Study Companion: S06 — Resultants of Coplanar Forces

A body is usually loaded by many forces at once. Before asking whether it stays put or what it does, it helps to replace them all by the simplest set of forces that does the same thing to the body: the resultant. In a plane, the resultant is a single force on a definite line, or, when the forces cancel, a single couple. This lesson finds that force, places its line, and shows why two quite different-looking systems can count as the same one.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/6 ("Resultants"), with Sample Problem 2/9. Marghitu, Dupac and Madsen, Statics with MATLAB, section 2.4 ("Equipollent Force Systems"). Harper, Solving Statics Problems in MATLAB, the "2D Resultants" problem (a thrust-vectored aircraft). Marghitu's word equipollent means the same as Meriam's equivalent: equal resultant force and equal moment. Sign convention throughout: counter-clockwise is positive, +\mathbf k out of the page. 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 four-engine airliner cruises straight and level, each engine pushing 90\,\mathrm{kN} forward. The inboard engines are 10\,\mathrm m from the centreline and the outboard ones 20\,\mathrm m. Engine 3, the inboard engine on the starboard (right) wing, suddenly fails.

The three remaining thrusts add to 270\,\mathrm{kN} forward. That is easy. But one 270\,\mathrm{kN} force on the centreline would push the aircraft straight, and pilots know that an engine failure makes the nose swing towards the dead engine. So the single force that replaces the three thrusts cannot be on the centreline. Where is it, and how much does it turn the aircraft? The answer comes in Nugget 06.

Nugget 01: The Force Polygon Gives the Size and Direction of the Resultant

Claim. The resultant force of any set of coplanar forces is their vector sum, \mathbf R = \Sigma\mathbf F. Its components are R_x = \Sigma F_x and R_y = \Sigma F_y, and the order of adding does not matter.

The picture. A plate (Meriam's Sample Problem 2/9) carries four forces: 40\,\mathrm N along +x, 80\,\mathrm N at 30^\circ above +x, 50\,\mathrm N straight up, and 60\,\mathrm N pointing up and to the left at 45^\circ. Draw the four arrows head to tail, keeping each one's length and direction.

The green arrow from the first tail to the last head is the sum. On the right the same four arrows are taken in a different order. The polygon changes shape, but its closing arrow is the same: same length, same direction. Adding vectors is commutative, so the polygon can be built in any order.

Check with numbers. By components:

R_x = 40 + 80\cos30^\circ - 60\cos45^\circ = 40 + 69.28 - 42.43 = 66.86\,\mathrm N

R_y = 0 + 80\sin30^\circ + 50 + 60\sin45^\circ = 40 + 50 + 42.43 = 132.43\,\mathrm N

So R = \sqrt{66.86^2 + 132.43^2} = 148.3\,\mathrm N and \theta = \tan^{-1}(132.43/66.86) = 63.2^\circ above +x. Measured off the drawing (1.4 px per newton), the closing arrow is 93.6\,\mathrm{px} across and 185.4\,\mathrm{px} up, the same numbers scaled.

The idea. Meriam (2/6) defines the resultant of a force system as the simplest force combination that can replace the original forces without altering the external effect on the rigid body. If the resultant is zero, the body is in equilibrium, and that is what statics studies. If it is not zero, dynamics equates it to mass times acceleration. Either way, finding the resultant comes first. The force polygon (head to tail, in any sequence) gives its magnitude and direction. Components turn the polygon into two sums.

The equation (mathematical identity: vector addition by components, Meriam's Eq. 2/9):

\mathbf R = \Sigma\mathbf F, \qquad R_x = \Sigma F_x, \quad R_y = \Sigma F_y, \qquad R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}, \quad \theta = \tan^{-1}\frac{\Sigma F_y}{\Sigma F_x}

Use the signs of R_x and R_y, not the calculator's \tan^{-1} alone, to put \theta in the right quadrant.

Worked number. A bracket is pulled by 500\,\mathrm N along +x and by 300\,\mathrm N at 120^\circ from +x. R_x = 500 + 300\cos120^\circ = 500 - 150 = 350\,\mathrm N and R_y = 300\sin120^\circ = 259.8\,\mathrm N, so R = 435.9\,\mathrm N at \tan^{-1}(259.8/350) = 36.6^\circ.

Real example. A tow truck freeing a car from a ditch uses two cables at different angles. The car feels one pull, the vector sum, and moves along its direction, not along either cable.

