Study Companion: S08 — Moment and Couple in Three Dimensions

In a plane, a moment is a number with a sign: clockwise or counter-clockwise. In space a force can turn a body about any of infinitely many axes, and the moment needs a direction as well as a size. The cross product, met in S04 for plane problems, now does all the work: \mathbf M_O = \mathbf r\times\mathbf F gives the moment's size, its axis and its sense in one calculation, with no perpendicular distances to find. This lesson builds that tool, uses the dot product of S07 to pick out the moment about any axis, and carries couples and force–couple systems into three dimensions.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/8 ("Moment and Couple"), with Sample Problems 2/11–2/15. Marghitu, Dupac and Madsen, Statics with MATLAB, sections 1.8 ("Vector Product") and 1.9 ("Scalar Triple Product"), 2.1 ("Moment of a Vector About a Point", including the moment about a line) and 2.2 ("Couples"), with Examples 2.2 and 2.3. Harper, Solving Statics Problems in MATLAB, the "3D Moment" problem (a guyed mast). Axes are right-handed with z up, as in S07; Meriam's Sample Problem 2/13 draws its mast with y up and is translated. 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 15\,\mathrm m mast is bolted to a base plate at O. A cable from its top A(0, 0, 15)\,\mathrm m to a ground anchor at B(12, 9, 0)\,\mathrm m is tensioned to 10\,\mathrm{kN} (Marghitu's Example 2.2; Meriam's Sample Problem 2/13 is the same mast).

The base bolts must resist the cable's turning effect. In a plane that would be "force times arm", but here the arm is a perpendicular distance in space from O to a slanting line, and the turning has a direction too. How big is the moment at the base, and about which axis? And if the base is rated for 100\,\mathrm{kN\cdot m}, how far from the mast may the anchor be placed? The answers come in Nugget 03.

Nugget 01: In Space the Moment Is a Vector Normal to the Plane

Claim. The moment of \mathbf F about a point O is the vector \mathbf M_O = \mathbf r\times\mathbf F, where \mathbf r runs from O to any point on the line of action. Its magnitude is Fd, and it points along the normal to the plane containing O and the line of \mathbf F, with the sense given by the right-hand rule.

The picture. Point O and the line of \mathbf F together fix a plane (Meriam's plane A).

Within that plane the situation is exactly S04's: \mathbf F turns the body about an axis through O perpendicular to the plane, with arm d. So the axis of the moment is the plane's normal, and its size is Fd. The cross product \mathbf r\times\mathbf F has both properties built in: it is perpendicular to both \mathbf r and \mathbf F (so normal to their plane), and its magnitude is rF\sin\alpha, where r\sin\alpha = d.

Check with numbers. For the mast, \mathbf T = 5.657\mathbf i + 4.243\mathbf j - 7.071\mathbf k\ \mathrm{kN} (S07: 10\,\mathrm{kN} times \overrightarrow{AB}/AB, with AB = \sqrt{144 + 81 + 225} = 21.21\,\mathrm m). With \mathbf r = \mathbf r_{OA} = 15\mathbf k\ \mathrm m, \mathbf M_O = 15\mathbf k\times(5.657\mathbf i + 4.243\mathbf j - 7.071\mathbf k) = 84.85\mathbf j - 63.64\mathbf i\ \mathrm{kN\cdot m}, using \mathbf k\times\mathbf i = \mathbf j and \mathbf k\times\mathbf j = -\mathbf i. Its size is \sqrt{63.64^2 + 84.85^2} = 106.07\,\mathrm{kN\cdot m}, so the perpendicular distance from O to the cable is d = 106.07/10 = 10.61\,\mathrm m. Now take \mathbf r = \mathbf r_{OB} = 12\mathbf i + 9\mathbf j, the other end of the cable: the result is the same vector. And \mathbf M_O\cdot\mathbf T = (-63.64)(5.657) + (84.85)(4.243) + 0 = 0: the moment is perpendicular to the force, as the picture says.

The idea. Meriam (2/8): in two dimensions the moment-arm rule is quick, but in three the perpendicular distance from a point to a line is tedious to find, and the cross product removes the need. The order matters: \mathbf F\times\mathbf r = -\mathbf M_O. Marghitu (2.1) proves that any point on the line will do. If B' is another point on the line, \mathbf r_{AB'} = \mathbf r_{AB} + \mathbf r_{BB'} with \mathbf r_{BB'} parallel to \mathbf F, and \mathbf r_{BB'}\times\mathbf F = \mathbf 0. The moment is zero exactly when the line passes through the point (or \mathbf F = \mathbf 0). Meriam adds that the moment of a force about a point is a sliding vector: it belongs to the axis through that point.

The equation (definition; |\mathbf M_O| = Fd is a geometric relationship; Meriam's Eq. 2/14):

Worked number (Meriam's Sample Problem 2/11). A force F\mathbf j acts through the point (a, b, c). By the cross product with \mathbf r = a\mathbf i + c\mathbf k (a point on the line, reached without the b step), \mathbf M_O = (a\mathbf i + c\mathbf k)\times F\mathbf j = aF\mathbf k - cF\mathbf i. By inspection: about the y-axis the force is parallel, so no moment; about x its arm is c and it turns negatively; about z its arm is a and it turns positively. Same answer. Using \mathbf r = a\mathbf i + b\mathbf j + c\mathbf k also gives it, because b\mathbf j\times F\mathbf j = \mathbf 0.

