Study Companion: S07 — Rectangular Components in Three Dimensions

Cables, booms and brackets rarely lie in one plane. A guy wire on a mast pulls down, sideways and backwards all at once. To work with such a force it has to be split into three perpendicular parts, along x, y and z, exactly as S03 split a plane force into two. This lesson builds a force in space from what an engineer is usually given (two points on its line, or two angles) and then uses the dot product to ask how much of a force acts along any line we choose.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/7 ("Rectangular Components"), with Sample Problem 2/10. Marghitu, Dupac and Madsen, Statics with MATLAB, sections 1.6 ("Position Vector") and 1.7 ("Scalar Product of Vectors"), and the force vector of 2.3. Harper, Solving Statics Problems in MATLAB, the "3D Rect. Components" problem (a pole held by cables). Axes are right-handed throughout: turning from x to y advances a right-handed screw along +z. In the oblique drawings, x comes towards the reader, y points right and z up. 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 40\,\mathrm m radio mast stands with its base at O. A guy cable runs from the mast top A(0, 0, 40)\,\mathrm m to a ground anchor at B(24, -18, 0)\,\mathrm m, and its turnbuckle has been tightened to 10\,\mathrm{kN}.

The mast is designed to carry a downward push along its axis, and the other cables are meant to cancel any sideways pull. So the designer needs two numbers: how much of the 10\,\mathrm{kN} pushes down the mast, and how much pulls its top sideways? The cable is not in any of the coordinate planes, so S03's plane trigonometry does not apply directly. The answer comes in Nugget 04.

Nugget 01: Three Perpendicular Components Rebuild Any Force in Space

Claim. Any force in space is the sum of three perpendicular components, \mathbf F = F_x\mathbf i + F_y\mathbf j + F_z\mathbf k, with F_x = F\cos\theta_x, F_y = F\cos\theta_y, F_z = F\cos\theta_z, where \theta_x, \theta_y, \theta_z are the angles between \mathbf F and the positive axes.

The picture. Draw the cable's pull at A as the diagonal of a box whose edges run along the three axes.

Going along the three blue edges, one after another, gets from A to the tip of the orange arrow, so the three edges add up to the diagonal. Each edge is the shadow of \mathbf F on one axis. The diagonal and an edge, such as the vertical one, form a right triangle with the right angle at the edge's far end, so that edge is F\cos\theta_z. Pythagoras twice (first across the base of the box, then up its height) gives the length of the diagonal.

Check with numbers. The cable's pull turns out (Nugget 02) to have edges F_x = 4.8, F_y = -3.6 and F_z = -8.0\,\mathrm{kN}. Its length is \sqrt{4.8^2 + 3.6^2 + 8.0^2} = \sqrt{23.04 + 12.96 + 64} = \sqrt{100} = 10\,\mathrm{kN}, as it should be. The angles with the axes are \theta_x = \cos^{-1}(4.8/10) = 61.31^\circ, \theta_y = \cos^{-1}(-0.36) = 111.10^\circ and \theta_z = \cos^{-1}(-0.8) = 143.13^\circ. A negative component simply means an angle greater than 90^\circ.

The idea. Meriam (2/7) calls l = \cos\theta_x, m = \cos\theta_y and n = \cos\theta_z the direction cosines of \mathbf F. Dividing F^2 = F_x^2 + F_y^2 + F_z^2 by F^2 shows that they always satisfy l^2 + m^2 + n^2 = 1, so only two of them are free. Written as \mathbf F = F(l\mathbf i + m\mathbf j + n\mathbf k), the force is its magnitude F times a unit vector \mathbf n_F = l\mathbf i + m\mathbf j + n\mathbf k that carries only direction: the scalar components of \mathbf n_F are the direction cosines of the line of action. Marghitu (2.3) writes the same force as \mathbf F = F_x\mathbf i + F_y\mathbf j + F_z\mathbf k and adds that two forces add by adding like components, \mathbf R = (F_{1x} + F_{2x})\mathbf i + (F_{1y} + F_{2y})\mathbf j + (F_{1z} + F_{2z})\mathbf k.

The equation (definition of the components and direction cosines; the magnitude and l^2 + m^2 + n^2 = 1 are geometric relationships, Pythagoras; Meriam's Eqs. 2/11 and 2/12):

Worked number. A force has \theta_x = 60^\circ and \theta_y = 45^\circ and a magnitude of 4\,\mathrm{kN}. Then l = 0.5, m = 0.7071 and n^2 = 1 - 0.25 - 0.5 = 0.25, so n = \pm0.5. The two angles do not fix the force: \theta_z is 60^\circ (pointing up) or 120^\circ (pointing down). Taking the upward one, \mathbf F = 2\mathbf i + 2.828\mathbf j + 2\mathbf k\ \mathrm{kN}.