Plausible wrong answer. "R = 40 + 80 + 50 + 60 = 230\,\mathrm N." Adding sizes ignores direction. The forces partly cancel sideways, and the true resultant is 148.3\,\mathrm N.

Check. Two 100\,\mathrm N forces, one along +x and one along +y. What is their resultant? (141.4\,\mathrm N at 45^\circ.)

Narration spine. Four orange arrows sit on the plate. They lift off and snap head to tail. A green arrow closes the chain, with readouts 66.9 and 132.4 on its legs. Shuffle the order: the chain wriggles into a new shape, and the green arrow does not move.

Bridge. The polygon threw away one thing: where each force acts. Moving the arrows around to chain them lost their lines of action. The resultant's size and direction are known, but not where it acts, and for a rigid body that matters as much as the size.

Nugget 02: Keeping the Lines Finds Where the Resultant Acts

Claim. Sliding forces only along their own lines (transmissibility) and adding them two at a time where those lines cross gives the resultant on its correct line of action.

The picture. Three forces: F_1 = 200\,\mathrm N along +x on the line y = 3\,\mathrm m; F_2 = 300\,\mathrm N up on the line x = 1\,\mathrm m; and F_3 = 141.4\,\mathrm N at 45^\circ through (3, 0)\,\mathrm m, that is (100, 100)\,\mathrm N.

Slide F_2 and F_3 along their dashed lines until both tails sit where the lines cross, at P. There the parallelogram gives \mathbf R_1 = \mathbf F_2 + \mathbf F_3, on a line through P. Slide \mathbf R_1 along its own line until it meets the line of F_1, at Q, and add again. The final arrow is \mathbf R, and it is on the right line because every move was along a line of action.

Check with numbers. The line of F_3 is y = x - 3, which meets x = 1 at P(1, -2)\,\mathrm m. \mathbf R_1 = (0 + 100, 300 + 100) = (100, 400)\,\mathrm N, so its line climbs 4\,\mathrm m for every 1\,\mathrm m across: y = -2 + 4(x - 1). It meets y = 3 at x = 2.25\,\mathrm m, so Q = (2.25, 3)\,\mathrm m. Then \mathbf R = (100 + 200, 400) = (300, 400)\,\mathrm N: 500\,\mathrm N at 53.13^\circ. Its line through Q has slope 400/300 = 4/3, and 3/2.25 = 4/3 too, so the line passes through the origin O. Check that with moments about O (M = xF_y - yF_x): F_1 gives -3(200) = -600, F_2 gives 1(300) = 300, F_3 gives 3(100) - 0 = 300\,\mathrm{N\cdot m}. They add to zero, exactly as a resultant through O requires.

The idea. Meriam (2/6): graphically, the correct line of action of \mathbf R is found by preserving the forces' lines of action and adding them by the parallelogram law, using the principle of transmissibility (S03). This is why the polygon of Nugget 01 is not enough on its own: it gives \mathbf R as a free vector. The construction here gives it as a sliding vector on one line.

The equation (geometric relationship): there is no new formula. The construction is two applications of the parallelogram law, \mathbf R_1 = \mathbf F_2 + \mathbf F_3 at P and \mathbf R = \mathbf F_1 + \mathbf R_1 at Q, with P and Q found as intersections of lines of action.

Assumption: rigid body, so sliding a force along its line changes no external effect.

Worked number. Two forces: 400\,\mathrm N down on the line x = 0 and 300\,\mathrm N to the right on the line y = 2\,\mathrm m. Their lines meet at (0, 2)\,\mathrm m, so the resultant, 500\,\mathrm N at 53.13^\circ below +x, passes through (0, 2)\,\mathrm m.

Real example. Before calculators, bridge designers found the resultant of the loads on a roof truss exactly this way, with a ruler and set square on the drawing board. The method is called graphical statics.

Plausible wrong answer. "The resultant acts at the centre of the three points of application." Nothing in the construction refers to that point. Here the resultant passes through O, which is not the centroid of (0, 3), (1, 0) and (3, 0).

Check. If two forces are parallel, why does this construction fail? (Their lines never meet. Nugget 06 handles parallel forces with moments instead.)

Narration spine. Three arrows on their dashed lines. Two of them glide down their lines to the crossing P, and a parallelogram blooms into a green \mathbf R_1. \mathbf R_1 glides up its own line to meet the line of F_1 at Q. Another parallelogram, and the final green \mathbf R lies on a line that runs straight through O.