Real example. A torque wrench on a wheel nut only measures the moment about the stud's axis. Push the handle partly along the stud and the cross product puts some of the moment on a different axis, bending the stud instead of turning the nut. Mechanics are taught to pull square to the handle for that reason.

Plausible wrong answer. "M = F \times (distance from O to A) = 10 \times 15 = 150\,\mathrm{kN\cdot m}." 15\,\mathrm m is the distance to a point on the line, not to the line. The perpendicular distance is 10.61\,\mathrm m, and the moment 106.07\,\mathrm{kN\cdot m}.

Check. A force passes through O. What is its moment about O? (Zero: take \mathbf r = \mathbf 0.)

Narration spine. A translucent plane slides into place through O and the force's line. Inside it, the familiar 2D picture: arm d, a curled arrow. Then the curl stands up into a green arrow normal to the plane. Slide \mathbf r's tip along the force's line: the green arrow does not change.

Bridge. The cross product gives the right vector, but how is it computed, and why does each component make physical sense?

Nugget 02: The Determinant Is Three Moment-Arm Rules at Once

Claim. In components, \mathbf r\times\mathbf F = (r_yF_z - r_zF_y)\mathbf i + (r_zF_x - r_xF_z)\mathbf j + (r_xF_y - r_yF_x)\mathbf k. Each bracket is the plane moment-arm rule about one coordinate axis.

The picture. Look straight down the x-axis at a force acting at A(r_x, r_y, r_z).

In this view the x-axis is a point, and the force shows two components. F_z acts at horizontal distance r_y from the axis and turns counter-clockwise: +r_yF_z. F_y acts at height r_z and turns clockwise: -r_zF_y. F_x is parallel to the axis and has no moment about it. So the moment about the x-axis is r_yF_z - r_zF_y, the \mathbf i-component of \mathbf r\times\mathbf F. Looking down y and down z gives the other two components by the same rule.

Check with numbers. Meriam's Sample Problem 2/12: a turnbuckle tightens cable AB to 2.4\,\mathrm{kN}. The cable's direction is 0.8\mathbf i + 1.5\mathbf j - 2\mathbf k (length 2.625\,\mathrm m), so \mathbf T = 0.731\mathbf i + 1.371\mathbf j - 1.829\mathbf k\ \mathrm{kN}, acting at A with \mathbf r_{OA} = 1.6\mathbf i + 2\mathbf k\ \mathrm m.

so \mathbf M_O = -2.74\mathbf i + 4.39\mathbf j + 2.19\mathbf k\ \mathrm{kN\cdot m}, of size \sqrt{2.74^2 + 4.39^2 + 2.19^2} = 5.62\,\mathrm{kN\cdot m}. Check the first component by the picture: about x, F_z = -1.829 at r_y = 0 gives nothing, and F_y = 1.371 at height r_z = 2 gives -2(1.371) = -2.74. It matches.

The idea. Marghitu (1.8) defines the cross product as \mathbf v_1\times\mathbf v_2 = |\mathbf v_1||\mathbf v_2|\sin(\mathbf v_1, \mathbf v_2)\,\mathbf n, where \mathbf n is the unit vector along which a right-handed screw advances when turned from \mathbf v_1 to \mathbf v_2. It is zero for parallel vectors, it reverses with the order (\mathbf v_2\times\mathbf v_1 = -\mathbf v_1\times\mathbf v_2) and it distributes over sums. A set \mathbf i, \mathbf j, \mathbf k is right-handed when \mathbf i\times\mathbf j = \mathbf k, and then \mathbf j\times\mathbf k = \mathbf i, \mathbf k\times\mathbf i = \mathbf j, while each unit vector crossed with itself is \mathbf 0. Expanding (a_x\mathbf i + a_y\mathbf j + a_z\mathbf k)\times(b_x\mathbf i + b_y\mathbf j + b_z\mathbf k) by the distributive law and these nine products gives the determinant, expanded along its first row (Marghitu Eq. 1.14). Meriam (2/8) stresses the symmetry: in each component the subscripts run in cyclic order x \to y \to z \to x. In a left-handed frame every sign would flip, which is why S07 insisted on right-handed axes.

The equation (mathematical identity: the determinant expansion of the cross product, Meriam's Eq. 2/15):

Worked number (Marghitu's Example 2.3). A 1\,\mathrm{kN} force acts straight down through E(0, 3, 2)\,\mathrm m. Its moment about A(1, 0, 0)\,\mathrm m: take B(0, 3, 0), where the line meets the floor, so \mathbf r_{AB} = -1\mathbf i + 3\mathbf j\ \mathrm m, and \mathbf M_A = (-\mathbf i + 3\mathbf j)\times(-\mathbf k) = -3\mathbf i - 1\mathbf j\ \mathrm{kN\cdot m} (since \mathbf i\times\mathbf k = -\mathbf j and \mathbf j\times\mathbf k = \mathbf i). With \mathbf r_{AE} = -\mathbf i + 3\mathbf j + 2\mathbf k the extra 2\mathbf k\times(-\mathbf k) = \mathbf 0, and the answer is the same.

rA = [1 0 0];  rB = [0 3 0];  rE = [0 3 2];  F = [0 0 -1];   % m, kN
MB = cross(rB - rA, F)        % [-3 -1 0] kN m
ME = cross(rE - rA, F)        % the same

Real example. Robot arm controllers compute exactly this determinant thousands of times a second: the torque each joint motor must supply is the moment of the payload's weight about that joint's axis.

