Study Companion: S12 — Equilibrium in Three Dimensions

S11 solved bodies whose forces lie in one plane: three equations, three unknowns. Most real machines are not planar. A shaft in two bearings, a mast held by cables, a frame on a ball joint: their forces point every way in space, and S11's three equations no longer describe them. This lesson extends equilibrium to space. A free body in space can move in six independent ways, so equilibrium becomes six equations. The work is in keeping six equations manageable: recognising special force systems, viewing a space problem as three plane problems, and taking moments about a well-chosen axis so that one equation contains one unknown. The lesson ends with a different kind of support: a particle held on a smooth surface or curve, whose reaction is found from the surface's own equation.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 3/4 ("Equilibrium Conditions", Section B), with Sample Problems 3/5–3/7, and the Chapter Review (3/5). Marghitu, Dupac and Madsen, Statics with MATLAB, section 4.1 (the six equations) and section 4.6 ("Particle on a Smooth Surface and on a Smooth Curve"), with Examples 4.1, 4.3–4.5 and 4.8. Harper, Solving Statics Problems in MATLAB, "3D Equilibrium" (Sample Problem 3/7). Axes are right-handed; vectors use \mathbf i, \mathbf j, \mathbf k as in S07–S09.

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

A hand-cranked hoist (Meriam's Sample Problem 3/6). A drum of radius 100\,\mathrm{mm} is fixed to a shaft carried in a radial bearing B and a thrust bearing A, 150\,\mathrm{mm} apart; the drum is 100\,\mathrm{mm} beyond B. At the other end a crank of radius 250\,\mathrm{mm} carries a handle whose grip is 175\,\mathrm{mm} beyond A. A hand pushes the grip with 200\,\mathrm N, at 60^\circ below the horizontal and 45^\circ round from the shaft's direction: a push that is neither along nor across the shaft.

Two questions. How heavy a load m can the 200\,\mathrm N push hold steady? And how hard is each bearing pushed, which is what decides whether they last? The push has components along all three axes, so no single side view shows the whole problem. It is answered in Nugget 03 and revisited in the Synthesis.

Nugget 01: Six Equations, Because a Body in Space Can Move Six Ways

Claim. A rigid body in space can translate along three axes and rotate about three. Equilibrium forbids all six, so it is six independent scalar equations: three force sums and three moment sums.

The picture.

Each force sum stops one translation; each moment sum about an axis stops one rotation about it. None of the six implies another, so all six are needed. A planar body (S11) is the special case where forces along z and moments about x and y are zero by construction: three equations are left.

Check with numbers. Meriam's examples of independence: a car speeding up on a straight level road has \Sigma F_x \ne 0 (it accelerates along x), while the other two force sums are zero. Its engine's flywheel, spinning up about the x-axis, has \Sigma M_x \ne 0 along with \Sigma F_x \ne 0, while the remaining four equations about its mass-centre axes hold. Any subset can hold without the rest.

The idea. Meriam (3/4) writes S10's vector conditions, \Sigma\mathbf F = \mathbf 0 and \Sigma\mathbf M = \mathbf 0, as six scalar equations, Eqs. 3/3. They are necessary and sufficient for complete equilibrium. The axes may be chosen freely, but must be right-handed when vector notation is used. In vector form, write each force in \mathbf i, \mathbf j, \mathbf k; \Sigma\mathbf F = \mathbf 0 holds only if each coefficient is zero, which gives the three force equations. For moments, take each force's \mathbf r\times\mathbf F about a convenient point O, with \mathbf r from O to any point on the force's line; setting the \mathbf i, \mathbf j, \mathbf k coefficients of the sum to zero gives \Sigma M_x = \Sigma M_y = \Sigma M_z = 0. Marghitu (4.3) states the same six equations (Eq. 4.6) and notes that moments about further points add no independent equations: a free-body diagram in space yields at most six, so at most six unknown forces or couples.

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

Forces in N, moments in N·m, about axes through any convenient point.

Worked number (Meriam's Sample Problem 3/5). A uniform 7\,\mathrm m steel shaft of mass 200\,\mathrm{kg} stands on a ball-and-socket joint at A in the floor; its upper end B rests against two smooth vertical walls. B is 2\,\mathrm m and 6\,\mathrm m from A horizontally, so its height is h = \sqrt{7^2 - 2^2 - 6^2} = 3\,\mathrm m. The walls push normal to themselves, B_x and B_y; the weight W = 200 \times 9.81 = 1962\,\mathrm N acts at the middle; A has three components. Moments about A remove A entirely: \mathbf r_{AB} = -2\mathbf i - 6\mathbf j + 3\mathbf k\ \mathrm m and \mathbf r_{AG} = -1\mathbf i - 3\mathbf j + 1.5\mathbf k\ \mathrm m, and

So B_y = 1962\,\mathrm N, B_x = 654\,\mathrm N, and the \mathbf k equation merely checks: -2(1962) + 6(654) = 0. The force sums give A_x = 654, A_y = 1962, A_z = 1962\,\mathrm N (in the senses drawn), so A = \sqrt{654^2 + 1962^2 + 1962^2} = 2850\,\mathrm N.

Real example. A flight simulator stands on six hydraulic legs, a hexapod. Six legs because a cockpit has six motions to control: three slides and three rotations. With fewer legs some motion would be uncontrolled; with more, the legs would fight each other.

Plausible wrong answer. "Take moments about several more points to get more equations for more unknowns." A body in space never yields more than six independent equations, whatever points are used.

Check. Why did the \mathbf k moment equation in Sample Problem 3/5 give no new information? (Nugget 02: every force crosses the line AB.)

Narration spine. A block floats in space. It slides along x, y, z and spins about each axis in turn; each motion lights one equation in a column of six. Cut to the shaft against the walls: a single vector equation at A splits into three coloured coefficients, two giving B_x = 654 and B_y = 1962, the third collapsing to 0 = 0.