Bridge. The drawing works, but it needs lines that intersect on the page, and its accuracy is that of the pencil. An algebraic route that gives the same line by arithmetic would be better, and S05 already has the tool: moving a force to a point costs a couple.

Nugget 03: Move Every Force to One Point, Then Add Forces and Couples

Claim. Any coplanar system is equivalent to one force \mathbf R = \Sigma\mathbf F at a chosen point O plus one couple M_O = \Sigma M_O, the sum of the moments of all the forces about O, plus any applied couples.

The picture. Meriam's plate again, now with the forces in place and a 140\,\mathrm{N\cdot m} counter-clockwise couple. O is at the lower right. The 40\,\mathrm N and 80\,\mathrm N forces act on lines through O.

Move each force to O. By S05, each move adds a couple equal to that force's moment about O. The forces now all act at O and add as in Nugget 01 to \mathbf R. The couples are free vectors normal to the plane, so they simply add as signed numbers, together with the applied 140\,\mathrm{N\cdot m}. The result, on the right, is one force and one couple at O.

Check with numbers. \mathbf R = (66.86, 132.43)\,\mathrm N, as in Nugget 01. For the moments, use M = xF_y - yF_x with O as origin:

40\,\mathrm N and 80\,\mathrm N: their lines pass through O, so 0;
50\,\mathrm N up at (-5, 4)\,\mathrm m: (-5)(50) = -250\,\mathrm{N\cdot m};
60\,\mathrm N at (-7, 4)\,\mathrm m, components (-42.43, 42.43)\,\mathrm N: (-7)(42.43) - 4(-42.43) = -297.0 + 169.7 = -127.3\,\mathrm{N\cdot m};
the couple: +140\,\mathrm{N\cdot m}.

M_O = 140 - 250 - 127.3 = -237.3\,\mathrm{N\cdot m}, that is 237\,\mathrm{N\cdot m} clockwise. Meriam writes it with arms read off the drawing: M_O = 140 - 50(5) + 60\cos45^\circ(4) - 60\sin45^\circ(7) = -237\,\mathrm{N\cdot m}.

The idea. Meriam's algebraic method (2/6) has three steps. (1) Choose a convenient point and move all forces to it, each bringing its transfer couple. (2) Add the forces to form \mathbf R and the couples to form M_O: this is the force–couple system at O. (3) Place \mathbf R alone (Nugget 04). Choosing O on the lines of several forces, as here, removes their moments from the sum. Meriam notes that with a clockwise-positive convention M_O would be +237\,\mathrm{N\cdot m}: the sign follows the convention, the sense does not.

Marghitu (2.4) states the same reduction as a theorem. However complicated a system of forces and moments, it can be represented by a single force at any given point P and a single couple, with the force equal to the sum of the forces and the couple equal to the sum of the moments about P. The force does not depend on the choice of P; the couple does.

The equation (mathematical identity, the S05 force–couple transfer applied to every force; Meriam's first two Eqs. 2/10):

\mathbf R = \Sigma\mathbf F, \qquad M_O = \Sigma M = \Sigma(Fd)

where each Fd carries its sign, and applied couples join the sum.

Worked number. A beam 3\,\mathrm m long, with O at its left end, carries 2\,\mathrm{kN} down at 1\,\mathrm m, 4\,\mathrm{kN} down at 3\,\mathrm m and a 1.5\,\mathrm{kN\cdot m} clockwise couple. At O: \mathbf R = 6\,\mathrm{kN} down and M_O = 1(-2) + 3(-4) - 1.5 = -15.5\,\mathrm{kN\cdot m}, clockwise.

Real example. This is what a cantilever's wall anchor must supply in reverse. The bolts at a balcony's root carry the balcony's total load as a force and the sum of its moments as a couple. Designers size the bolts from exactly this force–couple pair.

Plausible wrong answer. "Forget the 140\,\mathrm{N\cdot m} couple when finding \mathbf R, and forget it when finding M_O too, since it is not a force." It is right to leave it out of \mathbf R (a couple has zero resultant force). It is wrong to leave it out of M_O: it is turning, and turning is what M_O adds up.

Check. If O is moved to the top-left corner (-7, 4), which force drops out of the moment sum? (The 60\,\mathrm N, whose line now passes through the moment centre.)

Narration spine. Each force on the plate slides sideways to O and drops a small curled arrow as it goes. The arrows at O chain into \mathbf R; the curls, with the 140\,\mathrm{N\cdot m} one, merge into a single clockwise curl reading 237.