Plausible wrong answer. Expanding the middle term as +(r_xF_z - r_zF_x)\mathbf j. The determinant's middle term carries a minus sign, which is why the cyclic form (r_zF_x - r_xF_z)\mathbf j is safer. Getting it wrong flips M_y: -4.39 instead of +4.39 in Sample Problem 2/12.

Check. Find \mathbf i\times\mathbf k and \mathbf k\times\mathbf j. (-\mathbf j and -\mathbf i: against the cyclic order, so negative.)

Narration spine. View down the x-axis: the axis shrinks to a dot, the force shows F_y and F_z with their arms, and a readout builds r_yF_z - r_zF_y. The camera swings to look down y, then z, and the other two components build the same way. The three readouts drop into the rows of a determinant.

Bridge. The base plate does not care about the moment "about O" as a whole. Its bolts resist bending about the horizontal axes, and a swivel would care only about the vertical axis. How much of \mathbf M_O acts about a chosen axis?

Nugget 03: The Moment About an Axis Is a Triple Product

Claim. The moment of \mathbf F about an axis \lambda through O, with unit vector \mathbf n, is the component of \mathbf M_O along \mathbf n: M_\lambda = \mathbf r\times\mathbf F\cdot\mathbf n, a triple scalar product.

The picture. Look down on the mast's foot from above. The moment \mathbf M_O = -63.64\mathbf i + 84.85\mathbf j\ \mathrm{kN\cdot m} lies flat in the ground plane, perpendicular to the plan direction of the cable OB.

To find the turning about any axis through O, project \mathbf M_O onto that axis, exactly as S07 projected a force onto a line. About the cable's own plan direction OB the projection is zero: a pull in the vertical plane through OB cannot turn the mast about OB. Perpendicular to OB it is the full 106.07\,\mathrm{kN\cdot m}.

Check with numbers. About the x-axis: M_x = -63.64\,\mathrm{kN\cdot m}; about y: 84.85\,\mathrm{kN\cdot m}; about the vertical z-axis: 0, because the cable's line passes through the mast, which is that axis. About the axis \mathbf n = (\mathbf i + \mathbf j)/\sqrt2: M_\lambda = (-63.64 + 84.85)/\sqrt2 = 15.0\,\mathrm{kN\cdot m}. As a determinant with rows \mathbf r, \mathbf F, \mathbf n, it gives 15.0 again. Now the puzzle: the base must resist 106.07\,\mathrm{kN\cdot m}, about the horizontal axis through O perpendicular to OB, and nothing about the vertical. That exceeds the 100\,\mathrm{kN\cdot m} rating. Meriam's Sample Problem 2/13 finds the component 84.9\,\mathrm{kN\cdot m} about one horizontal axis three ways: as \mathbf M_O\cdot\mathbf k in his y-up frame; as T_{xy}d = 9.06 \times 9.37; and as T_x \times 15 = 5.66 \times 15.

The idea. Meriam (2/8): if \mathbf n is a unit vector along \lambda, then \mathbf M_O\cdot\mathbf n is the magnitude of the moment about \lambda, and \mathbf M_\lambda = (\mathbf r\times\mathbf F\cdot\mathbf n)\mathbf n is that moment as a vector. No brackets are needed in \mathbf r\times\mathbf F\cdot\mathbf n, because \mathbf r\times(\mathbf F\cdot\mathbf n) would cross a vector with a scalar and mean nothing. Marghitu (1.9) calls [\mathbf a, \mathbf b, \mathbf c] = \mathbf a\cdot\mathbf b\times\mathbf c the scalar triple product and shows that the dot and cross may swap places, that a cyclic change of order leaves it unchanged, that swapping two factors changes its sign, and that it is zero when the three vectors are parallel to one plane. It equals the determinant of the three vectors' components. Marghitu (2.1) reads off the special cases of a moment about a line: zero if the line is parallel to the force; zero if the line meets the force's line (choose \mathbf r = \mathbf 0 at the crossing); and d\,F if the two lines are perpendicular, d being their shortest distance. (Marghitu calls this moment a free vector; only its component along the axis matters physically.)

The equation (mathematical identity, the S07 projection applied to \mathbf M_O; Meriam's Eqs. 2/16 and 2/17):

where \alpha, \beta, \gamma are the direction cosines of \mathbf n.

Worked number (Harper's parametric mast). Let the anchor be at horizontal distance \rho from the mast foot, in any direction. The cable has length \sqrt{\rho^2 + 15^2}, the horizontal part of \mathbf T is 10\rho/\sqrt{\rho^2 + 225}, and its arm about O is the mast height 15\,\mathrm m (the vertical part passes through O). So

Setting M_O = 100: 22\,500\rho^2 = 10\,000(\rho^2 + 225), so \rho^2 = 180 and \rho = 13.42\,\mathrm m. The puzzle's anchor is at \rho = \sqrt{12^2 + 9^2} = 15\,\mathrm m, too far. Harper's advice for the construction crew: keep the anchor within a circle of radius 13.42\,\mathrm m (he quotes 44\,\mathrm{ft}) around the mast. Along a line 9\,\mathrm m out to the side, as in his figure, that allows |x_B| \le \sqrt{180 - 81} = 3\sqrt{11} = 9.95\,\mathrm m.

Real example (one you can see). Turn a door handle and push the door: the hinge line is the axis. Pushing on the hinge side, or pushing up or down along the door's face, does nothing, because those forces meet the hinge line or are parallel to it. Only a push with a component across the door and an arm from the hinges opens it.

