Study Companion: S09 — Resultants in Three Dimensions; the Wrench

S06 reduced any set of forces in a plane to a single force on a definite line, or to a couple. In space that promise breaks. A system of forces in three dimensions always reduces to one force and one couple at a point, but the couple need not lie square to the force, and then no single force can replace the system. What remains is a force together with a twist about its own line: a wrench, the same action as pushing on a screwdriver while turning it. This lesson closes Part 2. It reduces any force system in space, sorts out when a single force is possible, and finds the wrench when it is not.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/9 ("Resultants"), with Sample Problems 2/16–2/19, and the Chapter 2 Review. Marghitu, Dupac and Madsen, Statics with MATLAB, section 2.4 ("Equipollent Force Systems", the wrench), with Example 2.5. Harper has no 3D resultant problem. Axes are right-handed with z up, as in S07 and S08. Marghitu's equipollent means Meriam's equivalent: same resultant force and same moment about a point (S06). 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

A test rig loads an L-shaped bracket with three hydraulic rams: 200\,\mathrm N along x through a point on the x-axis, 400\,\mathrm N along y at 0.3\,\mathrm m up the vertical plate, and 400\,\mathrm N along z at (0.5, 0.4, 0)\,\mathrm m on the base. (This is Meriam's Sample Problem 2/19 with SI numbers: forces ten times the book's pounds, lengths in tenths of a metre.)

Running three rams is expensive. Could one ram, pushing along a single line, load the bracket exactly as the three do? In a plane, S06 says yes, always, unless the forces cancel. In space the answer is surprising: no single ram can do it, and the best one ram can do also needs a twist about its own axis. Why, and along which line? The answer comes in Nugget 05.

Nugget 01: Every System Reduces to One Force and One Couple at Any Point

Claim. Any system of forces and couples in space is equivalent to a force \mathbf R = \Sigma\mathbf F at a chosen point O plus a couple \mathbf M = \Sigma(\mathbf r\times\mathbf F), the moments about O plus any applied couples. \mathbf R is the same for every choice of O; \mathbf M is not.

The picture. Move each force to O, paying its couple (S08, Nugget 06).

After the moves, every force acts at O, so they add as concurrent forces to \mathbf R. Every couple is a free vector, so they add to \mathbf M. The whole system has become two vectors at O.

Check with numbers. For the bracket, with \mathbf r to each force's point:

• 200\mathbf i at (0.1, 0, 0): 0.1\mathbf i\times200\mathbf i = \mathbf 0;
• 400\mathbf j at (0, 0, 0.3): 0.3\mathbf k\times400\mathbf j = -120\mathbf i;
• 400\mathbf k at (0.5, 0.4, 0): (0.5\mathbf i + 0.4\mathbf j)\times400\mathbf k = -200\mathbf j + 160\mathbf i.

So \mathbf R = 200\mathbf i + 400\mathbf j + 400\mathbf k\ \mathrm N (600\,\mathrm N) and \mathbf M_O = 40\mathbf i - 200\mathbf j\ \mathrm{N\cdot m}. Moving the reference point changes the couple by S08's transfer formula, \mathbf M_P = \mathbf M_O + \mathbf r_{PO}\times\mathbf R, but leaves \mathbf R alone.

The idea. Meriam (2/9) moves each force \mathbf F_1, \mathbf F_2, \dots to an arbitrary point O, introducing the couple \mathbf M_1 = \mathbf r_1\times\mathbf F_1 and so on, and adds. The magnitude and direction of \mathbf R do not depend on O; those of \mathbf M do. In dynamics the mass centre is the usual choice, because \mathbf R then governs the change in linear motion and \mathbf M the change in angular motion. In statics the body is in complete equilibrium when both are zero. Marghitu (2.4) states the same as a theorem: however complicated a system of forces and moments, it is equipollent to a single force at a given point and a single couple.

The equation (mathematical identity: the S08 force–couple transfer applied to every force; Meriam's Eqs. 2/20 and 2/21):

Worked number (Meriam's Sample Problem 2/18). A part carries a 500\,\mathrm N force at B, a 600\,\mathrm N and a 700\,\mathrm N force, and a negative wrench whose 25\,\mathrm{N\cdot m} couple points against the 500\,\mathrm N force. Reduced to the point A: \mathbf R = 928\mathbf i + 871\mathbf j + 621\mathbf k\ \mathrm N, 1416\,\mathrm N. The moments about A are \mathbf M_{500} = 18.95\mathbf i - 5.59\mathbf j - 16.90\mathbf k (a cross product with \mathbf r_{AB} = 0.08\mathbf i + 0.12\mathbf j + 0.05\mathbf k\ \mathrm m), \mathbf M_{600} = 36.0\mathbf i + 24.0\mathbf k and \mathbf M_{700} = 10.5\mathbf i - 71.4\mathbf j - 18.19\mathbf k (both by inspection), and the wrench's own couple adds -16.07\mathbf i - 13.54\mathbf j - 13.54\mathbf k, all in N·m. Total: \mathbf M = 49.4\mathbf i - 90.5\mathbf j - 24.6\mathbf k\ \mathrm{N\cdot m}, 106.0\,\mathrm{N\cdot m}.