Bridge. A force plus a couple at O is simpler, but it is still two things. S05 showed that a force and a couple in its plane combine into one force, shifted. Where does this one go?

Nugget 04: The Principle of Moments Places the Single Force

Claim. The single resultant acts on the line, parallel to \mathbf R, whose distance d from O satisfies Rd = M_O, on the side that gives M_O its sense. In vector form, every point \mathbf r on that line satisfies \mathbf r\times\mathbf R = \mathbf M_O.

The picture. On the left of the figure, nothing is drawn yet but O. The resultant must make a clockwise moment of 237\,\mathrm{N\cdot m} about O.

All lines at distance d from O with the direction of \mathbf R are tangents to a circle of radius d about O, and only two are parallel to \mathbf R: at A and at B. A force pointing up and to the right, on the line through A (left of O), turns clockwise about O. On the line through B it would turn counter-clockwise. So the physics of the force–couple system picks A.

Check with numbers. d = |M_O|/R = 237.3/148.3 = 1.600\,\mathrm m. Where does the line cut the x-axis? At such a point C, only R_y has a moment about O (the x-component's line is the axis itself), so R_y\,b = |M_O| and b = 237.3/132.43 = 1.792\,\mathrm m, to the left. The vector form gives the whole line at once: with \mathbf r = x\mathbf i + y\mathbf j,

(x\mathbf i + y\mathbf j)\times(66.86\mathbf i + 132.43\mathbf j) = -237.3\mathbf k \quad\Rightarrow\quad 132.43x - 66.86y = -237.3

Setting y = 0 gives x = -1.792\,\mathrm m, which agrees with b, and setting x = 0 gives y = 3.549\,\mathrm m. As a last test, take moments about another point, the corner (-7, 0)\,\mathrm m. The original system gives +689.7\,\mathrm{N\cdot m}. The single force at C(-1.792, 0) gives (5.208)(132.43) - 0 = +689.7\,\mathrm{N\cdot m}. They agree, as they must.

The idea. Meriam (2/6) summarises the method as \mathbf R = \Sigma\mathbf F, M_O = \Sigma(Fd) and Rd = M_O. The last equation says that the moment of the resultant about any point equals the sum of the moments of the original forces about the same point. That is Varignon's theorem (S04), extended from concurrent forces to any system. Meriam calls it the principle of moments. He applies Rd = M_O to magnitudes and lets the sense of the force–couple system decide the side; the vector form \mathbf r\times\mathbf R = \mathbf M_O carries the sign automatically. Meriam's advice is to master both: the scalar form is more physical, the vector form is foolproof.

The equation (mathematical identity, from S05's reverse construction; Meriam's third Eq. 2/10):

Rd = |M_O|, \qquad\text{or in full}\qquad xR_y - yR_x = M_O

Worked number (Harper's thrust-vectoring study). An experimental aircraft has two engines of thrust T, both 10\,\mathrm m behind O and 3\,\mathrm m either side of the x-axis. The upper thrust points along +x. The lower one can be swung up by an angle \theta. Then R_x = T(1 + \cos\theta) and R_y = T\sin\theta. Moments about O: the upper thrust at (-10, 3) gives -3T; the lower at (-10, -3) gives (-10)(T\sin\theta) - (-3)(T\cos\theta). So

M_O = T(3\cos\theta - 3 - 10\sin\theta), \qquad x = \frac{M_O}{R_y} = \frac{3\cos\theta - 3 - 10\sin\theta}{\sin\theta}

is where the resultant crosses the x-axis (only R_y has a moment there).

\theta	R/T	M_O/T (m)	x (m)
5^\circ	1.998	-0.883	-10.13
10^\circ	1.992	-1.782	-10.26
15^\circ	1.983	-2.690	-10.39
20^\circ	1.970	-3.601	-10.53
25^\circ	1.953	-4.507	-10.67
30^\circ	1.932	-5.402	-10.80

Swinging the thrust barely changes R (it is 2T\cos(\theta/2)). But the line of action stays about 10\,\mathrm m behind O and drifts aft, so a small vertical force far behind O makes a large nose-up or nose-down couple. As \theta \to 0, x \to -10\,\mathrm m, the engines' own station. (Harper's script writes the last arm as 3 in one line, then uses the correct 10 in the bracket.)