Plausible wrong answer. "The moment about the z-axis is the z-component of the force times something." Here the force has a large z-component (-7.07\,\mathrm{kN}), but its moment about the z-axis is zero, because the force's line crosses that axis at A.

Check. A force is parallel to an axis. What is its moment about that axis? (Zero: \mathbf r\times\mathbf F is perpendicular to \mathbf F, hence to the axis.)

Narration spine. Top view: the green moment arrow lies flat, square to the cable's shadow. A blue axis spins slowly around O, and the green arrow's shadow on it grows and shrinks, vanishing when the axis lines up with OB. Then the anchor slides outwards, the moment readout climbs past 100 at 13.42\,\mathrm m, and a circle of that radius is drawn on the ground.

Bridge. The mast usually carries several cables, all pulling at the same point A. Must each moment be found and added, or is there a shortcut?

Nugget 04: Varignon Holds in Space

Claim. For forces that meet at one point A, the sum of their moments about O equals the moment of their resultant: \Sigma(\mathbf r\times\mathbf F) = \mathbf r\times\mathbf R.

The picture. A second cable from the mast top runs to C(-9, 12, 0)\,\mathrm m, also at 10\,\mathrm{kN}. Top view of the moments at the foot:

Each cable's moment is perpendicular to its own plan direction, OB or OC. They add as vectors. Because both cables act at A, the same \mathbf r = \mathbf r_{OA} appears in both cross products, and the distributive law lets it be factored out: the sum is \mathbf r_{OA}\times(\mathbf T_1 + \mathbf T_2).

Check with numbers. \mathbf T_2 = 10(-9\mathbf i + 12\mathbf j - 15\mathbf k)/21.21 = -4.243\mathbf i + 5.657\mathbf j - 7.071\mathbf k\ \mathrm{kN}. Separately: \mathbf M_1 = -63.64\mathbf i + 84.85\mathbf j and \mathbf M_2 = 15\mathbf k\times\mathbf T_2 = -84.85\mathbf i - 63.64\mathbf j\ \mathrm{kN\cdot m}, sum -148.49\mathbf i + 21.21\mathbf j\ \mathrm{kN\cdot m}. Together: \mathbf R = 1.414\mathbf i + 9.900\mathbf j - 14.142\mathbf k\ \mathrm{kN}, and 15\mathbf k\times\mathbf R = -148.49\mathbf i + 21.21\mathbf j\ \mathrm{kN\cdot m}. The same, of size 150.0\,\mathrm{kN\cdot m} (the two moments are at right angles, since OB and OC are).

The idea. Meriam (2/8) extends Varignon's theorem from S04: for concurrent forces, \mathbf r\times\mathbf F_1 + \mathbf r\times\mathbf F_2 + \cdots = \mathbf r\times(\mathbf F_1 + \mathbf F_2 + \cdots), by the distributive law of the cross product. Marghitu (1.8) states that law in exactly this form and names it Varignon's theorem. He goes further (2.1): for any system of forces, not only concurrent ones, moments about two points A and P are related by \mathbf M_A = \mathbf M_P + \mathbf r_{AP}\times\mathbf R. For concurrent forces, choosing P at the meeting point makes \mathbf M_P = \mathbf 0 and recovers Varignon. This transfer formula is the 3D form of S06's equipollence argument.

The equation (mathematical identity: distributivity of the cross product; Meriam's Eq. 2/18):

Worked number. A third cable is to be added at A to cancel the base moment of the first two completely. By Varignon it must make \mathbf r_{OA}\times\mathbf T_3 = 148.49\mathbf i - 21.21\mathbf j, that is 15(-T_{3y}\mathbf i + T_{3x}\mathbf j) = 148.49\mathbf i - 21.21\mathbf j, so its horizontal part must be T_{3x} = -1.414, T_{3y} = -9.900\,\mathrm{kN}: equal and opposite to the first two cables' horizontal pull, as common sense expects.

Real example. Every guyed mast and tower is designed this way: all the guys at one level meet at one point, so the designer sums the tensions first and takes one cross product, rather than one per cable.

Plausible wrong answer. "Varignon only works in a plane." The derivation uses nothing but the distributive law, which holds for cross products in space.

Check. If the forces are not concurrent, can the sum of moments still be written as \mathbf r\times\mathbf R for a single \mathbf r? (Not in general. That is the question S09 answers, with the wrench.)

Narration spine. Top view, two green moment arrows at right angles. They slide head to tail into a blue sum. Meanwhile the two cable forces at A merge into one resultant, its moment is computed once, and a second blue arrow lands exactly on the first.

Bridge. The cross product also describes a pair of forces that S05 met in the plane: the couple. In space, couples on different planes have different directions. How do they combine?

Nugget 05: Couples in Space Are Free Vectors That Add

Claim. A couple in space has moment \mathbf M = \mathbf r\times\mathbf F, the same about every point, directed normal to the plane of its two forces. Couples on different planes add as vectors.

The picture. Meriam's Sample Problem 2/14: a block carries two couples. Two 30\,\mathrm N forces 60\,\mathrm{mm} apart, in planes parallel to y-z and tilted at 60^\circ, give M_1 = 30 \times 0.06 = 1.80\,\mathrm{N\cdot m}. Two 25\,\mathrm N forces 100\,\mathrm{mm} apart give M_2 = 25 \times 0.10 = 2.50\,\mathrm{N\cdot m} along -z.

Each couple's vector is normal to its plane, with the right-hand rule's sense. Because each is a free vector, they can be drawn from a common point and added by the parallelogram rule. The sum is one couple, which a single pair of forces \mathbf F and -\mathbf F in a plane normal to it can produce.