Real example. A six-axis load cell under a wind-tunnel model reports exactly these six numbers: three force components and three moment components about its own centre. Every aerodynamic load on the model, however spread out, is measured as one \mathbf R and one \mathbf M.

Plausible wrong answer. "The resultant couple is the same about every point, because couples are free vectors." Each transfer couple is free, but its value depends on where the force was moved to. Change O and every \mathbf r changes, and so does \mathbf M.

Check. A system reduces at O to \mathbf R = \mathbf 0 and \mathbf M = 12\mathbf k\ \mathrm{N\cdot m}. What does it reduce to at any other point? (The same couple: with \mathbf R = \mathbf 0 the transfer term \mathbf r\times\mathbf R vanishes.)

Narration spine. A body bristling with arrows. One at a time, each arrow slides to O, dropping a curled couple as it goes. The arrows at O chain into a green \mathbf R; the curls stack into a blue \mathbf M. A probe moves O: the green arrow stays fixed while the blue one swings.

Bridge. One force plus one couple is simpler, but still two things. For some systems one of the two vanishes at once.

Nugget 02: Couples Alone, Concurrent Forces and Parallel Forces Reduce Further

Claim. Concurrent forces reduce to a single force through their common point. Forces that add to zero reduce to a single couple. Parallel forces reduce to a single parallel force, placed by \mathbf r\times\mathbf R = \mathbf M_O.

The picture. Each case is Nugget 01's pair with one member forced to zero or forced into a convenient direction.

• Concurrent: take O at the meeting point; every \mathbf r is zero, so \mathbf M = \mathbf 0.
• Zero resultant: \mathbf R = \mathbf 0 leaves only \mathbf M, and since \mathbf R = \mathbf 0, the transfer formula gives the same \mathbf M at every point: a pure couple.
• Parallel: with all forces along \mathbf j, say, every moment \mathbf r\times F\mathbf j has no \mathbf j-component, so \mathbf M is perpendicular to \mathbf R, and a line can be found where \mathbf R alone has moment \mathbf M.

Check with numbers. Meriam's Sample Problem 2/16 (US units): a box carries three pairs of equal and opposite forces and three couples. \mathbf R = \mathbf 0, and \mathbf M_O = [50(16) - 700]\mathbf i + [80(12) - 960]\mathbf j + [100(10) - 1000]\mathbf k = 100\mathbf i\ \mathrm{lb\text{-}in.}, which is 11.30\,\mathrm{N\cdot m}: the resultant is a couple. Meriam's Sample Problem 2/17: four forces along y on a plate, 200\,\mathrm N at (-0.5, 0, 0), 500\,\mathrm N at O, -300\,\mathrm N at (0, 0, -0.35) and -50\,\mathrm N at (0.5, 0, 0.35)\,\mathrm m. Then \mathbf R = 350\mathbf j\ \mathrm N and \mathbf M_O = -87.5\mathbf i - 125\mathbf k\ \mathrm{N\cdot m}. Place \mathbf R through (x, y, z): (x\mathbf i + y\mathbf j + z\mathbf k)\times350\mathbf j = 350x\mathbf k - 350z\mathbf i. Matching: 350x = -125 and -350z = -87.5, so x = -0.357\,\mathrm m and z = 0.250\,\mathrm m. The coordinate y drops out, as transmissibility says it must.

The idea. Meriam (2/9) lists the special systems. For concurrent forces only the first of Eqs. 2/20 is needed. For parallel forces the resultant's magnitude is the algebraic sum of the forces, and its line comes from the principle of moments, \mathbf r\times\mathbf R = \mathbf M_O, with \mathbf r from O to the line. Coplanar forces are S06's case. And if \mathbf R = \mathbf 0, the resultant, if it exists, is a couple. Marghitu (2.4) lists the same cases: concurrent forces are equipollent to one force through their common point; parallel forces with a non-zero sum are equipollent to one force.

The equation (mathematical identity, the general reduction with the special values substituted):

Worked number. A floor slab carries three columns' loads straight down: 2\,\mathrm{kN} at (0, 0), 3\,\mathrm{kN} at (4, 0) and 5\,\mathrm{kN} at (4, 3)\,\mathrm m. \mathbf R = -10\mathbf k\ \mathrm{kN} and \mathbf M_O = -15\mathbf i + 32\mathbf j\ \mathrm{kN\cdot m}. From (x\mathbf i + y\mathbf j)\times(-10\mathbf k) = -10y\mathbf i + 10x\mathbf j: x = 3.2\,\mathrm m and y = 1.5\,\mathrm m. That is the load-weighted average position, as in S06.

Real example. A foundation engineer does exactly this for a building's columns: the single resultant of all column loads should fall near the centre of the raft foundation, or the raft tilts.