Real example. A crane's load cell on a slewing jib reports three numbers, the force components along the jib's own axes. The controller rebuilds the total load from them by exactly \sqrt{F_x^2 + F_y^2 + F_z^2}.

Plausible wrong answer. "The direction angles are \tan^{-1} of ratios, as in the plane." In space each angle is measured from its own axis to the force, and the cosine is the right function: \theta_z = \cos^{-1}(F_z/F). There is no single "angle of the force".

Check. Can a force make 30^\circ with both the x- and y-axes? (l^2 + m^2 = 0.75 + 0.75 = 1.5 > 1: no.)

Narration spine. An orange arrow floats in space. A transparent box grows around it, edges aligned with the axes. The three edges light up blue one after another and are walked tip to tail from A to the arrow's head. Readouts of 4.8, -3.6 and -8.0 combine under a root sign into 10.

Bridge. Components are easy to use once known. But nobody gives a cable's direction cosines; they give the positions of its two ends.

Nugget 02: Two Points on the Line Give the Direction

Claim. If a force of magnitude F acts along the line from A to B, then \mathbf F = F\,\overrightarrow{AB}/AB: the vector from A to B, divided by its length, is the unit vector of the force.

The picture. Meriam's Sample Problem 2/10: a 100\,\mathrm N force acts at O along the line through A(3, 4, 5)\,\mathrm m.

Walking from O to A takes 3\,\mathrm m along x, 4\,\mathrm m along y and 5\,\mathrm m along z: those legs are the components of the position vector \overrightarrow{OA} = 3\mathbf i + 4\mathbf j + 5\mathbf k\ \mathrm m. The force points the same way, so its components are in the same proportion 3 : 4 : 5. Dividing by the length |OA| strips the metres away and leaves the unit vector; multiplying by F gives newtons.

Check with numbers. |OA| = \sqrt{9 + 16 + 25} = \sqrt{50} = 7.071\,\mathrm m, so \mathbf n_{OA} = 0.424\mathbf i + 0.566\mathbf j + 0.707\mathbf k and \mathbf F = 42.4\mathbf i + 56.6\mathbf j + 70.7\mathbf k\ \mathrm N. For the puzzle's cable, \overrightarrow{AB} = (24 - 0)\mathbf i + (-18 - 0)\mathbf j + (0 - 40)\mathbf k = 24\mathbf i - 18\mathbf j - 40\mathbf k\ \mathrm m, AB = \sqrt{576 + 324 + 1600} = \sqrt{2500} = 50\,\mathrm m, and \mathbf T = 10(0.48\mathbf i - 0.36\mathbf j - 0.8\mathbf k) = 4.8\mathbf i - 3.6\mathbf j - 8.0\mathbf k\ \mathrm{kN}, the box of Nugget 01. The force on the mast points from A towards B, because a cable pulls the body it is attached to towards the cable.

The idea. Meriam (2/7): in most problems the direction of a force is described either by two points on its line or by two angles. For two points, the components are the scalar coefficients of \mathbf i, \mathbf j and \mathbf k in F\,\overrightarrow{AB}/AB. Marghitu (1.6) supplies the tool behind it, the position vector. The position of P relative to O is \mathbf r_{OP} = x_P\mathbf i + y_P\mathbf j + z_P\mathbf k, and of M relative to P it is the difference \mathbf r_{PM} = (x_M - x_P)\mathbf i + (y_M - y_P)\mathbf j + (z_M - z_P)\mathbf k. Its length is the distance PM. The order matters: \mathbf r_{PM} points from P to M, so "head minus tail".

The equation (geometric relationship: a unit vector along a line, Meriam's Fig. 2/17):

Worked number (second view, Marghitu 2.3 by components). A second guy cable on the same mast runs to C(-30, 0, 0)\,\mathrm m with 8\,\mathrm{kN}: \overrightarrow{AC} = -30\mathbf i - 40\mathbf k\ \mathrm m, of length 50\,\mathrm m, so \mathbf T_2 = 8(-0.6\mathbf i - 0.8\mathbf k) = -4.8\mathbf i - 6.4\mathbf k\ \mathrm{kN}. The two cables together pull the top with \mathbf R = (4.8 - 4.8)\mathbf i + (-3.6)\mathbf j + (-8.0 - 6.4)\mathbf k = -3.6\mathbf j - 14.4\mathbf k\ \mathrm{kN}: the x-pulls cancel, 14.4\,\mathrm{kN} pushes down the mast and 3.6\,\mathrm{kN} is left pulling along -y, for a third cable to balance.