Check with numbers. M_1 lies in the y-z plane at 30^\circ above +y: M_y = 1.80\sin60^\circ = 1.559 and its z-part 1.80\cos60^\circ = 0.900\,\mathrm{N\cdot m}. Adding M_2: M_z = -2.50 + 0.900 = -1.600\,\mathrm{N\cdot m}. So M = \sqrt{1.559^2 + 1.600^2} = 2.23\,\mathrm{N\cdot m} at \theta = \tan^{-1}(1.559/1.600) = 44.3^\circ from -z towards +y. If the replacement forces act on the two block faces parallel to y-z, 100\,\mathrm{mm} apart, each is F = 2.23/0.10 = 22.3\,\mathrm N, at 44.3^\circ.

The idea. Meriam (2/8) repeats S05's argument with vectors: for \mathbf F at A and -\mathbf F at B, the moment about any O is \mathbf r_A\times\mathbf F + \mathbf r_B\times(-\mathbf F) = (\mathbf r_A - \mathbf r_B)\times\mathbf F = \mathbf r\times\mathbf F. O has disappeared, so the moment is the same about all points, of size Fd. A couple's moment is a free vector; the moment of a single force about a point is a sliding vector, tied to the axis through that point. Couple vectors obey every rule of vectors, so in Meriam's Fig. 2/26 two couples add to one, and that one can be built from a new pair of forces. Marghitu (2.2) gives the same result and calls the couple's moment its torque.

The equation (mathematical identity: the moment centre cancels; Meriam's Eq. 2/19):

Worked number. A gearbox casing carries a 30\,\mathrm{N\cdot m} couple along +y from its input shaft and a 40\,\mathrm{N\cdot m} couple along -z from its output shaft. The mounting must resist their sum, \sqrt{30^2 + 40^2} = 50\,\mathrm{N\cdot m}, at \tan^{-1}(30/40) = 36.87^\circ from -z towards +y. Two mounting bolts 0.2\,\mathrm m apart, in a plane normal to that vector, would each carry 50/0.2 = 250\,\mathrm N.

Real example. A car's engine and gearbox apply couples about different axes (crankshaft along the car, wheels across it). The engine mounts are placed and angled so that the combined couple vector is resisted with the least vibration.

Plausible wrong answer. "Add the two couples as numbers: 1.80 + 2.50 = 4.30\,\mathrm{N\cdot m}." They act about different axes; only their vector sum, 2.23\,\mathrm{N\cdot m}, is the single equivalent couple.

Check. Two couples, 5\,\mathrm{N\cdot m} along +x and 5\,\mathrm{N\cdot m} along -x, act on different faces of a block. What is the result? (Zero: free vectors, equal and opposite.)

Narration spine. A block with two pairs of forces. Each pair condenses into a green arrow normal to its plane. The arrows slide, as free vectors, to a common tail and add into a blue arrow. Finally the blue arrow unfolds into a single pair of 22.3\,\mathrm N forces on the block's faces.

Bridge. S05 used a couple to move a force sideways in the plane. The same trade works in space, and the cross product supplies the couple.

Nugget 06: Moving a Force in Space Costs the Couple r Cross F

Claim. A force \mathbf F acting through A is equivalent to the same force at any point B plus the couple \mathbf M = \mathbf r\times\mathbf F, where \mathbf r runs from B to any point on the original line of action.

The picture. Meriam's Sample Problem 2/15: a control lever on a fixed shaft along y is pushed at A along -x, into the page in this view along x.

Add \mathbf F and -\mathbf F at O on the shaft. The original force and -\mathbf F form a couple, whose vector is the moment of the original force about O: \mathbf r\times\mathbf F, normal to the plane containing \mathbf r and \mathbf F. Since \mathbf F is along x, that plane contains the x-axis and OA, and \mathbf M lies in the y-z plane, perpendicular to OA.

Check with numbers. The book's data are 40\,\mathrm{lb} (177.9\,\mathrm N) and \mathbf r = 8\mathbf j + 5\mathbf k\ \mathrm{in.} (0.2032\mathbf j + 0.127\mathbf k\ \mathrm m). Then \mathbf M = (0.2032\mathbf j + 0.127\mathbf k)\times(-177.9\mathbf i) = -22.60\mathbf j + 36.16\mathbf k\ \mathrm{N\cdot m} (Meriam: -200\mathbf j + 320\mathbf k\ \mathrm{lb\text{-}in.}), using \mathbf j\times\mathbf i = -\mathbf k and \mathbf k\times\mathbf i = \mathbf j. By the S05 route: the force moves a distance d = \sqrt{0.2032^2 + 0.127^2} = 0.2396\,\mathrm m (9.43 in.), so M = 177.9 \times 0.2396 = 42.6\,\mathrm{N\cdot m} (377 lb-in.), and \sqrt{22.60^2 + 36.16^2} = 42.6 agrees. The vector makes \tan^{-1}(5/8) = 32.0^\circ with the z-axis. Its \mathbf j-part twists the shaft; its \mathbf k-part bends it.

The idea. Meriam (2/8) repeats the force–couple construction of S05 in three dimensions: the couple vector is simply the moment of the original force about the point to which it is moved, and \mathbf r runs from B to any point on the original line through A. Marghitu (2.4, met in S06) states it as an equipollence: \mathbf F at I is equipollent to \mathbf F at P with the couple \mathbf r_{PI}\times\mathbf F. Doing this for every force of a system gives S09's reduction to one force and one couple.