Real example. A crane driver reads a load-moment indicator. It computes the resultant of the boom's weight, the hook load and the counterweight, and where its line falls relative to the tipping edge. If that line moves outside the outrigger base, the crane tips.

Plausible wrong answer. "d = 1.600\,\mathrm m, so put \mathbf R through the point 1.600\,\mathrm m to the left of O on the x-axis." d is measured perpendicular to \mathbf R, not along an axis. The axis crossing is b = 1.792\,\mathrm m.

Check. If M_O had been +237\,\mathrm{N\cdot m}, where would the line cut the x-axis? (x = +1.792\,\mathrm m, through the tangent point B.)

Narration spine. A dashed circle of radius 1.600 around O. The green \mathbf R arrow, held at 63.2^\circ, rolls around the circle as a tangent. A moment readout flips from clockwise to counter-clockwise as it passes. It stops at A, where the readout says 237 clockwise, and the line extends to cut the axis at C.

Bridge. The method works for any coplanar system. Some systems are special, though: their forces meet at one point, or run parallel, or cancel completely. And the method's own logic, equal force and equal moment, raises a question: does "equivalent" mean the body cannot tell the difference?

Nugget 05: Equipollent Systems Match in Force and in Moment About One Point

Claim. Two force systems that have the same resultant force and the same moment about one point have the same moment about every point. They are equipollent. For a rigid body that means the same external effect, but not the same internal effect.

The picture. Two rods, each loaded by two equal and opposite forces F on one line. One rod is pulled at both ends, the other pushed.

Both systems have \mathbf R = \mathbf 0, and both have zero moment about every point, because each pair is collinear. As systems of forces they are identical. Yet one rod stretches and the other shortens, and a slender rod pushed hard buckles. Equipollence describes the forces; it does not describe what the forces do inside the body.

Check with numbers. For the plate of Nugget 03, the original system, the force–couple system at O and the single force on the line 132.43x - 66.86y = -237.3 all have \mathbf R = (66.86, 132.43)\,\mathrm N. About O all three give -237.3\,\mathrm{N\cdot m}. About the corner (-7, 0) all three give +689.7\,\mathrm{N\cdot m}. For the force–couple system that last number is M_O + \mathbf r\times\mathbf R with \mathbf r from (-7, 0) to O, that is -237.3 + 7(132.43) = 689.7\,\mathrm{N\cdot m}.

The idea. Marghitu (2.4) defines two systems \{S\} and \{S'\} as equipollent if and only if their resultants are equal and there is at least one point P about which their moments are equal. One point is enough because of the transfer formula (S05): moments about any other point A are \mathbf M_A = \mathbf M_P + \mathbf r_{AP}\times\mathbf R, and every term on the right is the same for both systems. Equipollence is transitive: two systems equipollent to a third are equipollent to each other. That is why Meriam can pass from the plate to the force–couple system to the single force, and why the last of these is "the" resultant.

Marghitu also lists what any system can reduce to: nothing at all (a null system), a single force, a single couple, or a wrench, a force with a couple parallel to it. In a plane the couple is always normal to the plane and the force always lies in it, so the wrench cannot occur, and only the first three remain. The wrench is the subject of S09.

The equation (definition, with the transfer formula as the mathematical identity that justifies one point):

\{S\} \equiv \{S'\} \iff \mathbf R = \mathbf R' \text{ and } \mathbf M_P = \mathbf M'_P \text{ for one point } P, \qquad \mathbf M_A = \mathbf M_P + \mathbf r_{AP}\times\mathbf R

Assumption: the body is rigid. Equipollence says nothing about stress or deformation.

Worked number (Marghitu's Problem 2.6, new numbers). A beam AB of length 1.2\,\mathrm m carries downward forces of 40\,\mathrm N, 80\,\mathrm N and 60\,\mathrm N at 0.2, 0.5 and 0.9\,\mathrm m from A. Equipollent single force: 180\,\mathrm N down at \bar x = (40 \times 0.2 + 80 \times 0.5 + 60 \times 0.9)/180 = 102/180 = 0.567\,\mathrm m. Equipollent pair at A and B: moments about A give 1.2F_B = 102, so F_B = 85\,\mathrm N and F_A = 95\,\mathrm N, both down.

Real example. A shelf carrying a row of books and the same shelf carrying a single box of equal weight at the books' centre of gravity load the wall brackets identically. The shelf board itself bends differently under the two.