A = [0 0 40];  B = [24 -18 0];  C = [-30 0 0];        % m
T1 = 10*(B-A)/norm(B-A)       % [4.8 -3.6 -8.0] kN
T2 =  8*(C-A)/norm(C-A)       % [-4.8 0 -6.4] kN
R  = T1 + T2                  % [0 -3.6 -14.4] kN

Real example (one you can see). Look at the guy wires of a radio mast or a wind-measuring tower: they leave the mast at several heights and fan out to anchors set at equal angles around it, usually three at 120^\circ. That spacing is chosen so that the horizontal components cancel and only the downward ones add.

Plausible wrong answer. "The direction is \overrightarrow{BA}, from the anchor up to the mast." That is the direction of the force the cable exerts on the anchor. On the mast it pulls from A towards B.

Check. A force of 26\,\mathrm N acts from (0, 0, 0) towards (3, 4, 12)\,\mathrm m. What are its components? (6\mathbf i + 8\mathbf j + 24\mathbf k\ \mathrm N, since the length is 13\,\mathrm m.)

Narration spine. Two dots, O and A. A dashed staircase of three legs connects them, labelled 3, 4 and 5\,\mathrm m. The straight arrow OA appears, shrinks to a unit arrow (the division by 7.071), then stretches to 100\,\mathrm N while the leg labels turn into 42.4, 56.6 and 70.7.

Bridge. Two points are the usual description for cables and members. For a force applied by a hand, a jet or a sloping push, the natural description is two angles instead.

Nugget 03: Two Angles Give the Direction

Claim. If a force makes an elevation angle \phi with the x-y plane and its shadow on that plane makes an azimuth angle \theta with the x-axis, then F_z = F\sin\phi, F_x = F\cos\phi\cos\theta and F_y = F\cos\phi\sin\theta.

The picture. A force of 2\,\mathrm{kN} leaves O at \phi = 30^\circ above the horizontal plane, and its shadow lies at \theta = 40^\circ from x.

The construction is two plane resolutions in turn. In the vertical plane that contains \mathbf F, split \mathbf F into a vertical part F\sin\phi and a horizontal part F_{xy} = F\cos\phi (green). Then, in the horizontal plane, split F_{xy} into x and y parts with the angle \theta, as in S03.

Check with numbers. F_{xy} = 2\cos30^\circ = 1.732\,\mathrm{kN} and F_z = 2\sin30^\circ = 1.000\,\mathrm{kN}. Then F_x = 1.732\cos40^\circ = 1.327\,\mathrm{kN} and F_y = 1.732\sin40^\circ = 1.113\,\mathrm{kN}. Rebuilding: \sqrt{1.327^2 + 1.113^2 + 1.000^2} = 2.000\,\mathrm{kN}. The puzzle's cable can be described the same way: its shadow on the ground points along (24, -18), an azimuth of \tan^{-1}(-18/24) = -36.87^\circ, and it falls 40\,\mathrm m over a horizontal run of 30\,\mathrm m, an elevation of -53.13^\circ. Then F_x = 10\cos53.13^\circ\cos36.87^\circ = 4.8\,\mathrm{kN}, the same as by two points.

The idea. Meriam (2/7) resolves \mathbf F first into horizontal and vertical components, F_{xy} = F\cos\phi and F_z = F\sin\phi, and then F_{xy} into x and y. The choice of axes is free (convenience decides), but they must be right-handed, because the cross product of S08 is defined by the right-hand rule: rotate from x to y through 90^\circ, and +z is the direction a right-handed screw advances. Two angles are enough, as Nugget 01 showed: with three direction cosines tied by l^2 + m^2 + n^2 = 1, a direction in space has two degrees of freedom.

The equation (geometric relationship: two right triangles, Meriam's Fig. 2/18):

Assumption: \phi is measured from the x-y plane and \theta from the x-axis. If a problem measures \phi from the vertical instead, \sin and \cos swap.

Worked number. A climber's rope pulls on an anchor bolt with 5\,\mathrm{kN} at 35^\circ above the horizontal, its shadow 50^\circ from x. F_{xy} = 5\cos35^\circ = 4.096\,\mathrm{kN}, F_z = 5\sin35^\circ = 2.868\,\mathrm{kN}, F_x = 4.096\cos50^\circ = 2.633\,\mathrm{kN} and F_y = 4.096\sin50^\circ = 3.138\,\mathrm{kN}.