Bridge. In Sample Problem 3/5 one of six equations was automatic. As in a plane, some force systems need fewer.

Nugget 02: Special Force Systems in Space Need Fewer Equations

Claim. Forces concurrent at a point need three equations; forces that all cross one line need five; parallel forces need three; only a general system needs all six.

The picture. Meriam's four categories in three dimensions (Fig. 3/9).

Concurrent forces have zero moment about every axis through their common point. Forces that all cross one line have zero moment about that line, since each has zero arm to it (S08: a force meeting an axis has no moment about it). Parallel forces along x have no y or z components and no moment about x (each is parallel to that axis), so \Sigma F_y, \Sigma F_z and \Sigma M_x vanish by themselves.

Check with numbers. In Sample Problem 3/5 the wall forces act at B, the floor force at A, and the weight at G on the line AB: every force crosses the line AB. Category 2, five equations, so the sixth (the moment about AB) is automatic. Each force acts at a point of AB, so its moment about A is perpendicular to \mathbf r_{AB} = -2\mathbf i - 6\mathbf j + 3\mathbf k, and so is the total: -2M_x - 6M_y + 3M_z = 0. Once M_x = M_y = 0, M_z = 0 follows, which is why -2B_y + 6B_x = 0 only checked.

The idea. Meriam (3/4): Category 1, forces concurrent at O, needs the three force equations and no moment equations. Category 2, forces concurrent with a line, needs all equations except the moment about that line. Category 3, parallel forces (along x), needs one force equation, \Sigma F_x = 0, and two moment equations about the axes normal to the forces, \Sigma M_y = \Sigma M_z = 0. Category 4, a general system, needs all six. Meriam adds that these observations are usually evident while solving a problem.

The equation (mathematical identity: Eqs. 3/3 with the automatic ones removed):

Worked number. A uniform square plate, 0.5\,\mathrm m by 0.5\,\mathrm m and 30\,\mathrm{kg}, hangs level from three vertical wires at A(0, 0), B(0.5, 0) and C(0.25, 0.5)\,\mathrm m. Parallel forces: three equations, three tensions. Weight 294.3\,\mathrm N at (0.25, 0.25). Moments about the x-axis (through A and B): T_C \times 0.5 = 294.3 \times 0.25, T_C = 147.15\,\mathrm N. Moments about the y-axis (through A): T_B \times 0.5 + 147.15 \times 0.25 = 294.3 \times 0.25, T_B = 73.58\,\mathrm N. Force sum: T_A = 294.3 - 147.15 - 73.58 = 73.58\,\mathrm N.

Real example. A three-legged stool on a level floor is category 3: three vertical floor forces, three equations, always determinate. That is why it never rocks, while a four-legged one can.

Plausible wrong answer. "Three wires and six equations: the plate is over-determined, so the wire forces are inconsistent." Three of the six are automatically satisfied for parallel forces; three remain for three unknowns.

Check. A mast on a ball joint at A is held by two cables from its top B and loaded at B. Which category, and how many unknowns can be found? (Every force crosses the line AB: category 2, five.)

Narration spine. Four panels, each with a column of six equations beside it. In the first, the arrows meet at a point and the three moment equations grey out. In the second, a dashed line threads every arrow and one moment equation greys out. In the third, the parallel arrows grey out three. The plate on three wires then shows its three tensions appearing one per equation.

Bridge. Six equations in six unknowns can be solved as a linear system, but by hand that is laborious. Meriam's hoist shows a way to split them.

Nugget 03: Three Views Turn One Space Problem into Three Plane Problems

Claim. Project the free-body diagram onto the three coordinate planes. Each view shows a planar problem in two force components and one moment, and the six equations split among the three views: two or three at a time, often one unknown at a time.

The picture. The hoist of the puzzle, seen from the end, from above and from the side (Meriam's orthographic projections). The push resolves into 200\sin60^\circ = 173.2\,\mathrm N down (-y) and 200\cos60^\circ = 100\,\mathrm N horizontal, which splits again into 70.7\,\mathrm N along x and 70.7\,\mathrm N along z.

In the end view, the bearing forces pass through the shaft axis and drop out of the moment about it: only the weight on the drum and the push on the crank remain. In the top view only x and z components appear; in the side view only y and z. Each view is an S11 problem.

Check with numbers (Meriam's solution, lengths in mm).

• End view, moments about the shaft axis: 100(9.81m) - 250(173.2) = 0, so m = 44.1\,\mathrm{kg} (433.0\,\mathrm N).
• Top view, moments about A: 150B_x + 175(70.7) - 250(70.7) = 0, so B_x = 35.4\,\mathrm N; \Sigma F_x = 0: A_x + 35.4 - 70.7 = 0, A_x = 35.4\,\mathrm N.
• Side view, moments about A: 150B_y + 175(173.2) - 250(433.0) = 0, so B_y = 519.6 \approx 520\,\mathrm N; \Sigma F_y = 0: A_y = 173.2 + 433.0 - 519.6 = 86.6\,\mathrm N; \Sigma F_z = 0: A_z = 70.7\,\mathrm N, the thrust.
• Radial loads: A_r = \sqrt{35.4^2 + 86.6^2} = 93.5\,\mathrm N and B = \sqrt{35.4^2 + 519.6^2} = 521\,\mathrm N.

(Meriam prints A_y = 86.8\,\mathrm N; its own equation gives 86.6\,\mathrm N unrounded and 85.8\,\mathrm N with its rounded 520 and 44.1. Its A_r = 93.5\,\mathrm N matches 86.6.)