Plausible wrong answer. "Parallel forces in 3D need three equations to place the resultant, one for each coordinate." The coordinate along the forces is free (transmissibility), so only two are found.

Check. Three forces meet at (1, 2, 3)\,\mathrm m and add to \mathbf R. What is the resultant couple about that point? (Zero.)

Narration spine. Three panels. In the first, arrows radiate from one point and merge. In the second, paired arrows cancel and leave a single curl. In the third, a plate's parallel arrows add to one, which slides across the plate until its moment readouts match -87.5 and -125, at x = -0.357, z = 0.250.

Bridge. In the plane and in these special cases, \mathbf R and \mathbf M came out perpendicular. Is that why a single force was possible? And what happens when they are not?

Nugget 03: A Single Force Exists Only When R Is Perpendicular to M

Claim. A system with \mathbf R \ne \mathbf 0 can be replaced by a single force exactly when \mathbf R\cdot\mathbf M = 0. The force is \mathbf R, acting through the point \mathbf r = \mathbf R\times\mathbf M/R^2.

The picture. A single force \mathbf R on some line has a moment \mathbf r\times\mathbf R about O, and a cross product is always perpendicular to both its factors.

So whatever line it is put on, a single force can only produce moments perpendicular to itself. If the system's \mathbf M has any part along \mathbf R, no single force can match it.

Check with numbers. Sample Problem 2/17: \mathbf R\cdot\mathbf M = (0)(-87.5) + 350(0) + 0(-125) = 0, so a single force exists, as found. The puzzle's bracket: \mathbf R\cdot\mathbf M = 200(40) + 400(-200) + 400(0) = 8000 - 80\,000 = -72\,000\,\mathrm{N^2\cdot m} \ne 0. No single ram can load the bracket as the three do. A case that passes: if \mathbf R = 2\mathbf i + \mathbf k\ \mathrm{kN} and \mathbf M = \mathbf i + 3\mathbf j - 2\mathbf k\ \mathrm{kN\cdot m}, then \mathbf R\cdot\mathbf M = 2 + 0 - 2 = 0, and the formula gives \mathbf r = (2\mathbf i + \mathbf k)\times(\mathbf i + 3\mathbf j - 2\mathbf k)/5 = (-3\mathbf i + 5\mathbf j + 6\mathbf k)/5 = -0.6\mathbf i + \mathbf j + 1.2\mathbf k\ \mathrm m. Check: \mathbf r\times\mathbf R = (1 \cdot 1 - 1.2 \cdot 0)\mathbf i + (1.2 \cdot 2 - (-0.6)(1))\mathbf j + (-0.6 \cdot 0 - 1 \cdot 2)\mathbf k = \mathbf i + 3\mathbf j - 2\mathbf k. It matches \mathbf M.

The idea. The necessity is the picture above. For sufficiency, the point formula is a step the sources skip (both find the line by solving \mathbf r\times\mathbf R = \mathbf M component by component instead). Step the sources skip: with \mathbf r = \mathbf R\times\mathbf M/R^2, the vector triple product of S02 (Marghitu 1.10) gives \mathbf r\times\mathbf R = -\mathbf R\times(\mathbf R\times\mathbf M)/R^2 = -[(\mathbf R\cdot\mathbf M)\mathbf R - (\mathbf R\cdot\mathbf R)\mathbf M]/R^2 = \mathbf M - (\mathbf R\cdot\mathbf M)\mathbf R/R^2, which equals \mathbf M exactly when \mathbf R\cdot\mathbf M = 0. This \mathbf r is the foot of the perpendicular from O to the line; adding any multiple of \mathbf R gives other points on the same line. In a plane, \mathbf M is always along \mathbf k and \mathbf R always in the plane, so the test always passes, which is why S06 never met a wrench.

The equation (mathematical identity):

Worked number. The scalar \mathbf R\cdot\mathbf M does not depend on the reference point: moving from O to P adds \mathbf r_{PO}\times\mathbf R to \mathbf M, which is perpendicular to \mathbf R and adds nothing to the dot product. For a system with \mathbf R = 100\mathbf i - 200\mathbf j + 300\mathbf k\ \mathrm N and \mathbf M_O = 300\mathbf i - 600\mathbf j - 100\mathbf k\ \mathrm{N\cdot m}, \mathbf R\cdot\mathbf M_O = 30\,000 + 120\,000 - 30\,000 = 120\,000. At P(1, 2, 0)\,\mathrm m, \mathbf M_P = -300\mathbf i - 300\mathbf j + 300\mathbf k\ \mathrm{N\cdot m} and \mathbf R\cdot\mathbf M_P = -30\,000 + 60\,000 + 90\,000 = 120\,000 again. So "can this be one force?" has one answer, whichever point the system was reduced to.