Real example. Surveyors and antenna installers point things with exactly these two angles: azimuth (compass bearing) and elevation. A satellite dish's mount is set with one scale for each.

Plausible wrong answer. "F_x = F\cos\theta." That skips the first step: only the horizontal part F\cos\phi is split by \theta. With \phi = 30^\circ it would give 1.532\,\mathrm{kN} instead of 1.327\,\mathrm{kN}.

Check. A force is straight up (\phi = 90^\circ). What do F_x and F_y come to, whatever \theta is? (Zero, because \cos90^\circ = 0: the shadow vanishes.)

Narration spine. The force arrow tilts up from the floor plane. A green shadow drops onto the floor, and an arc marks \phi between arrow and shadow. The shadow then splits, with an arc \theta at its base, into x and y parts. Readouts fill in: 1.732, then 1.327 and 1.113, with 1.000 for the height.

Bridge. Every force can now be written in components. The puzzle, though, does not ask for components along x, y or z. It asks how much of the force acts along the mast and how much across it, and in general how much acts along any line.

Nugget 04: The Dot Product Projects a Force onto Any Line

Claim. The component of \mathbf F along a line with unit vector \mathbf n is the dot product F_n = \mathbf F\cdot\mathbf n. As a vector it is (\mathbf F\cdot\mathbf n)\mathbf n.

The picture. Meriam's force again, 100\,\mathrm N along OA, and a line OB through B(6, 6, 2)\,\mathrm m.

Drop a perpendicular from the tip of \mathbf F onto the line OB. The part of the line it cuts off, from O to the foot, is the projection: F\cos\alpha, where \alpha is the angle between \mathbf F and the line. The dot product is designed to give exactly that number.

Check with numbers. \overrightarrow{OB} = 6\mathbf i + 6\mathbf j + 2\mathbf k\ \mathrm m, length \sqrt{76} = 8.718\,\mathrm m, so \mathbf n_{OB} = 0.688\mathbf i + 0.688\mathbf j + 0.229\mathbf k. Then

and as a vector \mathbf F_{OB} = 84.4\,\mathbf n_{OB} = 58.1\mathbf i + 58.1\mathbf j + 19.35\mathbf k\ \mathrm N. The projection on the x-y plane is found the same way, through its angle: \cos\theta_{xy} = \sqrt{3^2 + 4^2}/\sqrt{50} = 0.707, so F_{xy} = 70.7\,\mathrm N. Now the puzzle: the mast's axis is \mathbf k, so the cable pushes down the mast with -\mathbf T\cdot\mathbf k = 8.0\,\mathrm{kN}. The sideways part is what is left, \mathbf T - (\mathbf T\cdot\mathbf k)\mathbf k = 4.8\mathbf i - 3.6\mathbf j\ \mathrm{kN}, of size \sqrt{4.8^2 + 3.6^2} = 6.0\,\mathrm{kN}. Check: 8.0^2 + 6.0^2 = 100 = 10^2.

The idea. Meriam (2/7) defines the dot product (scalar product) of two vectors as \mathbf P\cdot\mathbf Q = PQ\cos\alpha, the product of their magnitudes and the cosine of the angle between them. It can be read as P\cos\alpha (the projection of \mathbf P on \mathbf Q) times Q, or the other way round. With a unit vector, |\mathbf n| = 1, it is the projection itself: F_n = \mathbf F\cdot\mathbf n. In particular F_x = \mathbf F\cdot\mathbf i. To make a vector of it, multiply by \mathbf n again: \mathbf F_n = (\mathbf F\cdot\mathbf n)\mathbf n, which can be written \mathbf F\cdot\mathbf n\mathbf n without brackets because "\mathbf n\mathbf n" means nothing on its own.

Why can the projection be computed by multiplying components? Because \mathbf i, \mathbf j and \mathbf k have unit length and are mutually perpendicular: \mathbf i\cdot\mathbf i = \mathbf j\cdot\mathbf j = \mathbf k\cdot\mathbf k = 1 and every mixed product such as \mathbf i\cdot\mathbf j is 0. Marghitu (1.7) lists the rules that let the product be multiplied out term by term: it is commutative, \mathbf a\cdot\mathbf b = \mathbf b\cdot\mathbf a, and distributive, \mathbf a\cdot(\mathbf b + \mathbf c) = \mathbf a\cdot\mathbf b + \mathbf a\cdot\mathbf c. Expanding (a_x\mathbf i + a_y\mathbf j + a_z\mathbf k)\cdot(b_x\mathbf i + b_y\mathbf j + b_z\mathbf k) gives nine terms; six vanish and three survive. Marghitu then proves that every vector is \mathbf v = (\mathbf i\cdot\mathbf v)\mathbf i + (\mathbf j\cdot\mathbf v)\mathbf j + (\mathbf k\cdot\mathbf v)\mathbf k: dotting \mathbf v = v_x\mathbf i + v_y\mathbf j + v_z\mathbf k with \mathbf i kills all but v_x.