The idea. Meriam notes the hoist has no plane of symmetry, so it is a general space problem. A single pictorial view would do, but three orthogonal projections make each set of components visible. The correct senses of A_x and B_x can be seen from the top view (the push's resultant there passes between A and B); those of A_y and B_y are assumed until the moments are known. Meriam's hint: picture the views as images projected onto the faces of a clear box around the body. The problem is determinate only because each bearing is assumed to carry no couple about axes normal to the shaft (S10, Nugget 06); without that assumption it would be statically indeterminate.

The equation (mathematical identity: Eqs. 3/3 regrouped by view):

Each force equation appears in two views; use it once.

Worked number. The answer to the puzzle: the push holds m = 44.1\,\mathrm{kg}. The radial bearing B, between the drum and the crank but close to the drum, carries 521\,\mathrm N, more than twice the hand's push; the thrust bearing A carries only 93.5\,\mathrm N radially plus the 70.7\,\mathrm N thrust.

Real example. Every gearbox shaft is designed this way. A helical gear's tooth force has radial, tangential and axial components; engineers resolve it into three views exactly as here to size each bearing (Marghitu's Problems 4.6–4.10 are such shafts).

Plausible wrong answer. "The bearings share the drum's weight, half each." B_y = 520\,\mathrm N and A_y = 86.6\,\mathrm N: the drum overhangs beyond B, so B takes more than the weight and A only the difference, as in S11's switchgear shaft.

Check. Which view gives the mass m in one step, and why? (The end view: both bearing forces pass through the shaft axis, so the moment about it holds only m.)

Narration spine. The hoist in perspective, inside a transparent box. Its shadow is cast onto the end face: the drum and crank, with two arrows; m appears. Then onto the top face: B_x, A_x. Then the side face: B_y, A_y, A_z. The three faces unfold flat like a net, each with its own little equation.

Bridge. Projections work well along the coordinate axes. When the unknowns cross a slanting line, a moment about that line is quicker still.

Nugget 04: A Moment About an Axis Can Eliminate Five Unknowns

Claim. The moment of a force about an axis is zero if the force's line meets the axis. Choose an axis that every unknown but one meets, take moments about it, and that one unknown comes out of a single equation, even when the axis runs obliquely.

The picture. Meriam's Sample Problem 3/7: a welded tubular frame on a ball joint at A, held by a loose ring at B and a cable CD, loaded with 2\,\mathrm{kN}.

There are six unknowns: A_x, A_y, A_z at the ball joint, B_x, B_z at the ring (it slides freely along y), and T. Every reaction at A passes through A, and both ring forces pass through B. So all five meet the axis AB, and the moment about AB contains only T and the known 2\,\mathrm{kN}.

Check with numbers. A is the origin; B = (0, 4.5, 6), C = (-3, 0, 6), D = (-1, 2.5, 0), E = (2.5, 0, 6)\,\mathrm m. The axis has \mathbf n = (4.5\mathbf j + 6\mathbf k)/7.5 = \frac15(3\mathbf j + 4\mathbf k). The cable runs from C to D, 2\mathbf i + 2.5\mathbf j - 6\mathbf k, length \sqrt{46.25} = 6.80\,\mathrm m (Meriam writes \sqrt{46.2}), so \mathbf T = \frac{T}{\sqrt{46.2}}(2\mathbf i + 2.5\mathbf j - 6\mathbf k). Moment of \mathbf T about AB: (\mathbf r_1\times\mathbf T)\cdot\mathbf n with \mathbf r_1 = \overrightarrow{AD} = -\mathbf i + 2.5\mathbf j (any point on the cable's line will do); of the load: (\mathbf r_2\times\mathbf F)\cdot\mathbf n with \mathbf r_2 = 2.5\mathbf i + 6\mathbf k, \mathbf F = 2\mathbf j\ \mathrm{kN}. Carrying out the products and multiplying through by 5, as Meriam does:

with components T_x = 0.833, T_y = 1.042, T_z = -2.50\,\mathrm{kN}.

The idea. S08 gave the moment of \mathbf F about an axis through A with unit vector \mathbf n as the triple product (\mathbf r\times\mathbf F)\cdot\mathbf n. Meriam's point is that vector notation frees us to take moments about any axis, so we may choose one that eliminates five unknowns. The rest follow one at a time: \Sigma M_z = 0 about A: 2(2.5) - 4.5B_x - 1.042(3) = 0, B_x = 0.417\,\mathrm{kN}; \Sigma M_x = 0: 4.5B_z - 2(6) - 1.042(6) = 0, B_z = 4.06\,\mathrm{kN}; and the force sums A_x = -1.250, A_y = -3.04, A_z = -1.556\,\mathrm{kN}, the minus signs meaning the components act opposite to the arrows drawn. Harper (3.3) solves the same problem the direct way, moments about the point A (three equations in T, B_x, B_z, solved together) and then forces, and gets the same values.

The equation (geometric relationship: S08's moment about an axis, applied to every force):

Worked number. Check the result against the axis equation term by term. The load's moment about AB is (\mathbf r_2\times\mathbf F)\cdot\mathbf n: \mathbf r_2\times 2\mathbf j = -12\mathbf i + 5\mathbf k, and \cdot\,\frac15(3\mathbf j + 4\mathbf k) = 4\,\mathrm{kN\cdot m}. The cable's: \mathbf r_1\times\mathbf T = (-15\mathbf i - 6\mathbf j - 7.5\mathbf k)T/\sqrt{46.2}, and \cdot\,\mathbf n = -(18 + 30)T/(5\sqrt{46.2}) = -1.412T. So 1.412T = 4 and T = 2.83\,\mathrm{kN}.

Real example. A door on two hinges: take moments about the hinge axis and both hinge reactions vanish, leaving only the push on the handle and the door closer's force. That is how a door-closer's required torque is found.