Real example. A ship's propeller delivers thrust along its shaft and a reaction torque about the same shaft. Its \mathbf R and \mathbf M are parallel, so \mathbf R\cdot\mathbf M \ne 0: no single force could represent a propeller, which is why ships carry a rudder offset against the torque.

Plausible wrong answer. "Choose a cleverer reference point and \mathbf M will become perpendicular to \mathbf R." The dot product \mathbf R\cdot\mathbf M is the same at every point, so no choice of point helps.

Check. \mathbf R = 3\mathbf j\ \mathrm{kN} and \mathbf M = 6\mathbf i + 2\mathbf j\ \mathrm{kN\cdot m}. Single force? (\mathbf R\cdot\mathbf M = 6 \ne 0: no.)

Narration spine. A green \mathbf R and a blue \mathbf M at O. A ghost single force tries every line in space; its moment arrow sweeps around, always staying in the plane square to \mathbf R. When \mathbf M lies in that plane the ghost finds a line and locks on. Tilt \mathbf M out of the plane, and the ghost searches forever.

Bridge. When the test fails, the part of \mathbf M along \mathbf R cannot be removed. The part across \mathbf R still can. Doing that gives the simplest system there is.

Nugget 04: Otherwise the Simplest System Is a Wrench

Claim. Split \mathbf M into \mathbf M_1 along \mathbf R and \mathbf M_2 across it. Moving \mathbf R sideways by d = M_2/R cancels \mathbf M_2. What remains, \mathbf R on a unique axis with the parallel couple \mathbf M_1, is a wrench.

The picture. Meriam's Fig. 2/30, in four steps.

(a) The force–couple system at O. (b) \mathbf M split into \mathbf M_1 along \mathbf R and \mathbf M_2 perpendicular to it. (c) \mathbf M_2, a couple, is replaced by two forces \mathbf R and -\mathbf R a distance d = M_2/R apart, with -\mathbf R at O cancelling the original \mathbf R (S05, Nugget 05, in reverse). (d) What is left is \mathbf R on a new line and the free couple \mathbf M_1 along it. The wrench's axis lies in the plane through O normal to the plane of \mathbf R and \mathbf M.

Check with numbers. The bracket: \mathbf n_R = \mathbf R/R = \frac13\mathbf i + \frac23\mathbf j + \frac23\mathbf k. \mathbf M\cdot\mathbf n_R = (40 - 400)/3 = -120\,\mathrm{N\cdot m}, so \mathbf M_1 = -120\,\mathbf n_R = -40\mathbf i - 80\mathbf j - 80\mathbf k\ \mathrm{N\cdot m}, pointing against \mathbf R. Then \mathbf M_2 = \mathbf M - \mathbf M_1 = 80\mathbf i - 120\mathbf j + 80\mathbf k\ \mathrm{N\cdot m}, of size 164.9\,\mathrm{N\cdot m}, and d = 164.9/600 = 0.275\,\mathrm m. The foot of the perpendicular from O to the axis is \mathbf r = \mathbf R\times\mathbf M/R^2 = 0.222\mathbf i + 0.044\mathbf j - 0.156\mathbf k\ \mathrm m, whose length is 0.275\,\mathrm m, as it should be.

The idea. Meriam (2/9): when the resultant couple \mathbf M is parallel to the resultant force \mathbf R, the resultant is a wrench. It is positive if the couple and force point the same way and negative if they point opposite ways. A screwdriver driving a right-handed screw applies a positive wrench: push along the screw and turn clockwise as seen from behind. Any general force system can be represented by a wrench on a unique line of action. It is the simplest form of the resultant, though of limited use, because a convenient reference point such as the mass centre is usually not on the wrench axis. Marghitu (2.4) gives the same three steps (choose a point, split \mathbf M into parallel and normal parts, move \mathbf F so that \mathbf r_{IP}\times\mathbf F = \mathbf M_n) and concludes that any system is equipollent to a null system, a single force, a single couple or a wrench. The ratio p = \mathbf R\cdot\mathbf M/R^2, a length, is the wrench's pitch: the couple \mathbf M_1 = p\mathbf R is p times the force. Its sign tells the wrench's type.

The equation (mathematical identity: projection of \mathbf M on \mathbf R, then the force–couple trade):

Worked number (Sample Problem 2/18 finished as a wrench). With \mathbf R = 928\mathbf i + 871\mathbf j + 621\mathbf k\ \mathrm N and \mathbf M = 49.4\mathbf i - 90.5\mathbf j - 24.6\mathbf k\ \mathrm{N\cdot m} at A: \mathbf R\cdot\mathbf M = -48\,298\,\mathrm{N^2\cdot m}, the angle between them is 108.8^\circ, and \mathbf M_1 = -34.1\,\mathrm{N\cdot m} along \mathbf R: a negative wrench, pitch -24.1\,\mathrm{mm}. M_2 = 100.4\,\mathrm{N\cdot m}, so the wrench axis runs 100.4/1416 = 70.9\,\mathrm{mm} from A.