The equation (definition of the dot product; the component formula is a mathematical identity that follows from the rules):

If \mathbf n has direction cosines \alpha, \beta, \gamma and \mathbf F has l, m, n, Meriam writes the projection as F_n = F(l\alpha + m\beta + n\gamma).

Worked number (Harper's pole and cables). A turnbuckle sets the tension in cable CD to T = 1.2\,\mathrm{kN}. The vector from C to D is 1.5\mathbf i + d\mathbf j - 4.5\mathbf k\ \mathrm m, where d is how far out along y the anchor D is placed, and a line GF on the structure has direction 2\mathbf i - 3\mathbf k. The projection of \mathbf T on GF is

The y-part of \mathbf T is perpendicular to GF, so moving the anchor out along y tips more of the tension across GF and less along it.

Real example. On a ski lift or a cable car, the haul rope's tension at each tower is split the same way, into a part along the tower and a part across it. The tower is built strong along its axis and braced against the rest.

Plausible wrong answer. "The projection of \mathbf F on OB is F_x + F_y + F_z along OB," or "multiply the magnitudes, 100 \times 1." The components must be weighted by \mathbf n's components, and the angle between the directions decides the result: 84.4\,\mathrm N, not 100\,\mathrm N or 169.7\,\mathrm N.

Check. Find the projection of \mathbf F = 3\mathbf i - 4\mathbf j + 12\mathbf k\ \mathrm N on the z-axis, and on the line through O and (1, 0, 0). (12\,\mathrm N and 3\,\mathrm N: dotting with \mathbf k and \mathbf i.)

Narration spine. The force arrow and a dashed line OB through space. A light from far above-left, perpendicular to the line, throws the arrow's shadow onto it: a green segment of length 84.4. Swing the line round, and the shadow grows, shrinks, reaches 100 when the line lines up with \mathbf F, and vanishes when the line is perpendicular to it.

Bridge. The projection hid an angle inside it: \mathbf F\cdot\mathbf n = F\cos\alpha. Read backwards, the dot product measures the angle between any two directions, and tells at once when two directions are perpendicular.

Nugget 05: The Dot Product Also Measures the Angle Between Two Directions

Claim. The angle between two vectors is \alpha = \cos^{-1}(\mathbf P\cdot\mathbf Q/PQ). In particular, two non-zero vectors are perpendicular exactly when \mathbf P\cdot\mathbf Q = 0.

The picture. Two guy cables leave the same mast top: AB to (24, -18, 0) and AC to (-30, 0, 0)\,\mathrm m.

The angle at A cannot be read off any coordinate plane, since neither cable lies in one with the other. But the dot product needs only components: multiply matching components and add, then divide by the two lengths.

Check with numbers. \overrightarrow{AB}\cdot\overrightarrow{AC} = (24)(-30) + (-18)(0) + (-40)(-40) = -720 + 0 + 1600 = 880\,\mathrm{m^2}. Both lengths are 50\,\mathrm m, so \cos\alpha = 880/2500 = 0.352 and \alpha = 69.39^\circ. For the angle between the puzzle's cable and the mast, dot \mathbf n_{AB} with -\mathbf k (down the mast): \cos\alpha = 0.8, so \alpha = 36.87^\circ. And between Meriam's force and the line OB: \cos\alpha = 84.36/100, so \alpha = 32.48^\circ.

The idea. Meriam (2/7) rearranges the definition: since \mathbf F\cdot\mathbf n = F\cos\alpha for a unit vector \mathbf n, the angle is \alpha = \cos^{-1}(\mathbf F\cdot\mathbf n/F), and for any two vectors \alpha = \cos^{-1}(\mathbf P\cdot\mathbf Q/PQ). If \mathbf F is perpendicular to the line, \cos\alpha = 0 and \mathbf F\cdot\mathbf n = 0, even though neither vector is zero. That is unlike ordinary multiplication, where a zero product needs a zero factor. The dot product also works for vectors that do not meet: treated as free vectors, each can be slid to meet the other without changing the product.