Plausible wrong answer. "Equivalent systems are interchangeable for every purpose." Only for external effects on a rigid body: support reactions, acceleration, equilibrium. For stresses and deflections inside the body the actual loads must be used.

Check. System I is 100\,\mathrm N up at x = 0 and 100\,\mathrm N up at x = 2\,\mathrm m. System II is 200\,\mathrm N up at x = 1\,\mathrm m. Are they equipollent? (Yes: equal \mathbf R, and about x = 1\,\mathrm m both have zero moment.)

Narration spine. The plate, the force–couple pair and the single green force sit side by side. A probe point wanders over the screen, and three moment readouts beneath them change together, always equal. Then two rods appear, pulled and pushed. Their readouts are identical zeros, but one rod stretches and the other bows.

Bridge. With equipollence settled, the general method can be applied to the three kinds of system that turn up most often. Each takes a shortcut, and one of them has no single force at all.

Nugget 06: Concurrent, Parallel, or Only a Couple

Claim. If all the lines of action meet at one point, the resultant passes through it. If the forces are parallel, one axis along them is enough. If the forces add to zero but are not concurrent, the resultant is a couple.

The picture. Three small systems.

On the left, every force passes through O, so each has zero moment about O, and so does their sum: M_O = 0 and d = 0. In the middle, all forces are vertical, so R_x = 0 automatically; only \Sigma F_y and the moments are needed. On the right, three forces drawn head to tail around a triangle close on themselves, so \mathbf F_1 + \mathbf F_2 = -\mathbf F_3 and \mathbf R = \mathbf 0. But they do not meet at a point, so their moments do not cancel: what is left is a pure couple.

Check with numbers. The puzzle is a parallel system. Take x forward, y to port and O on the centreline. With M = -yF_x for forces along x: engines at y = 20, 10 and -20\,\mathrm m give M_O = -(20 + 10 - 20)(90) = -900\,\mathrm{kN\cdot m}, clockwise seen from above. With R = 270\,\mathrm{kN}, Rd = 900 gives d = 3.33\,\mathrm m. A forward force turns clockwise when it is on the port side (-yF_x < 0 needs y > 0), so:

The single equivalent thrust is 270\,\mathrm{kN}, 3.33\,\mathrm m to port of the centreline. That answers the puzzle. About the centre of gravity on the centreline it yaws the nose to starboard, towards the dead engine, with 900\,\mathrm{kN\cdot m}. With all four engines running, M_O = 0 and the line is the centreline. For a couple, take a square plate of side 0.4\,\mathrm m with 50\,\mathrm N along each edge, all circulating counter-clockwise: \mathbf R = \mathbf 0, and about the centre each force has arm 0.2\,\mathrm m, so M = 4(50)(0.2) = 40\,\mathrm{N\cdot m}, the same as two couples 50 \times 0.4.

The idea. Meriam (2/6): for a concurrent system the moment sum about the common point is zero, so the line of \mathbf R passes through it. For a parallel system, choose a coordinate axis in the direction of the forces. And if the resultant force is zero, the resultant of the system need not be zero, because it may be a couple, like the three forces of his Fig. 2/15 with a clockwise couple M = F_3d. Marghitu (2.4) lists the same cases: concurrent forces are equipollent to a single force through the common point, and parallel forces whose sum is not zero to a single force parallel to them.

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

\text{concurrent at } O:\ M_O = 0 \;\Rightarrow\; d = 0; \qquad \text{parallel to } x:\ R = \Sigma F_x,\ \ \bar y = -\frac{M_O}{R}; \qquad \mathbf R = \mathbf 0:\ \text{resultant} = M_O

Worked number. For parallel forces the line's position is a weighted average. Equal thrusts at y = 20, 10 and -20\,\mathrm m give \bar y = (20 + 10 - 20)/3 = 3.33\,\mathrm m, as above. Unequal ones are weighted by their size: \bar y = \Sigma(yF)/\Sigma F.

Real example (one you can see). On an engine-out take-off the pilot holds full rudder against the yaw, and the aircraft flies slightly crabbed. The rudder supplies a sideways force far aft whose moment cancels the 900\,\mathrm{kN\cdot m} of the puzzle. Twin-engine aircraft have a published minimum control speed for exactly this reason: below it the rudder's force is too small.

Plausible wrong answer. "If the forces add to zero, the body is in equilibrium." Not if they leave a couple: the square plate with four edge forces has \mathbf R = \mathbf 0 and spins under 40\,\mathrm{N\cdot m}. Equilibrium needs zero force and zero moment (S11).