Real example (one you can see). Watch someone drive a screw by hand: the palm pushes along the screwdriver and the wrist turns it. That combined action, force and twist about one line, is a positive wrench. Drawing a cork with a corkscrew while pulling out is a negative one.

Plausible wrong answer. "Moving \mathbf R far enough removes the whole couple." Moving \mathbf R only creates couples perpendicular to \mathbf R (they are \mathbf r\times\mathbf R). The parallel part \mathbf M_1 survives every move.

Check. \mathbf R = 500\mathbf k\ \mathrm N and \mathbf M = 300\mathbf i + 400\mathbf k\ \mathrm{N\cdot m} at O. What is the wrench? (\mathbf M_1 = 400\mathbf k, positive; \mathbf M_2 = 300\mathbf i; the axis is parallel to z through 0.6\mathbf j\ \mathrm m, because 0.6\mathbf j\times500\mathbf k = 300\mathbf i.)

Narration spine. \mathbf R and \mathbf M at O at an oblique angle. \mathbf M splits into a part along \mathbf R and a part across it. The across part unfolds into a pair of forces, one landing on \mathbf R and annihilating it. The survivor sits on a new dashed axis, the parallel couple riding along it like a screwdriver's twist.

Bridge. The formulas place the axis by its nearest point to O. In practice the question is often where the axis meets a surface, such as the bracket's base plate, so a ram can be mounted there.

Nugget 05: The Wrench Axis Is Found Where It Crosses a Plane

Claim. If the wrench axis crosses a given plane at P, then the moment of the original forces about P is the wrench couple itself, parallel to \mathbf R. Writing that condition with direction cosines gives three equations for P's two coordinates and the couple M.

The picture. The bracket again, with the wrench axis drawn.

At any point of the wrench axis, \mathbf R has no moment, so the system's moment there is just \mathbf M_1, which points along \mathbf R. So: find the point P(x, y, 0) of the base plane about which the three rams' moments point along \mathbf R.

Check with numbers. Moments about P(x, y, 0) of the three forces: 200\mathbf i gives 200y\mathbf k; 400\mathbf j at (0, 0, 0.3) gives -120\mathbf i - 400x\mathbf k; 400\mathbf k at (0.5, 0.4, 0) gives 400(0.4 - y)\mathbf i - 400(0.5 - x)\mathbf j. In total

Assume a positive wrench, \mathbf M_P = M(\frac13\mathbf i + \frac23\mathbf j + \frac23\mathbf k):

The last two give -200 + 400x = 200y - 400x, so 4x - y = 1. Twice the first equals the second: 80 - 800y = -200 + 400x, so x + 2y = 0.7. Then y = 4x - 1, 9x = 2.7, x = 0.3\,\mathrm m and y = 0.2\,\mathrm m, and M = 3(40 - 80) = -120\,\mathrm{N\cdot m}. The negative M means the assumption was wrong in sign: the wrench is negative. That answers the puzzle. One ram of 600\,\mathrm N along the line through P(0.3, 0.2, 0)\,\mathrm m with direction (\frac13, \frac23, \frac23), plus a 120\,\mathrm{N\cdot m} twist about that line against the push, loads the bracket exactly as the three rams do. The nearest-point formula of Nugget 04 agrees: \mathbf r + t\mathbf R reaches z = 0 at t = 0.156/400, giving (0.3, 0.2, 0). (Meriam's original numbers, in US units: \mathbf R = 20\mathbf i + 40\mathbf j + 40\mathbf k lb, M = -120\,\mathrm{lb\text{-}in.} (-13.56\,\mathrm{N\cdot m}), P(3, 2) in., that is (76.2, 50.8)\,\mathrm{mm}.)

The idea. Meriam's Sample Problem 2/19 sets the direction cosines of the wrench couple equal to those of \mathbf R, assuming a positive wrench, and lets a negative M reveal a negative one. Marghitu's Example 2.5 does the same for three forces on a rectangular plate in the x-z plane and solves the three equations with MATLAB's solve. The method works because the moment about a point of the axis contains no \mathbf M_2: that part has been used up in placing the axis.

The equation (mathematical identity: on the wrench axis \mathbf M_P = \mathbf M_1 \parallel \mathbf R):

Worked number (second view, Marghitu's Example 2.5). A plate in the x-z plane, a = 4\,\mathrm{ft} (1.219\,\mathrm m) by b = 6\,\mathrm{ft} (1.829\,\mathrm m), carries F_A = 900\,\mathrm{lb} along -z at (a/2, 0, 0), F_B = 500\,\mathrm{lb} along +y at (a, 0, b) and F_C = 300\,\mathrm{lb} along +x at (0, 0, b). Then \mathbf R = 300\mathbf i + 500\mathbf j - 900\mathbf k\ \mathrm{lb}, 1072.4\,\mathrm{lb} (4770\,\mathrm N). The direction-cosine equations give M = -839.3\,\mathrm{lb\cdot ft} (-1137.9\,\mathrm{N\cdot m}), a negative wrench, through T(2.591, 0, 5.530)\,\mathrm{ft}, which is (0.790, 0, 1.686)\,\mathrm m.