The equation (mathematical identity: adding \mathbf F - \mathbf F = \mathbf 0 at B; Meriam's Fig. 2/27):

Assumption: rigid body; external effects only.

Worked number. A 180\,\mathrm N push along -x on a lever arm at (0, 0.2, 0.125)\,\mathrm m relative to a point O on a shaft along y: \mathbf M = (0.2\mathbf j + 0.125\mathbf k)\times(-180\mathbf i) = -22.5\mathbf j + 36\mathbf k\ \mathrm{N\cdot m}. The shaft carries 180\,\mathrm N of shear, 22.5\,\mathrm{N\cdot m} of torque about its own axis (the -\mathbf j part) and 36\,\mathrm{N\cdot m} of bending.

Real example. A bicycle pedal is this lever. The rider's push, moved to the crank axle, becomes a force on the bearings, a torque that drives the chain, and a bending couple that the crank arm and axle must survive.

Plausible wrong answer. "Moving the force to the shaft only adds a torque about the shaft." The couple \mathbf r\times\mathbf F has components about other axes too: here 36\,\mathrm{N\cdot m} of bending, larger than the twisting part.

Check. If the same force acted at (0, 0.2, 0), on the shaft's level, what would the couple be? (0.2\mathbf j\times(-180\mathbf i) = 36\mathbf k\ \mathrm{N\cdot m}: pure bending, no twist.)

Narration spine. The lever seen end-on along x, the push a cross-in-circle at A. Ghost forces appear at O. The original and one ghost fold into a green couple arrow, swinging round to stand perpendicular to OA. Its two shadows on y and z are labelled "twist" and "bend".

Synthesis

The mast's cable exerts \mathbf M_O = -63.64\mathbf i + 84.85\mathbf j\ \mathrm{kN\cdot m} about the base: 106.07\,\mathrm{kN\cdot m} about the horizontal axis square to the cable's plan direction, and nothing about the vertical. The base's 100\,\mathrm{kN\cdot m} rating caps the anchor distance at 13.42\,\mathrm m. In space, one cross product, \mathbf r\times\mathbf F, replaces the hunt for perpendicular distances, and one dot product with an axis then picks out the turning that matters for that axis. The same two operations handle couples, which are free cross products that add like vectors, and the force–couple trade, which costs the couple \mathbf r\times\mathbf F. With them, any system of forces in space can be moved to one point. What it reduces to there is S09's question.

• Central idea: the moment in space is a vector, \mathbf M_O = \mathbf r\times\mathbf F; its component along any axis is the moment about that axis.
• Key relationship: M_\lambda = \mathbf r\times\mathbf F\cdot\mathbf n; for concurrent forces \Sigma\mathbf r\times\mathbf F = \mathbf r\times\mathbf R; couples \mathbf M = \mathbf r\times\mathbf F add as vectors.
• Most important assumption: right-handed axes; every sign in the determinant depends on \mathbf i\times\mathbf j = \mathbf k.
• Open question: a system's resultant force and resultant couple at a point are generally not perpendicular in space. What is the simplest equivalent then? (S09: the wrench.)

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). S08_moment_3d.py takes Example 2.3's moment from two points on the line. It does the guyed mast cable by cable and then by Varignon with the sum, finds the third cable's horizontal part, sweeps Harper's anchor distance, and splits a lever force's moment into torque and bending.

Part 1: Marghitu Example 2.3, 1 kN down through E
  r_AB x F = [-3. -1. -0.] kN m
  r_AE x F = [-3. -1. -0.] kN m
Part 2: the 15 m mast, two 10 kN cables from A
  M1 = [-63.64  84.85   0.  ], M2 = [-84.85 -63.64   0.  ] kN m
  r x (T1 + T2) = [-148.49   21.21    0.  ] kN m, |M| = 150.0
  third cable's horizontal part = [-1.414 -9.899] kN
Part 3: Harper's sweep, anchor at distance rho
  rho =      5.00 m: M_O = 47.4 kN m
  rho =     10.00 m: M_O = 83.2 kN m
  rho =     13.42 m: M_O = 100.0 kN m
  rho =     15.00 m: M_O = 106.1 kN m
  rho =     20.00 m: M_O = 120.0 kN m
  rho = 1000000.00 m: M_O = 150.0 kN m
Part 4: couple vectors, and a lever force moved to a shaft
  gearbox: C = [  0.  30. -40.] N m, |C| = 50 N m, 36.87 deg from -z
  lever: M = [  0.  -22.5  36. ] N m, torque -22.5, bending 36.0 N m

The -0 is a floating-point zero with a sign; it is zero. The third cable's -9.899 is the -9.900\,\mathrm{kN} of Nugget 04 to one more figure (9.8995).