Plausible wrong answer. "The moment about the axis AB is the size of the moment about A." It is only its component along \mathbf n. The load's moment about A has magnitude 13\,\mathrm{kN\cdot m}; about the axis it is 4\,\mathrm{kN\cdot m}.

Check. Which forces have zero moment about an axis? (Those whose lines meet it, and those parallel to it.)

Narration spine. The frame in space. The axis AB glows. One by one the five reaction arrows at A and B slide along their lines until they touch the axis and fade (zero moment). Only the cable and the load remain; their moment vectors about A are projected onto the axis, and the two projections balance at T = 2.83\,\mathrm{kN}.

Bridge. Every example so far had just enough supports, properly placed. In space that is harder to see.

Nugget 05: Six Supports Are Not Always Enough, and Cables Only Pull

Claim. Six equations find six unknown reactions only if the supports stop every motion. Six links can still leave a rotation free, if they all meet one line. And a cable, which can only pull, holds a body only if the equations give it a positive tension.

The picture. Two blocks each held by six links, each link a two-force member (one unknown).

On the left, three links fix the corner A, and the other three, placed away from A, stop rotation about the three axes through A. On the right, three links fix A and three fix E: every link force passes through A or E, so none has a moment about the line AE. A load with a moment about AE turns the block, though six links and six equations match.

Check with numbers (Marghitu's Example 4.8). A 100\,\mathrm{kg} square plate, 0.6\,\mathrm m by 0.6\,\mathrm m, hangs from a vertical shaft in bearings A (bottom, also carrying the vertical load) and B (top, 0.6\,\mathrm m above). A 120\,\mathrm{N\cdot m} couple about the vertical +z axis would turn it; a cable from its free bottom corner D to a floor point P 0.5\,\mathrm m out prevents that. Six unknowns, six equations. The moment about the shaft axis has only the couple and the cable: 120 - 0.6 \times (0.5/0.781)T = 0, so T = 312.41\,\mathrm N, a pull. The other results: F_{Ax} = -200, F_{Ay} = 730.5, F_{Az} = 981, F_{Bx} = 0, F_{By} = -490.5\,\mathrm N. (Marghitu prints T = -312.410\,\mathrm N, but his own F_{Ax} = -200\,\mathrm N needs \Sigma F_x = -200 + 0.640T = 0, so T = +312.41\,\mathrm N. A negative value would mean the cable pushing.) Reverse the couple and T becomes -312.41\,\mathrm N: impossible for a cable. It goes slack and the plate swings round. The same six supports are adequate for one load and not for the other.

The idea. Meriam (3/4): the six equations are necessary and sufficient for equilibrium, but whether they determine the reactions depends on the constraints. Fig. 3/10 shows four cases: (a) corner A fixed by links 1, 2, 3 and rotation about their axes stopped by links 4, 5, 6: complete fixity, adequate constraints; (b) the same number of links, but none resists a moment about the axis AE: incomplete fixity, partial constraints; (c) no resistance to a force along y: also partial; (d) a seventh link added to a fully fixed body: excessive fixity, redundant constraints, statically indeterminate. With few exceptions, Meriam's bodies are adequately and determinately supported. An analytical test for adequacy exists but is beyond the book. Marghitu's one-sided constraint (4.6) adds a further condition: a support described by an inequality (a cable, a contact) is adequate only while its force has the allowed sign.

The equation (definition, as in S11 with six equations):

Worked number. A shaft on two bearings, each modelled as a ball joint (three force components each), carries a load and an input couple about its axis: six force unknowns, nothing to resist the couple, and two supports fighting over the axial load. The count is six, but the shaft spins: improper (and redundant along the axis at the same time). Replace one ball joint by a radial bearing (two components) and add a clamp that resists twisting (one couple): six unknowns again, now adequate. That is the switchgear shaft of S10: radial bearing, thrust bearing, latch.

Real example. A camera on a tripod head with its pan lock released: the head's bearings stop every motion of the camera except turning about the vertical pan axis, because every bearing force meets that axis. However firmly the head is built, a light touch swings the camera round. The pan lock is the missing sixth constraint.

Plausible wrong answer. "Six links, six equations, so it is fixed." The right-hand block has exactly that and turns about AE.

Check. A body is held by a ball joint at A and three cables whose lines all pass through one point Q on a line through A. Adequate? (No: nothing resists rotation about the line AQ.)

Narration spine. Two blocks with six links each. A twisting arrow is applied to both about the diagonal AE. The left block holds; on the right, all six links glow where they touch the dashed line AE, and the block rotates. Then the hanging plate: the couple arrow flips, the cable's tension readout dips through zero, the cable goes slack and the plate swings.

Bridge. Supports that are nearly improper are not safe either: as a support approaches an inadequate position, the reactions grow without bound.

Nugget 06: Nearly Improper Supports Carry Huge Forces

Claim. As a support moves towards a position where the constraints become partial, the equations stay solvable but the reactions grow without limit. A plot of the reactions against the support's position shows the danger before it happens.

The picture. Harper's version of Sample Problem 3/7: the ring B moved to (0, c, 6)\,\mathrm m, with c from 0 to 10\,\mathrm m (c = 4.5\,\mathrm m is Meriam's frame).

As c shrinks, the ring approaches the post's axis above the ball joint. The load and the cable try to turn the frame about the x-axis through A, and the only reaction with a moment about that axis is the ring's vertical force B_z, with arm c. At c = 0 the arm is gone: turning about the x-axis would move B along y, which the ring allows, so nothing resists it and the constraints are partial (Nugget 05). Just before that, the equations still have a solution, but B_z, and with it A_z, must be enormous.