FA = [0 0 -900]; rA = [2 0 0];        % lb, ft (Marghitu's units)
FB = [0 500 0];  rB = [4 0 6];
FC = [300 0 0];  rC = [0 0 6];
R  = FA + FB + FC;  uR = R/norm(R);
syms x z M
rT = [x 0 z];
MT = cross(rA-rT,FA) + cross(rB-rT,FB) + cross(rC-rT,FC);
s  = solve(MT(1)==M*uR(1), MT(2)==M*uR(2), MT(3)==M*uR(3), [x z M]);
double([s.x s.z s.M])                 % 2.591  5.530  -839.25

Real example. Robot and machine-tool designers describe a tool load as a wrench about the tool's axis: drilling is a positive wrench along the bit. Fixtures are designed so their clamps resist exactly that thrust and twist.

Plausible wrong answer. "P is where the line of the biggest force meets the plane." The wrench axis depends on all three forces; here it passes through (0.3, 0.2, 0), on none of the original lines.

Check. Why does the assumption "positive wrench" not cause an error when the wrench is negative? (The unknown M carries its own sign; a negative value simply reverses the couple.)

Narration spine. The bracket with its three rams. A probe point slides over the base plate, and the moment arrow about it swings. At (0.3, 0.2) the moment arrow snaps into line with \mathbf R, pointing the opposite way. The three rams fade, and a single ram with a twisting arrow takes their place.

Synthesis

The three rams cannot be replaced by one push: their \mathbf R and \mathbf M have \mathbf R\cdot\mathbf M = -72\,000\,\mathrm{N^2\cdot m}, which no choice of point can remove. The best single actuator pushes 600\,\mathrm N along the axis through (0.3, 0.2, 0)\,\mathrm m and twists 120\,\mathrm{N\cdot m} against the push. Every force system in space is one of four things: nothing, a single force, a single couple, or a wrench. The dot product \mathbf R\cdot\mathbf M, the same at every point, decides between a single force and a wrench. All of Part 2 has been building this classification: the moment and couple (S04, S05) and their 3D vector forms (S07, S08) reduce any loading to \mathbf R and \mathbf M. Part 3 asks the opposite question: when both are zero, what must the supports be?

• Central idea: any system reduces to \mathbf R and \mathbf M at a point; it is a single force only if \mathbf R\cdot\mathbf M = 0, and otherwise a wrench.
• Key relationship: \mathbf R = \Sigma\mathbf F, \mathbf M = \Sigma\mathbf r\times\mathbf F; \mathbf M_1 = (\mathbf R\cdot\mathbf M/R^2)\mathbf R; axis through \mathbf R\times\mathbf M/R^2.
• Most important assumption: rigid body, as in S06: the wrench has the same external effect as the original forces, not the same internal stresses.
• Open question: complete equilibrium is \mathbf R = \mathbf 0 and \mathbf M = \mathbf 0. How many unknown support reactions can those six equations find? (Part 3, S10–S12.)

Computed views

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

Python (numpy). S09_wrench.py places the slab's single resultant, shows \mathbf R\cdot\mathbf M is the same about two points, and turns Sample Problem 2/18 into a wrench. It also does Marghitu's Example 2.5 without a symbolic solver: the conditions \mathbf M_T = M\,\mathbf u_R are linear in x, z and M, so one 3\times3 solve gives the axis.

Part 1: three column loads on a slab
  R = [  0.   0. -10.] kN, M_O = [-15.  32.   0.] kN m, acts at x = 3.2, y = 1.5 m
Part 2: R . M does not depend on the point
  M_P = [-300. -300.  300.], R . M_O = 120000, R . M_P = 120000
Part 3: Sample Problem 2/18 as a wrench
  |R| = 1416 N, R . M = -48259, angle 108.8 deg
  M1 = -34.1 N m, pitch -24.1 mm, M2 = 100.4 N m
  axis 70.9 mm from A
Part 4: Marghitu Example 2.5 (lb and ft, as the book)
  R = [ 300.  500. -900.] lb, |R| = 1072.4 lb
  axis through T(2.591, 0, 5.530) ft, M = -839.25 lb ft
  in SI: T(0.790, 0, 1.686) m, M = -1137.9 N m

Part 3 starts from \mathbf R and \mathbf M rounded to three figures, so its \mathbf R\cdot\mathbf M is -48\,259 against Nugget 04's -48\,298 from the unrounded components. Every quantity derived from it agrees to the figures shown.