The equation (mathematical identity, the definition solved for the angle; Meriam's Eqs. 2/13 and 2/13a):

Worked number. A force \mathbf F = 4\mathbf i + F_y\mathbf j - 2\mathbf k\ \mathrm{kN} must act perpendicular to a shaft along \mathbf i + 2\mathbf j + 3\mathbf k, so that it has no component pushing along the shaft. Setting \mathbf F\cdot(\mathbf i + 2\mathbf j + 3\mathbf k) = 4 + 2F_y - 6 = 0 gives F_y = 1\,\mathrm{kN}.

Real example. A bearing designer checks exactly this: a load on a shaft that has a non-zero dot product with the shaft's axis is partly a thrust, and needs a thrust bearing. A purely radial load has zero dot product with the axis.

Plausible wrong answer. "If \mathbf P\cdot\mathbf Q = 0, one of them must be zero." The cables \mathbf T and \mathbf T_2 are both large; a dot product of zero would only mean they are at 90^\circ.

Check. Are \mathbf a = 3\mathbf i - 2\mathbf j + 6\mathbf k and \mathbf b = 2\mathbf i + 6\mathbf j + \mathbf k perpendicular? (6 - 12 + 6 = 0: yes.)

Narration spine. Two cable arrows from the mast top. Their components appear in two columns, are multiplied pairwise (-720, 0, 1600) and summed to 880. The lengths 50 and 50 divide it to 0.352, and an arc opens between the cables to 69.4^\circ. Rotate one cable until the products cancel: the arc snaps to a right-angle mark.

Synthesis

The guy cable's 10\,\mathrm{kN} is 4.8\mathbf i - 3.6\mathbf j - 8.0\mathbf k\ \mathrm{kN}, written from its two end points. Dotting with the mast's axis splits it: 8.0\,\mathrm{kN} down the mast and 6.0\,\mathrm{kN} sideways, at 36.87^\circ to the mast. In space, a force is a magnitude times a unit vector, and every question about "how much of it acts along this line" is a dot product with that line's unit vector. Components along the axes are the special case of dotting with \mathbf i, \mathbf j and \mathbf k. Whatever the description of the direction (two points, two angles or three direction cosines), it ends as \mathbf n, and the rest is arithmetic. S08 adds the second product, the cross product, to find the moment of such a force.

• Central idea: a force in space is F\mathbf n_F, and its component along any direction is \mathbf F\cdot\mathbf n.
• Key relationship: \mathbf F = F\,\overrightarrow{AB}/AB; F_n = \mathbf F\cdot\mathbf n = F\cos\alpha; l^2 + m^2 + n^2 = 1.
• Most important assumption: right-handed, mutually perpendicular unit vectors; the component formula for the dot product rests on \mathbf i\cdot\mathbf i = 1 and \mathbf i\cdot\mathbf j = 0.
• Open question: the dot product says how much of a force acts along a line. How much does it turn a body about a point or a line in space? (S08.)

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). S07_components_3d.py builds a force three ways (direction cosines, azimuth and elevation, two points), sweeps Harper's projection of the cable tension on GF over the anchor offset d, and uses a zero dot product to fix F_y.

Part 1: three ways to give a 3D force
  cosines l = 0.5000, m = 0.7071, n = +/-0.5000: F = [2.    2.828 2.   ] kN
  climber's rope: F = [2.633 3.138 2.868] kN
  mast: T1 = [ 4.8 -3.6 -8. ], T2 = [-4.8  0.  -6.4], R = [  0.   -3.6 -14.4] kN
Part 2: Harper's projection of T = 1.2 kN on the line GF
  d =  0 m: T_GF = 1.158 kN
  d =  3 m: T_GF = 0.978 kN
  d =  5 m: T_GF = 0.797 kN
  d = 10 m: T_GF = 0.496 kN
  d = 20 m: T_GF = 0.267 kN
Part 3: no component along the shaft
  F_y = 1.0 kN, F . shaft = 0.0

MATLAB (Octave-compatible). S07_components_3d.m does the same with norm and dot.

cosines: n = +/-0.5000, F = [2.000 2.828 2.000] kN
climber's rope: F = [2.633 3.138 2.868] kN
mast: T1 = [4.80 -3.60 -8.00], T2 = [-4.80 0.00 -6.40], R = [0.00 -3.60 -14.40] kN
d =  0 m: T_GF = 1.158 kN
d =  3 m: T_GF = 0.978 kN
d =  5 m: T_GF = 0.797 kN
d = 10 m: T_GF = 0.496 kN
d = 20 m: T_GF = 0.267 kN
F_y = 1.0 kN, F . s = 0.0