MATLAB (Octave-compatible). S08_moment_3d.m does the same with cross and norm.

r_AB x F = [-3 -1 -0], r_AE x F = [-3 -1 -0] kN m
M1 + M2 = [-148.49 21.21 0.00], r x (T1 + T2) = [-148.49 21.21 0.00] kN m
|M| = 150.0 kN m; third cable horizontal part [-1.414 -9.899] kN
rho =      5.00 m: M_O = 47.4 kN m
rho =     10.00 m: M_O = 83.2 kN m
rho =     13.42 m: M_O = 100.0 kN m
rho =     15.00 m: M_O = 106.1 kN m
rho =     20.00 m: M_O = 120.0 kN m
rho = 1000000.00 m: M_O = 150.0 kN m
gearbox |C| = 50 N m; lever M = [0.0 -22.5 36.0] N m

Traps

• Using the distance to a point instead of to the line. 15\,\mathrm m to A gives 150\,\mathrm{kN\cdot m}; the true arm is 10.61\,\mathrm m and the moment 106.07\,\mathrm{kN\cdot m}. (From Nugget 01.)
• Reversing the order. \mathbf F\times\mathbf r = -\mathbf M_O: every component flips. (From Nugget 01.)
• Letting \mathbf r run to the moment centre. \mathbf r runs from the moment centre to the line of action. (From Nugget 01.)
• Dropping the middle minus sign. The \mathbf j-component is r_zF_x - r_xF_z. (From Nugget 02.)
• Using a non-unit axis vector. M_\lambda = \mathbf r\times\mathbf F\cdot\mathbf n needs |\mathbf n| = 1; with \mathbf i + \mathbf j instead of (\mathbf i + \mathbf j)/\sqrt2 the mast gives 21.2 instead of 15.0\,\mathrm{kN\cdot m}. (From Nugget 03.)
• Expecting a moment about an axis the force meets. The cable crosses the vertical axis at A, so its moment about that axis is zero, whatever its size. (From Nugget 03.)
• Adding couples as numbers. 1.80 + 2.50 \ne 2.23\,\mathrm{N\cdot m}; couple vectors add as vectors. (From Nugget 05.)
• Forgetting the bending part of a transfer couple. Moving the lever force to the shaft adds 36\,\mathrm{N\cdot m} of bending as well as 22.5\,\mathrm{N\cdot m} of torque. (From Nugget 06.)

Self-check

1. A force \mathbf F = 4\mathbf i + 5\mathbf j - 2\mathbf k\ \mathrm N acts through the point \mathbf r = 2\mathbf i - \mathbf j + 3\mathbf k\ \mathrm m. Find its moment about O and the perpendicular distance from O to its line.
2. A 500\,\mathrm N force acts straight down through (0.3, 0.4, 0)\,\mathrm m. Find its moment about O by inspection, then check by cross product.
3. A 60\,\mathrm N force acts along the line from A(1, 0, 2) towards B(3, 3, 8)\,\mathrm m. Find its moment about O using \mathbf r_{OA} and again using \mathbf r_{OB}.
4. A cable from A(0, 0, 3)\,\mathrm m to B(2, 1.5, 0)\,\mathrm m carries 3\,\mathrm{kN}. Find the moment about O of the cable force acting on A, and its magnitude.
5. For the puzzle's mast, find the moment of the cable about the horizontal axis through O along \mathbf i + \mathbf j.
6. In Harper's mast problem, how far may the anchor be from the mast if the base moment must not exceed 80\,\mathrm{kN\cdot m}?
7. Two 10\,\mathrm{kN} cables from the mast top (0, 0, 15)\,\mathrm m run to B(12, 9, 0) and C(-9, 12, 0)\,\mathrm m. Find the total moment about O two ways.
8. A casing carries a 30\,\mathrm{N\cdot m} couple along +y and a 40\,\mathrm{N\cdot m} couple along -z. Find the single equivalent couple and the forces of a pair 0.2\,\mathrm m apart that could replace it.
9. A 180\,\mathrm N force along -x acts at (0, 0.2, 0.125)\,\mathrm m. Replace it by a force–couple system at O, and check the couple's size by Fd.
10. For the force of Question 1, find the moment about the axis through O and D(2, 1, 2)\,\mathrm m, by the triple-product determinant.
11. A 100\,\mathrm N force acts along +z through (1, 1, 0)\,\mathrm m. Without a full calculation, find its moment about (a) the z-axis, (b) the line through (1, 0, 0) parallel to y, (c) the x-axis.
12. A 2\,\mathrm{kN} force acts straight down through E(0, 2, 3)\,\mathrm m. Find its moment about A(1.5, 0, 0)\,\mathrm m.

Fully worked solutions

1. \mathbf M_O = \mathbf r\times\mathbf F: \mathbf i[(-1)(-2) - (3)(5)] + \mathbf j[(3)(4) - (2)(-2)] + \mathbf k[(2)(5) - (-1)(4)] = -13\mathbf i + 16\mathbf j + 14\mathbf k\ \mathrm{N\cdot m}. |\mathbf M_O| = \sqrt{169 + 256 + 196} = 24.92\,\mathrm{N\cdot m}; F = \sqrt{16 + 25 + 4} = 6.708\,\mathrm N; d = 24.92/6.708 = 3.715\,\mathrm m.

2. By inspection: about x the arm is 0.4\,\mathrm m and a downward force at y = 0.4 turns negatively about x: -200\,\mathrm{N\cdot m}; about y the arm is 0.3\,\mathrm m and it turns positively: +150\,\mathrm{N\cdot m}; about z it is parallel: 0. Cross product: (0.3\mathbf i + 0.4\mathbf j)\times(-500\mathbf k) = -150(\mathbf i\times\mathbf k) - 200(\mathbf j\times\mathbf k) = 150\mathbf j - 200\mathbf i. So \mathbf M_O = -200\mathbf i + 150\mathbf j\ \mathrm{N\cdot m}.