MATLAB (Octave-compatible). S09_wrench.m does the same, with the backslash operator in place of Marghitu's symbolic solve.

slab: R = -10 kN, M_O = [-15 32 0] kN m, at (3.2, 1.5) m
R . M_O = 120000, R . M_P = 120000
SP 2/18: M1 = -34.1 N m, pitch -24.1 mm, M2 = 100.4 N m, axis 70.9 mm from A
Ex 2.5: |R| = 1072.4 lb, T(2.591, 0, 5.530) ft, M = -839.25 lb ft

Traps

• Thinking the couple is the same at every point. \mathbf R is; \mathbf M changes by \mathbf r\times\mathbf R when the point moves. (From Nugget 01.)
• Leaving applied couples out of \mathbf M. Sample Problem 2/16's three couples are what reduce \mathbf M to 100\mathbf i lb-in. (From Nugget 01.)
• Solving for the coordinate along parallel forces. It drops out; any point on the line will do. (From Nugget 02.)
• Assuming a single force always exists, as in the plane. The bracket has \mathbf R\cdot\mathbf M \ne 0. (From Nugget 03.)
• Hoping a better reference point removes \mathbf M_1. \mathbf R\cdot\mathbf M is the same at every point. (From Nugget 03.)
• Moving \mathbf R by M/R instead of M_2/R. Only the perpendicular part is traded for a shift: 164.9/600 = 0.275\,\mathrm m, not 204/600. (From Nugget 04.)
• Reading a negative M as an error. It means a negative wrench: the couple points against \mathbf R. (From Nugget 05.)

Self-check

1. Forces of 300\,\mathrm N along +z at (2, 0, 0)\,\mathrm m, 200\,\mathrm N along -y at (0, 0, 1.5)\,\mathrm m and 100\,\mathrm N along +x at (0, 1, 0)\,\mathrm m act on a body. Reduce them to a force–couple system at O. Is a single force possible?
2. A block carries 400\,\mathrm N along +x at (0, 0.2, 0) and 400\,\mathrm N along -x at O, and 300\,\mathrm N along +z at (0.5, 0, 0) and 300\,\mathrm N along -z at O (all in metres). Find the resultant.
3. A slab carries downward column loads of 2\,\mathrm{kN} at (0, 0), 3\,\mathrm{kN} at (4, 0) and 5\,\mathrm{kN} at (4, 3)\,\mathrm m. Find the resultant and where it acts.
4. A system reduces to \mathbf R = 2\mathbf i + \mathbf k\ \mathrm{kN} and \mathbf M = \mathbf i + 3\mathbf j - 2\mathbf k\ \mathrm{kN\cdot m} at O. Show that a single force is possible and find a point on its line.
5. A system reduces at O to \mathbf R = 500\mathbf k\ \mathrm N and \mathbf M = 300\mathbf i + 400\mathbf k\ \mathrm{N\cdot m}. Find the wrench: its type, its couple and its axis.
6. For Meriam's Sample Problem 2/18 (\mathbf R = 928\mathbf i + 871\mathbf j + 621\mathbf k\ \mathrm N, \mathbf M = 49.4\mathbf i - 90.5\mathbf j - 24.6\mathbf k\ \mathrm{N\cdot m} at A), find the wrench couple and the distance of its axis from A.
7. A plate in the x-z plane, a = 1.2\,\mathrm m by b = 1.8\,\mathrm m, carries 3\,\mathrm{kN} along -z at (0.6, 0, 0), 2\,\mathrm{kN} along +y at (1.2, 0, 1.8) and 1\,\mathrm{kN} along +x at (0, 0, 1.8)\,\mathrm m. Find the wrench and where its axis crosses the plate.
8. A hand drives a right-handed screw: it pushes 150\,\mathrm N along the screw and turns it with 3\,\mathrm{N\cdot m}. Is the wrench positive or negative? What about pulling a cork with a corkscrew?
9. For the system of Question 1, find the couple at P(1, 2, 0)\,\mathrm m and check that \mathbf R\cdot\mathbf M is unchanged.
10. What force and couple must the supports apply at O to hold the body of Question 1 in equilibrium?

Fully worked solutions

1. \mathbf R = 100\mathbf i - 200\mathbf j + 300\mathbf k\ \mathrm N, 374.2\,\mathrm N. Moments: 2\mathbf i\times300\mathbf k = -600\mathbf j; 1.5\mathbf k\times(-200\mathbf j) = 300\mathbf i; \mathbf j\times100\mathbf i = -100\mathbf k. \mathbf M_O = 300\mathbf i - 600\mathbf j - 100\mathbf k\ \mathrm{N\cdot m}, 678.2\,\mathrm{N\cdot m}. \mathbf R\cdot\mathbf M_O = 30\,000 + 120\,000 - 30\,000 = 120\,000 \ne 0: no single force; it is a positive wrench of pitch 120\,000/140\,000 = 0.857\,\mathrm m.

2. The forces cancel: \mathbf R = \mathbf 0. Moments about O: 0.2\mathbf j\times400\mathbf i = -80\mathbf k and 0.5\mathbf i\times300\mathbf k = -150\mathbf j; the forces at O give none. The resultant is the couple \mathbf M = -150\mathbf j - 80\mathbf k\ \mathrm{N\cdot m}, 170\,\mathrm{N\cdot m}, the same at every point.