Traps

• Using \tan^{-1} for direction angles in space. Each angle is \cos^{-1} of a component over the magnitude, measured from its own axis. (From Nugget 01.)
• Forgetting that two direction angles leave a sign open. \theta_x = 60^\circ, \theta_y = 45^\circ allows \theta_z = 60^\circ or 120^\circ. (From Nugget 01.)
• Tail minus head. \overrightarrow{AB} = \mathbf r_B - \mathbf r_A; reversing it reverses the force. A cable pulls the body towards the cable's far end. (From Nugget 02.)
• Forgetting to divide by the length. 10\,\overrightarrow{AB} has magnitude 500\,\mathrm{kN}, not 10; only \overrightarrow{AB}/AB is a unit vector. (From Nugget 02.)
• Splitting the force by θ before φ. F_x = F\cos\phi\cos\theta, not F\cos\theta. (From Nugget 03.)
• Dotting with a non-unit vector. \mathbf F\cdot\overrightarrow{OB} is 8.718 times too big; use \mathbf n_{OB}. (From Nugget 04.)
• Confusing the scalar and vector projection. \mathbf F\cdot\mathbf n = 84.4\,\mathrm N is a number; (\mathbf F\cdot\mathbf n)\mathbf n is a vector along OB. (From Nugget 04.)
• Reading a zero dot product as a zero vector. It means perpendicular. (From Nugget 05.)

Self-check

1. A force has components F_x = 120\,\mathrm N, F_y = -90\,\mathrm N and F_z = 200\,\mathrm N. Find its magnitude and its direction angles.
2. A cable pulls on a bracket at A(1, 2, 0)\,\mathrm m with 2.6\,\mathrm{kN}, towards B(-3, 5, 12)\,\mathrm m. Write the force as a vector.
3. A 5\,\mathrm{kN} force has elevation 35^\circ above the x-y plane and azimuth 50^\circ from x. Find its components.
4. A force makes 60^\circ with the x-axis and 45^\circ with the y-axis. What angle does it make with the z-axis?
5. Two cables pull on the top of a mast at A(0, 0, 40)\,\mathrm m: 10\,\mathrm{kN} towards B(24, -18, 0) and 8\,\mathrm{kN} towards C(-30, 0, 0)\,\mathrm m. Find their resultant and the angle it makes with the vertical.
6. A 600\,\mathrm N force acts from O towards A(2, 3, 6)\,\mathrm m. Find (a) its components, (b) its projection on the x-y plane, (c) its scalar and vector projections on the line from O to B(4, 4, -2)\,\mathrm m.
7. Find the angle between the two cables of Question 5.
8. A force \mathbf F = 4\mathbf i + F_y\mathbf j - 2\mathbf k\ \mathrm{kN} must have no component along the direction \mathbf i + 2\mathbf j + 3\mathbf k. Find F_y.
9. In Harper's pole problem (Nugget 04), find T_{GF} when d = 6\,\mathrm m, and the distance d at which T_{GF} = 0.6\,\mathrm{kN}.
10. Three forces act at one point: \mathbf F_1 = 3\mathbf i + 4\mathbf j - 2\mathbf k, \mathbf F_2 = -\mathbf i + 5\mathbf j + 2\mathbf k and \mathbf F_3 = 6\mathbf i - 3\mathbf j + 4\mathbf k, all in newtons. Find the resultant, its magnitude and its direction cosines.
11. Split \mathbf F = 200\mathbf i - 100\mathbf j + 50\mathbf k\ \mathrm N into a part along \mathbf n = 0.6\mathbf i + 0.8\mathbf j and a part perpendicular to it. Find the angle between \mathbf F and \mathbf n.

Fully worked solutions

1. F = \sqrt{120^2 + 90^2 + 200^2} = \sqrt{62\,500} = 250\,\mathrm N. \theta_x = \cos^{-1}(120/250) = 61.31^\circ, \theta_y = \cos^{-1}(-90/250) = 111.10^\circ, \theta_z = \cos^{-1}(200/250) = 36.87^\circ. Check: 0.48^2 + 0.36^2 + 0.8^2 = 1.