Check with numbers. Harper's moments about the point A (three equations in T, B_x, B_z, solved together) give

and the force sums give A_x = -30/(15 + 2c), A_y = -(55 + 4c)/(15 + 2c), A_z = 6(6c - 55)/[c(15 + 2c)]. At c = 4.5\,\mathrm m these are Meriam's T = 2.83, B_z = 4.06 and A_z = -1.556\,\mathrm{kN}. At c = 1\,\mathrm m, holding the same 2\,\mathrm{kN} load needs A = 17.73\,\mathrm{kN} and B = 20.83\,\mathrm{kN}: ten times the load. The c in the denominators of A_z and B_z is the blow-up.

The idea. Harper plots T, A and B against c (limiting the axis to 10 kN because A and B run off to infinity), explains the blow-up physically (the ring cannot support the frame when it sits on the z-axis), and then finds where A = B. Graphically, the curves cross near c = 5.3\,\mathrm m; zooming the plot to 5.3 \le c \le 5.4 gives 5.373\,\mathrm m. Then a trick: A - B has no minimum, but \lvert A - B\rvert has one, equal to zero, exactly where A = B; so [y, i] = min(abs(A - B)) on a fine grid returns c = 5.3730\,\mathrm m.

The equation (mathematical identity: Eqs. 3/3 solved with the ring position left as a parameter):

Worked number. Move the ring out to c = 6\,\mathrm m: T = 5\sqrt{185}/27 = 2.519\,\mathrm{kN}, B_x = 10/27 = 0.370\,\mathrm{kN}, B_z = 6(79)/(6 \times 27) = 2.926\,\mathrm{kN}. Every support force has dropped compared with c = 4.5\,\mathrm m: spreading the supports apart reduces them.

Real example (one you can see). Stand a broom upright on the floor and hold it by two fingers close to the floor: the fingers must squeeze very hard to keep it upright against a small push at the top. Move the upper finger higher up the handle and the squeeze needed drops. Supports close together, in line with the load's turning tendency, are nearly improper.

Plausible wrong answer. "Any position of the ring works as long as the equations can be solved." At c = 1\,\mathrm m they can, but the ring must carry over 20\,\mathrm{kN} to hold a 2\,\mathrm{kN} load.

Check. Why does \lvert A - B\rvert have a minimum where A - B does not? (Its minimum is zero, at the crossing, where A - B changes sign; A - B itself is steadily increasing.)

Narration spine. The frame with the ring sliding along the arm towards the post. Beside it, the graph draws itself as the ring moves; as the ring nears the post, the A and B curves shoot off the top and the reaction arrows on the frame swell. The ring slides back out; the curves cross at 5.373 m, where a marker drops to the axis.

Bridge. All the supports so far held rigid bodies. A last kind of support holds a particle on a surface, and its reaction is set by the surface's shape.

Nugget 07: On a Smooth Surface the Reaction Lies Along the Gradient

Claim. A particle held on a smooth surface f(x, y, z) = 0 feels a reaction normal to the surface, and the normal is the gradient: \mathbf N = \lambda\nabla f. With the surface equation added, equilibrium is four equations for x, y, z and \lambda: they find where the particle can rest, not only the force.

The picture. Marghitu's Example 4.5: a 1\,\mathrm{kg} bead on a smooth vertical circle x^2 + y^2 = R^2 (R = 1\,\mathrm m), under gravity and a horizontal force \mathbf F = kx\,\mathbf i (k = 20\,\mathrm{N/m}) pushing it away from the y-axis.

At the top and bottom the force \mathbf F is zero (x = 0) and the circle pushes the bead straight up against its weight. At the side positions the outward push kx and the weight together point along the inward normal, and the circle's reaction cancels them. Everywhere else, the sum of \mathbf F and the weight has a component along the circle that nothing can balance: the bead slides.

Check with numbers. At P_3: y = -mg/k = -9.8/20 = -0.49\,\mathrm m and x = \sqrt{1 - 0.49^2} = 0.8717\,\mathrm m. With \lambda = -k/2 = -10 and \nabla f = (2x, 2y) = (1.743, -0.98), the reaction is \mathbf N = -10(1.743, -0.98) = (-17.43, 9.80)\,\mathrm N, of size \sqrt{17.43^2 + 9.8^2} = 20.0\,\mathrm N = kR. Sum: \mathbf F + \mathbf G + \mathbf N = (17.43 - 17.43,\ -9.8 + 9.8) = \mathbf 0. At P_1 (0, 1), \lambda = mg/(2R) = 4.9 and \mathbf N = 4.9(0, 2) = (0, 9.8)\,\mathrm N.

The idea. Marghitu (4.6): a particle at rest on a smooth surface f(x, y, z) = 0 is under an ideal bilateral constraint (it can neither leave nor enter the surface); f \ge 0 would be a unilateral one. The constraint removes one degree of freedom, leaving two. The reaction is perpendicular to the surface, and so collinear with \nabla f (Eq. 4.8). The vector equilibrium equation is \mathbf F + \lambda\nabla f = \mathbf 0 (Eq. 4.9), three scalar equations (4.10), and with f = 0 they fix x, y, z and \lambda. On a smooth curve, the intersection of f_1 = 0 and f_2 = 0, one degree of freedom remains and the reaction is \lambda_1\nabla f_1 + \lambda_2\nabla f_2 (Eqs. 4.12–4.14): five equations for x, y, z, \lambda_1, \lambda_2. Here \lambda is a Lagrange multiplier, though Marghitu does not use the name. His examples, all solved symbolically in MATLAB:

