot hold. The constraints are partial (improper), even though there are three of them.

10. T = 26.17 - 4.369c = 15 gives c = 11.165/4.369 = 2.556\,\mathrm m from B.

11. W_L = 4474/(1.4224 + 0.40) = 2455\,\mathrm N, a mass of 2455/9.81 = 250.3\,\mathrm{kg}.

12. \Sigma M_A = 0: B \times 0.200 = 500 \times 0.280, B = 700\,\mathrm N against the spring. A = 700 - 500 = 200\,\mathrm N with the spring. The extra 20\,\mathrm{mm} of overhang raises both bearing loads by 50\,\mathrm N.