2. \overrightarrow{AB} = (-3 - 1)\mathbf i + (5 - 2)\mathbf j + (12 - 0)\mathbf k = -4\mathbf i + 3\mathbf j + 12\mathbf k\ \mathrm m, of length \sqrt{16 + 9 + 144} = 13\,\mathrm m. \mathbf F = 2.6(-4\mathbf i + 3\mathbf j + 12\mathbf k)/13 = -0.8\mathbf i + 0.6\mathbf j + 2.4\mathbf k\ \mathrm{kN}.

3. F_{xy} = 5\cos35^\circ = 4.096\,\mathrm{kN}, F_z = 5\sin35^\circ = 2.868\,\mathrm{kN}, F_x = 4.096\cos50^\circ = 2.633\,\mathrm{kN}, F_y = 4.096\sin50^\circ = 3.138\,\mathrm{kN}.

4. l = \cos60^\circ = 0.5, m = \cos45^\circ = 0.7071, so n^2 = 1 - 0.25 - 0.5 = 0.25 and n = \pm0.5. \theta_z = 60^\circ if the force points up, 120^\circ if it points down. Both are valid; the data do not decide.

5. \mathbf T_1 = 4.8\mathbf i - 3.6\mathbf j - 8.0\mathbf k\ \mathrm{kN} (Nugget 02) and \mathbf T_2 = -4.8\mathbf i - 6.4\mathbf k\ \mathrm{kN}. \mathbf R = -3.6\mathbf j - 14.4\mathbf k\ \mathrm{kN}, R = \sqrt{3.6^2 + 14.4^2} = 14.84\,\mathrm{kN}, at \tan^{-1}(3.6/14.4) = 14.04^\circ from the downward vertical, leaning towards -y.

6. (a) |OA| = \sqrt{4 + 9 + 36} = 7\,\mathrm m, so \mathbf F = 600(2\mathbf i + 3\mathbf j + 6\mathbf k)/7 = 171.43\mathbf i + 257.14\mathbf j + 514.29\mathbf k\ \mathrm N. (b) F_{xy} = 600\sqrt{2^2 + 3^2}/7 = 309.05\,\mathrm N. (c) |OB| = \sqrt{16 + 16 + 4} = 6\,\mathrm m, \mathbf n_{OB} = \frac{2}{3}\mathbf i + \frac{2}{3}\mathbf j - \frac{1}{3}\mathbf k. F_{OB} = (171.43)(\frac{2}{3}) + (257.14)(\frac{2}{3}) - (514.29)(\frac{1}{3}) = 114.29 + 171.43 - 171.43 = 114.29\,\mathrm N. As a vector: 114.29\,\mathbf n_{OB} = 76.19\mathbf i + 76.19\mathbf j - 38.10\mathbf k\ \mathrm N.

7. \overrightarrow{AB}\cdot\overrightarrow{AC} = 880\,\mathrm{m^2} and both lengths are 50\,\mathrm m: \alpha = \cos^{-1}(0.352) = 69.39^\circ.

8. \mathbf F\cdot(\mathbf i + 2\mathbf j + 3\mathbf k) = 4 + 2F_y - 6 = 0, so F_y = 1\,\mathrm{kN}.

9. T_{GF} = 19.8/(\sqrt{13}\sqrt{22.5 + 36}) = 19.8/(3.6056 \times 7.6485) = 0.718\,\mathrm{kN}. For 0.6\,\mathrm{kN}: \sqrt{22.5 + d^2} = 19.8/(0.6\sqrt{13}) = 9.1526, so d^2 = 83.77 - 22.5 = 61.27 and d = 7.83\,\mathrm m.

10. \mathbf R = (3 - 1 + 6)\mathbf i + (4 + 5 - 3)\mathbf j + (-2 + 2 + 4)\mathbf k = 8\mathbf i + 6\mathbf j + 4\mathbf k\ \mathrm N. R = \sqrt{64 + 36 + 16} = \sqrt{116} = 10.77\,\mathrm N. Direction cosines: 8/10.77 = 0.743, 6/10.77 = 0.557, 4/10.77 = 0.371.

11. \mathbf F\cdot\mathbf n = 200(0.6) - 100(0.8) + 0 = 40\,\mathrm N. Parallel part: 40\mathbf n = 24\mathbf i + 32\mathbf j\ \mathrm N. Perpendicular part: \mathbf F - 40\mathbf n = 176\mathbf i - 132\mathbf j + 50\mathbf k\ \mathrm N; check 176(0.6) - 132(0.8) = 105.6 - 105.6 = 0. F = \sqrt{52\,500} = 229.13\,\mathrm N, so the angle is \cos^{-1}(40/229.13) = 79.95^\circ. Check: 40^2 + 225.61^2 = 52\,500 = F^2.