3. \overrightarrow{AB} = 2\mathbf i + 3\mathbf j + 6\mathbf k, length 7\,\mathrm m, so \mathbf F = 17.14\mathbf i + 25.71\mathbf j + 51.43\mathbf k\ \mathrm N. With \mathbf r_{OA} = \mathbf i + 2\mathbf k: \mathbf M_O = \mathbf i[0 - 2(25.71)] + \mathbf j[2(17.14) - 1(51.43)] + \mathbf k[1(25.71) - 0] = -51.43\mathbf i - 17.14\mathbf j + 25.71\mathbf k\ \mathrm{N\cdot m}. With \mathbf r_{OB} = 3\mathbf i + 3\mathbf j + 8\mathbf k: \mathbf i[3(51.43) - 8(25.71)] + \mathbf j[8(17.14) - 3(51.43)] + \mathbf k[3(25.71) - 3(17.14)] = -51.43\mathbf i - 17.14\mathbf j + 25.71\mathbf k. The same, of size 60.0\,\mathrm{N\cdot m}.

4. \overrightarrow{AB} = 2\mathbf i + 1.5\mathbf j - 3\mathbf k, length \sqrt{15.25} = 3.905\,\mathrm m, so \mathbf T = 1.536\mathbf i + 1.152\mathbf j - 2.305\mathbf k\ \mathrm{kN}. \mathbf M_O = 3\mathbf k\times\mathbf T = 3(1.536)\mathbf j - 3(1.152)\mathbf i = -3.457\mathbf i + 4.609\mathbf j\ \mathrm{kN\cdot m}, of size 5.762\,\mathrm{kN\cdot m}.

5. \mathbf n = (\mathbf i + \mathbf j)/\sqrt2: M_\lambda = (-63.64 + 84.85)/\sqrt2 = 21.21/1.4142 = 15.0\,\mathrm{kN\cdot m}.

6. 150\rho/\sqrt{\rho^2 + 225} = 80: 22\,500\rho^2 = 6400\rho^2 + 1\,440\,000, so 16\,100\rho^2 = 1\,440\,000, \rho^2 = 89.44 and \rho = 9.46\,\mathrm m.

7. Separately: \mathbf M_1 = -63.64\mathbf i + 84.85\mathbf j, \mathbf M_2 = 15\mathbf k\times(-4.243\mathbf i + 5.657\mathbf j - 7.071\mathbf k) = -84.85\mathbf i - 63.64\mathbf j; sum -148.49\mathbf i + 21.21\mathbf j\ \mathrm{kN\cdot m}. By Varignon: \mathbf R = 1.414\mathbf i + 9.900\mathbf j - 14.142\mathbf k\ \mathrm{kN} and 15\mathbf k\times\mathbf R = 15(1.414)\mathbf j - 15(9.900)\mathbf i = -148.49\mathbf i + 21.21\mathbf j. Size 150.0\,\mathrm{kN\cdot m}.

8. \mathbf M = 30\mathbf j - 40\mathbf k\ \mathrm{N\cdot m}, M = 50\,\mathrm{N\cdot m} at \tan^{-1}(30/40) = 36.87^\circ from -z towards +y. A pair 0.2\,\mathrm m apart in a plane normal to \mathbf M: F = 50/0.2 = 250\,\mathrm N.

9. At O: the force -180\mathbf i\ \mathrm N and \mathbf M = (0.2\mathbf j + 0.125\mathbf k)\times(-180\mathbf i) = -36(\mathbf j\times\mathbf i) - 22.5(\mathbf k\times\mathbf i) = 36\mathbf k - 22.5\mathbf j\ \mathrm{N\cdot m}. |\mathbf M| = \sqrt{22.5^2 + 36^2} = 42.45\,\mathrm{N\cdot m}. Fd: the force moves \sqrt{0.2^2 + 0.125^2} = 0.2358\,\mathrm m, and 180 \times 0.2358 = 42.45\,\mathrm{N\cdot m}.

10. \mathbf n = (2\mathbf i + \mathbf j + 2\mathbf k)/3. M_\lambda = \begin{vmatrix} 2 & -1 & 3 \\ 4 & 5 & -2 \\ 2/3 & 1/3 & 2/3 \end{vmatrix} = 2\left(\frac{10}{3} + \frac{2}{3}\right) + 1\left(\frac{8}{3} + \frac{4}{3}\right) + 3\left(\frac{4}{3} - \frac{10}{3}\right) = 8 + 4 - 6 = 6\,\mathrm{N\cdot m}. Check with Question 1: (-13)(2/3) + 16(1/3) + 14(2/3) = (-26 + 16 + 28)/3 = 6.

11. (a) Zero: the force is parallel to the z-axis. (b) Zero: the force's line (x = 1, y = 1) meets that line at (1, 1, 0). (c) The x-axis is perpendicular to the force, at shortest distance 1\,\mathrm m (the y-coordinate), so |M_x| = 100\,\mathrm{N\cdot m}; by the cross product, (\mathbf i + \mathbf j)\times100\mathbf k = 100\mathbf i - 100\mathbf j, so M_x = +100\,\mathrm{N\cdot m}.

12. Use B(0, 2, 0) on the line: \mathbf r_{AB} = -1.5\mathbf i + 2\mathbf j\ \mathrm m. \mathbf M_A = (-1.5\mathbf i + 2\mathbf j)\times(-2\mathbf k) = 3(\mathbf i\times\mathbf k) - 4(\mathbf j\times\mathbf k) = -3\mathbf j - 4\mathbf i. So \mathbf M_A = -4\mathbf i - 3\mathbf j\ \mathrm{kN\cdot m}, size 5\,\mathrm{kN\cdot m}; \mathbf r_{AE} adds 3\mathbf k\times(-2\mathbf k) = \mathbf 0.