Check. Three forces on a body are concurrent at a point Q and add to zero. What is their resultant? (Nothing at all: the null system. Zero force and, about Q, zero moment.)

Narration spine. The airliner from above with three orange thrusts. They merge into one green arrow that starts on the centreline, then slides to port until a moment readout reaches 900 clockwise, stopping 3.33\,\mathrm m off. A fourth thrust reappears and the green arrow slides back onto the centreline. Finally a triangle of forces chases its tail and leaves only a curl.

Synthesis

The airliner's three thrusts are one 270\,\mathrm{kN} force on a line 3.33\,\mathrm m to port, the same as 270\,\mathrm{kN} on the centreline plus a 900\,\mathrm{kN\cdot m} clockwise couple. The two descriptions are equipollent, and choosing between them is a matter of convenience. Every coplanar system reduces in two moves: add the forces, add the moments about one point. The sum of forces fixes the size and direction of the resultant. The sum of moments fixes its line, through Rd = M_O or \mathbf r\times\mathbf R = \mathbf M_O. If the sum of forces is zero, what remains is a couple, and only if both sums vanish is there nothing left. That last case, zero force and zero moment, is equilibrium. Part 3 turns it from something to check into something to solve for.

Central idea: the resultant is the simplest system with the same external effect: one force on a definite line, or a couple, or nothing.
Key relationship: \mathbf R = \Sigma\mathbf F, M_O = \Sigma M_O, and Rd = |M_O| or xR_y - yR_x = M_O: the principle of moments.
Most important assumption: rigid body. Equipollent systems have the same external effect, not the same internal one: the pulled and pushed rods.
Open question: in three dimensions the couple need not be normal to the force. What then is the simplest system? (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). S06_coplanar_resultants.py has one function, reduce, that turns any set of plane forces into \mathbf R and M about a chosen point. It reduces Meriam's plate, places the single resultant on its line and checks it about the corner, runs Harper's thrust-vectoring sweep, and finds the equipollent systems on the beam.

Part 1: Meriam's plate, O at the lower right
  R = [ 66.86 132.43] N, 148.3 N at 63.2 deg, M_O = -237.3 N m
  d = 1.600 m, line 132.43 x - 66.86 y = -237.3
  cuts x at -1.792 m, y at 3.549 m
  about (-7, 0): system 689.7, single force 689.7 N m
  bracket: R = 435.9 N at 36.6 deg
  beam: R = [ 0 -6] kN, M_O = -15.5 kN m
Part 2: Harper's thrust vectoring, engines 10 m aft, 3 m either side
  theta   R/T   M_O/T (m)   x (m)
     5   1.998    -0.883    -10.13
    10   1.992    -1.782    -10.26
    15   1.983    -2.690    -10.39
    20   1.970    -3.601    -10.53
    25   1.953    -4.507    -10.67
    30   1.932    -5.402    -10.80
Part 3: equipollent systems on a 1.2 m beam
  single force 180 N at x = 0.567 m
  pair: F_A = 95 N, F_B = 85 N
  equal thrusts at y = 20, 10, -20 m: y_bar = 3.33 m

MATLAB (Octave-compatible). S06_coplanar_resultants.m stores the forces and their points as rows of two matrices, so every sum is one sum.

R = [66.86 132.43] N, 148.3 N at 63.2 deg, M_O = -237.3 N m
d = 1.600 m; the line cuts x at -1.792 m and y at 3.549 m
about (-7, 0): system 689.7 N m, single force 689.7 N m
 theta     R/T  M_O/T (m)    x (m)
     5   1.998     -0.883   -10.13
    10   1.992     -1.782   -10.26
    15   1.983     -2.690   -10.39
    20   1.970     -3.601   -10.53
    25   1.953     -4.507   -10.67
    30   1.932     -5.402   -10.80
single force 180 N at 0.567 m; pair F_A = 95 N, F_B = 85 N