3. \mathbf R = -10\mathbf k\ \mathrm{kN}. \mathbf M_O = 4\mathbf i\times(-3\mathbf k) + (4\mathbf i + 3\mathbf j)\times(-5\mathbf k) = 12\mathbf j + 20\mathbf j - 15\mathbf i = -15\mathbf i + 32\mathbf j\ \mathrm{kN\cdot m}. (x\mathbf i + y\mathbf j)\times(-10\mathbf k) = -10y\mathbf i + 10x\mathbf j, so y = 1.5\,\mathrm m and x = 3.2\,\mathrm m: 10\,\mathrm{kN} down through (3.2, 1.5)\,\mathrm m.

4. \mathbf R\cdot\mathbf M = 2 + 0 - 2 = 0, so yes. R^2 = 5; \mathbf R\times\mathbf M = (0 \cdot (-2) - 1 \cdot 3)\mathbf i + (1 \cdot 1 - 2 \cdot (-2))\mathbf j + (2 \cdot 3 - 0)\mathbf k = -3\mathbf i + 5\mathbf j + 6\mathbf k, so \mathbf r = -0.6\mathbf i + \mathbf j + 1.2\mathbf k\ \mathrm m. Check: \mathbf r\times\mathbf R = \mathbf i + 3\mathbf j - 2\mathbf k.

5. \mathbf R\cdot\mathbf M = 200\,000, R^2 = 250\,000, pitch 0.8\,\mathrm m (positive). \mathbf M_1 = 0.8 \times 500\mathbf k = 400\mathbf k\ \mathrm{N\cdot m}, \mathbf M_2 = 300\mathbf i\ \mathrm{N\cdot m}. \mathbf r = \mathbf R\times\mathbf M/R^2 = (500\mathbf k\times300\mathbf i)/250\,000 = 0.6\mathbf j\ \mathrm m. A positive wrench: 500\,\mathrm N along +z on the line through (0, 0.6, 0), with 400\,\mathrm{N\cdot m} about it in the same sense.

6. \mathbf R\cdot\mathbf M = 928(49.4) + 871(-90.5) + 621(-24.6) = -48\,300\,\mathrm{N^2\cdot m}, R = 1416\,\mathrm N. M_1 = -48\,300/1416 = -34.1\,\mathrm{N\cdot m}: a negative wrench. M_2 = \sqrt{106.0^2 - 34.1^2} = 100.4\,\mathrm{N\cdot m}, so the axis is 100.4/1416 = 0.0709\,\mathrm m = 70.9\,\mathrm{mm} from A.

7. \mathbf R = 1\mathbf i + 2\mathbf j - 3\mathbf k\ \mathrm{kN}, R = 3.742\,\mathrm{kN}. Moments about T(x, 0, z) equated to M\mathbf R/R (Marghitu's method) give x = 0.814\,\mathrm m, z = 1.671\,\mathrm m and M = -0.962\,\mathrm{kN\cdot m}: a negative wrench. Check by Nugget 04: at O, \mathbf M_O = 0.6\mathbf i\times(-3\mathbf k) + (1.2\mathbf i + 1.8\mathbf k)\times2\mathbf j + 1.8\mathbf k\times\mathbf i = 1.8\mathbf j + 2.4\mathbf k - 3.6\mathbf i + 1.8\mathbf j = -3.6\mathbf i + 3.6\mathbf j + 2.4\mathbf k\ \mathrm{kN\cdot m}, and \mathbf R\cdot\mathbf M_O/R = (-3.6 + 7.2 - 7.2)/3.742 = -0.962\,\mathrm{kN\cdot m}.

8. Driving a right-handed screw: push and turn are in the same sense, a positive wrench (150\,\mathrm N with 3\,\mathrm{N\cdot m}, pitch 0.02\,\mathrm m). Pulling a cork while turning the corkscrew in: the pull is outward and the twist drives inward, a negative wrench.

9. \mathbf M_P = \mathbf M_O + \mathbf r_{PO}\times\mathbf R with \mathbf r_{PO} = -\mathbf i - 2\mathbf j: (-\mathbf i - 2\mathbf j)\times(100\mathbf i - 200\mathbf j + 300\mathbf k) = -600\mathbf i + 300\mathbf j + 400\mathbf k, so \mathbf M_P = -300\mathbf i - 300\mathbf j + 300\mathbf k\ \mathrm{N\cdot m}. \mathbf R\cdot\mathbf M_P = -30\,000 + 60\,000 + 90\,000 = 120\,000, unchanged.

10. The supports must supply the negative of the resultant at O: -100\mathbf i + 200\mathbf j - 300\mathbf k\ \mathrm N and -300\mathbf i + 600\mathbf j + 100\mathbf k\ \mathrm{N\cdot m}. Then \mathbf R = \mathbf 0 and \mathbf M = \mathbf 0: complete equilibrium.