• Example 4.1, a 100\,\mathrm{kg} block on a 30^\circ incline: the plane's reaction has normal component mg\cos30^\circ = 849.571\,\mathrm N and tangential component mg\sin30^\circ = 490.5\,\mathrm N (the tangential part is friction, which a smooth plane could not supply).
• Example 4.3, a 10\,\mathrm{kg} particle on the plane x + y + z + 1.5 = 0, pulled by \mathbf F = k\,\overrightarrow{PA} (k = 10\,\mathrm{N/m}) towards A(0, 0, 5)\,\mathrm m, with g = 9.8\,\mathrm{m/s^2}: \nabla f = (1, 1, 1), the equations -10x + \lambda = 0, -10y + \lambda = 0, 10(5 - z) - 98 + \lambda = 0, x + y + z + 1.5 = 0 give P(1.1, 1.1, -3.7)\,\mathrm m and \lambda = 11, so \mathbf N = (11, 11, 11)\,\mathrm N, 19.05\,\mathrm N.
• Example 4.4, a 10\,\mathrm{kg} particle on the parabola y = -x^2 (in metres) with \mathbf F = -kx\,\mathbf i (k = 10\,\mathrm{N/m}): the equations 2\lambda x - kx = 0, \lambda = mg give x = 0, \mathbf N = 98\mathbf j\ \mathrm N, unless k = 2agm = 196\,\mathrm{N/m}, when every point of the curve is an equilibrium position.
• Example 4.5, the bead above: four positions.

The equation (geometric relationship for the normal, with the material model of a smooth constraint; Marghitu's Eqs. 4.8–4.9):

\lambda has units of N divided by the units of \nabla f (for f in \mathrm{m^2}, N/m).

Worked number. A 60\,\mathrm{kg} crate on a 20^\circ ramp (Example 4.1's method): the ramp's reaction has normal part 60 \times 9.81\cos20^\circ = 553.1\,\mathrm N and tangential part 60 \times 9.81\sin20^\circ = 201.3\,\mathrm N. If the ramp were smooth, \lambda\nabla f could supply only the normal part, and the crate would not be in equilibrium anywhere on it.

Real example. A ball rolling to rest in a smooth bowl ends where the bowl's normal is vertical, the bottom, because there gravity has no component along the surface. With a magnet pulling sideways, the rest point moves up the side to where weight plus magnet force point along the normal, as on the circle.

Plausible wrong answer. "The bead rests wherever the forces' sizes balance." The forces must point along the normal; at any other point the component along the curve accelerates the bead. On the circle that leaves only four points.

Check. On the circle with k < mg/R, how many equilibrium positions are there? (Two: y = -mg/k would lie below -R, off the circle, so only the top and bottom remain.)

Narration spine. The bead slides round the circle while the weight, the outward push and the curve's normal are drawn at each point; the part of the force along the curve shows as a small red arrow that vanishes at four points, where the bead stops. A surface appears with its gradient arrow standing up from it, labelled \lambda\nabla f.

Bridge. Back to the switchgear shaft, now with all six equations.

Synthesis

The hoist answered with three views: a 200\,\mathrm N push holds 44.1\,\mathrm{kg}, the radial bearing near the drum carries 521\,\mathrm N, the thrust bearing 93.5\,\mathrm N radially and 70.7\,\mathrm N along the shaft. The switchgear shaft of S10 and S11 is the same kind of body, and the six equations now check S11's shortcut. With z along the shaft, A at the origin, B at z = 0.20\,\mathrm m, the spring force \mathbf F = -491.0\mathbf i - 94.5\mathbf j\ \mathrm N at the crank pin (0.030, 0.052, 0.260)\,\mathrm m, unknowns A_x, A_y, B_x, B_y, B_z, M_L:

• \Sigma F_z = 0: B_z = 0. The thrust bearing carries nothing, because nothing pushes along the shaft.
• \Sigma M_z = 0: M_L = -22.7\,\mathrm{N\cdot m}, the latch couple, clockwise seen from +z.
• \Sigma M_x, \Sigma M_y, \Sigma F_x, \Sigma F_y: \mathbf B = 638.3\mathbf i + 122.8\mathbf j (650\,\mathrm N, against the spring) and \mathbf A = -147.3\mathbf i - 28.3\mathbf j (150\,\mathrm N, with the spring).

These are S11's numbers. S11's plane view was legitimate because the spring force has no z component and the bearing forces are radial: the six equations split into a plane problem, a torque balance and a trivially satisfied axial equation, just as the hoist's split into three views. In space, equilibrium is six equations; the art is in choosing views, axes and points so that they fall apart into small, separate problems, and in checking that the supports really stop all six motions. Part 3 is complete: any body, isolated and drawn, can now be put in equilibrium. Part 4 applies this to structures made of many bodies.

• Central idea: six independent equations for a body in space; fewer for special force systems; a moment about a well-chosen axis isolates one unknown.
• Key relationship: \Sigma\mathbf F = \mathbf 0, \Sigma\mathbf M = \mathbf 0; (\mathbf r\times\mathbf F)\cdot\mathbf n about an axis; \mathbf N = \lambda\nabla f on a smooth surface.
• Most important assumption: adequate, non-redundant constraints, with bearings carrying no couples and cables in tension.
• Open question: a truss or a machine is many bodies, each with its own six (or three) equations, joined by pins. How are they solved together? (Part 4.)

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.

By hand, the work is choosing moment axes cleverly. A computer needs no cleverness. Each unknown is a unit force at a point, its column in a 6\times6 matrix is [\mathbf F;\ \mathbf r\times\mathbf F], and one solve returns every support force at once.

Python (numpy). S12_equilibrium_3d.py solves Sample Problem 3/5, the three-wire plate, the hoist of the puzzle, and Sample Problem 3/7 over Harper's range of ring positions c.

Part 1: Sample Problem 3/5, shaft on a ball joint and two walls
  A = [ -654. -1962.  1962.] N, B_x = 654.0 N, B_y = 1962.0 N
Part 2: square plate on three vertical wires
  T_A, T_B, T_C = [ 73.57  73.57 147.15] N
Part 3: the hoist, shaft along z (mm and N), y up
  load 433.0 N, m = 44.1 kg
  A = [ 35.4  86.6 -70.7] N, radial 93.5 N
  B = [ 35.4 519.6] N, radial 520.8 N
Part 4: Sample Problem 3/7 and Harper's ring position c
  c = 4.5 m: T = 2.834 kN, A = [-1.25  -3.042 -1.556], B = [0.417 4.056] kN
   c (m)   T (kN)   |A| (kN)   |B| (kN)
    1.0    4.000     17.73     20.83
    2.0    3.579      7.72      9.96
    4.5    2.834      3.64      4.08
    6.0    2.519      3.21      2.95
   10.0    1.943      2.85      1.65
  T = 5 sqrt(185) / (15 + 2c) at c = 6: 2.519 kN

The plate's 73.57 is 73.575 exactly; it rounds down in floating point, where Nugget 02 rounds it up to 73.58.

MATLAB (Octave-compatible). S12_equilibrium_3d.m builds the same columns with an anonymous function and solves with the backslash operator.

SP 3/5: A = [-654.0 -1962.0 1962.0] N, B_x = 654.0 N, B_y = 1962.0 N
hoist: m = 44.1 kg, A radial 93.5 N, thrust 70.7 N, B 520.8 N
 c (m)   T (kN)  |A| (kN)  |B| (kN)
   1.0    4.000     17.73     20.83
   2.0    3.579      7.72      9.96
   4.5    2.834      3.64      4.08
   6.0    2.519      3.21      2.95
  10.0    1.943      2.85      1.65

Traps

• Expecting more than six equations. Moments about further points or axes are combinations of the six. (From Nugget 01.)
• Worrying when an equation reads 0 = 0. Forces that all cross one line make the moment about it automatic; it is a check. (From Nugget 02.)
• Using a force equation twice across views. \Sigma F_y appears in two projections; it is one equation. (From Nugget 03.)
• Giving each bearing two couples. The hoist becomes statically indeterminate; short bearings in pairs carry forces only. (From Nugget 03.)
• Confusing the moment about an axis with the moment about a point on it. It is the component along \mathbf n: 4\,\mathrm{kN\cdot m} about AB, not 13\,\mathrm{kN\cdot m}. (From Nugget 04.)
• Accepting a negative cable tension. A cable cannot push; a negative T means it goes slack and the body moves. (From Nugget 05.)
• Counting six supports as six constraints. Six links that all meet one line leave a rotation free. (From Nugget 05.)
• Placing supports close together. Reactions blow up as supports approach an improper position: 20.83\,\mathrm{kN} at c = 1\,\mathrm m for a 2\,\mathrm{kN} load. (From Nugget 06.)
• Letting a smooth surface supply a tangential force. \mathbf N = \lambda\nabla f is normal; a tangential part needs friction. (From Nugget 07.)

Self-check

1. A vertical 6\,\mathrm m mast stands on a ball joint at A. Two cables run from its top B to floor points C(-4, 3, 0) and D(-4, -3, 0)\,\mathrm m, and a 4\,\mathrm{kN} horizontal force along +x acts at B. Which category is this? Find the cable tensions and the joint reaction.
2. Repeat Meriam's Sample Problem 3/5 for a 150\,\mathrm{kg} shaft. Find the wall forces and the total force at A.
3. Name the category and the number of independent equations for: (a) a ring pulled by three cables in different directions; (b) a table on four legs; (c) a boom held by a ball joint and two cables from its tip, loaded at the tip; (d) a crankshaft in two bearings with forces on a crank.
4. A shaft along z has a bearing A at z = 0 (taking all axial load) and a radial bearing B at z = 0.15\,\mathrm m. A gear at z = 0.12\,\mathrm m carries, at its tooth (0, 0.1, 0.12)\,\mathrm m, a tangential force 4500\,\mathrm N along +x, a radial force 2500\,\mathrm N along -y and an axial force 1000\,\mathrm N along -z (Marghitu's Problem 4.7 numbers). A coupling applies the torque about z. Find the bearing forces and the torque.
5. The switchgear crank turns to 90^\circ (crank pin at (0, 0.06)\,\mathrm m in the end view), the spring still pulling 500\,\mathrm N towards the anchor (-0.24, 0)\,\mathrm m. What happens to the bearing forces and to the latch couple?
6. For Harper's frame with the ring at c = 6\,\mathrm m, find T, B_x and B_z.
7. Why do Harper's A and B become infinite as c \to 0, while T does not?
8. A block is held by six links, three at corner A and three at the opposite corner E. A load acts along a line parallel to AE but not on it. Can it be resisted? And a load whose line crosses AE obliquely, not through A or E?
9. Meriam's hoist assumes the bearings carry no couples. Count the unknowns without that assumption.
10. A 60\,\mathrm{kg} crate rests on a 20^\circ rough ramp. Find the normal and tangential components of the ramp's reaction.
11. A 2\,\mathrm{kg} particle on the smooth plane x + z - 1 = 0 is pulled towards A(0, 0, 2)\,\mathrm m by \mathbf F = 10\,\overrightarrow{PA}\ \mathrm N (\overrightarrow{PA} in m), with z up. Find its equilibrium position and the plane's reaction.
12. A 1\,\mathrm{kg} bead on a smooth vertical circle of radius 0.5\,\mathrm m is pushed away from the vertical axis by \mathbf F = 40x\,\mathbf i\ \mathrm N. Find all equilibrium positions and the reaction at each.

Fully worked solutions

1. Every force passes through B or A: all cross the line AB, category 2, five equations; unknowns T_C, T_D, A_x, A_y, A_z. By symmetry about the x-z plane, T_C = T_D = T. Unit vector from B(0, 0, 6) to C: (-4, 3, -6)/\sqrt{61}, \sqrt{61} = 7.810\,\mathrm m. Moments about B: every force except the joint force acts at B, so the joint force must have zero moment about B too, which puts it along the mast: A_x = A_y = 0. \Sigma F_x = 0: 4 - 2T(4/\sqrt{61}) = 0, T = \sqrt{61}/2 = 3.905\,\mathrm{kN}. \Sigma F_y = 0 by symmetry. \Sigma F_z = 0: A_z = 2T(6/\sqrt{61}) = 6.00\,\mathrm{kN} up. The mast is in pure compression.

2. Weight 150 \times 9.81 = 1471.5\,\mathrm N. The same moment equation with 1471.5 for 1962: -3B_y + 3(1471.5) = 0, B_y = 1471.5\,\mathrm N; 3B_x - 1471.5 = 0, B_x = 490.5\,\mathrm N. Forces: A_x = 490.5, A_y = 1471.5, A_z = 1471.5\,\mathrm N; A = \sqrt{490.5^2 + 1471.5^2 + 1471.5^2} = 2138\,\mathrm N. Everything scales with the weight.

3. (a) Concurrent at a point: 3. (b) Parallel: 3 (four unknown leg forces, so indeterminate to degree 1). (c) Concurrent with the boom's line: 5. (d) General: 6.

4. Moments about A, \mathbf r = 0.1\mathbf j + 0.12\mathbf k, \mathbf F = 4500\mathbf i - 2500\mathbf j - 1000\mathbf k: \mathbf r\times\mathbf F = (0.1(-1000) - 0.12(-2500))\mathbf i + (0.12 \times 4500)\mathbf j + (-0.1 \times 4500)\mathbf k = 200\mathbf i + 540\mathbf j - 450\mathbf k. B at 0.15\mathbf k: 0.15\mathbf k\times(B_x\mathbf i + B_y\mathbf j) = -0.15B_y\mathbf i + 0.15B_x\mathbf j. Torque M\mathbf k. \Sigma M_x: 200 - 0.15B_y = 0, B_y = 1333.3\,\mathrm N. \Sigma M_y: 540 + 0.15B_x = 0, B_x = -3600\,\mathrm N. \Sigma M_z: M = 450\,\mathrm{N\cdot m}. Forces: A_x = -4500 + 3600 = -900\,\mathrm N, A_y = 2500 - 1333.3 = 1166.7\,\mathrm N, A_z = 1000\,\mathrm N. Radial loads: A_r = \sqrt{900^2 + 1166.7^2} = 1473.5\,\mathrm N, B = \sqrt{3600^2 + 1333.3^2} = 3839\,\mathrm N.

5. The bearing forces are parallel to the spring force and their sizes come only from the distances along the shaft (S11): still 650\,\mathrm N at B and 150\,\mathrm N at A, now along the new spring direction. The latch couple changes: the spring's line from (0, 0.06) to (-0.24, 0) passes 0.24 \times 0.06/0.2474 = 0.0582\,\mathrm m from the axis, so M_L = 500 \times 0.0582 = 29.1\,\mathrm{N\cdot m}.

6. 15 + 2c = 27: T = 5\sqrt{185}/27 = 2.519\,\mathrm{kN}, B_x = 10/27 = 0.370\,\mathrm{kN}, B_z = 6(55 + 24)/(6 \times 27) = 2.926\,\mathrm{kN}.

7. A_z and B_z have c in the denominator. The moment about the x-axis through A is -(6\sqrt{185}/37)T - 12 + cB_z = 0: the load and cable's moment must be balanced by B_z on the arm c, so B_z \propto 1/c, and A_z follows from \Sigma F_z = 0. At c = 0 that rotation is unresisted (the ring lets B move along y). T comes from the moments about the y- and z-axes, which do not need the arm c, so it stays finite: 5\sqrt{185}/15 = 4.53\,\mathrm{kN} at c = 0.

8. A load parallel to AE has a moment about AE equal to zero, so the links can resist it (no rotation about AE is excited). A load whose line neither meets nor is parallel to AE has a moment about AE that no link can balance: the block turns. The constraints are partial.

9. Ball-and-socket-like thrust bearing A: 3 forces; radial bearing B: 2; plus 2 couples each about the axes normal to the shaft: 3 + 2 + 2 + 2 = 9 unknowns, with m also unknown: 10 for 6 equations. Without the assumption, indeterminate.

10. mg = 588.6\,\mathrm N; normal 588.6\cos20^\circ = 553.1\,\mathrm N, tangential 588.6\sin20^\circ = 201.3\,\mathrm N up the ramp (friction).

11. \nabla f = (1, 0, 1). Equations: -10x + \lambda = 0, -10y = 0, 10(2 - z) - 19.62 + \lambda = 0, x + z - 1 = 0. So y = 0, x = \lambda/10, z = 1 - \lambda/10, and 10(1 + \lambda/10) - 19.62 + \lambda = 0: 2\lambda = 9.62, \lambda = 4.81\,\mathrm N. Position (0.481, 0, 0.519)\,\mathrm m; \mathbf N = 4.81(1, 0, 1), 4.81\sqrt2 = 6.802\,\mathrm N.

12. f = x^2 + y^2 - 0.25; equations 40x + 2\lambda x = 0, -9.81 + 2\lambda y = 0. Either x = 0: y = \pm0.5\,\mathrm m, \lambda = \pm9.81, \mathbf N = (0, 9.81)\,\mathrm N at both top and bottom; or \lambda = -20: y = -9.81/40 = -0.2453\,\mathrm m, x = \pm\sqrt{0.25 - 0.0601} = \pm0.4357\,\mathrm m, \lvert\mathbf N\rvert = kR = 40 \times 0.5 = 20\,\mathrm N. Four positions, since kR = 20 > mg = 9.81.