Traps

Adding magnitudes. 40 + 80 + 50 + 60 = 230\,\mathrm N is not the resultant; components give 148.3\,\mathrm N. (From Nugget 01.)
Trusting \tan^{-1} for the quadrant. R_x = -3, R_y = -4 gives \tan^{-1}(4/3) = 53.1^\circ on a calculator, but \mathbf R points at 233.1^\circ. (From Nugget 01.)
Thinking the polygon gives the position. The polygon moves the arrows; it gives a free vector. The line needs the lines of action or the moments. (From Nugget 02.)
Leaving applied couples out of M_O. The 140\,\mathrm{N\cdot m} couple has no force but it belongs in the moment sum; without it M_O would be -377\,\mathrm{N\cdot m}. (From Nugget 03.)
Measuring d along an axis. d = 1.600\,\mathrm m is perpendicular to \mathbf R; the x-axis crossing is 1.792\,\mathrm m away. (From Nugget 04.)
Putting the force on the wrong side. Rd = |M_O| gives two tangent lines. Only one gives M_O its sense; the vector form picks it automatically. (From Nugget 04.)
Equivalent means interchangeable everywhere. A pulled rod and a pushed rod are equipollent; one is in tension, the other in compression. (From Nugget 05.)
Zero force means equilibrium. Four edge forces circulating round a square add to zero and leave a 40\,\mathrm{N\cdot m} couple. (From Nugget 06.)
Putting a parallel resultant at the middle. The engine-out thrust is 3.33\,\mathrm m to port, not on the centreline; unequal forces shift it further. (From Nugget 06.)

Self-check

An eye bolt carries 6\,\mathrm{kN} along +x, 8\,\mathrm{kN} along -y, and a cable tension T. Find T and its direction if the resultant is 15\,\mathrm{kN} straight down.
A square plate of side 0.6\,\mathrm m has its centre at O. (a) Forces of 200\,\mathrm N act along all four edges, circulating counter-clockwise. (b) Forces of 200\,\mathrm N act along the bottom edge (along +x) and the right edge (along +y) only. In each case find the force–couple system at O and, where possible, the single resultant.
A post carries horizontal forces of 400\,\mathrm N to the right at 0.5\,\mathrm m, 250\,\mathrm N to the right at 1.5\,\mathrm m and 150\,\mathrm N to the left at 2.0\,\mathrm m above its base. At what height does the resultant act?
A lever pivoted at O carries 400\,\mathrm N downward at 0.3\,\mathrm m to the left of O, 250\,\mathrm N horizontally to the right at 0.12\,\mathrm m above O, and a vertical force P at 0.5\,\mathrm m to the right of O. Find P if the resultant passes through the bearing O.
A twin-engine aircraft has engines 6\,\mathrm m either side of its centreline, each rated 110\,\mathrm{kN}. The starboard engine loses power and delivers only 66\,\mathrm{kN}. Find the resultant thrust and its line.
A plate has O at its lower left corner. Forces: 100\,\mathrm N along +x at (0, 3)\,\mathrm m; 150\,\mathrm N along -y at (4, 3)\,\mathrm m; 200\,\mathrm N at 30^\circ above +x at (4, 0)\,\mathrm m; and a couple of 80\,\mathrm{N\cdot m} clockwise. Find the resultant, its distance from O, the equation of its line and both axis intercepts.
A beam AB, 1.2\,\mathrm m long, carries downward forces 40\,\mathrm N, 80\,\mathrm N and 60\,\mathrm N at 0.2, 0.5 and 0.9\,\mathrm m from A. Find the single equipollent force and an equipollent pair of vertical forces at A and B.
Three forces of 60\,\mathrm N act along the sides of an equilateral triangle of side 0.5\,\mathrm m, head to tail around it. Find the resultant.
In Harper's thrust-vectoring aircraft (Nugget 04), at what angle \theta does the resultant cross the x-axis at x = -10.5\,\mathrm m? What is R/T then?
An exhaust pipe along the x-axis has weights 100\,\mathrm N at 0.6\,\mathrm m and 50\,\mathrm N at 1.5\,\mathrm m from O. A hanger at 0.2\,\mathrm m pulls up with 40\,\mathrm N. Find the upward forces in hangers at 1.0\,\mathrm m and 1.8\,\mathrm m that make the force–couple system at O zero.
A system reduces at O to \mathbf R = -300\mathbf i + 400\mathbf j\ \mathrm N and a couple of 600\,\mathrm{N\cdot m} counter-clockwise. Find the line of the single resultant, its intercepts and its distance from O.

Fully worked solutions

1. \mathbf R = (0, -15)\,\mathrm{kN}. x: 6 + T_x = 0, so T_x = -6\,\mathrm{kN}. y: -8 + T_y = -15, so T_y = -7\,\mathrm{kN}. T = \sqrt{6^2 + 7^2} = 9.220\,\mathrm{kN}, pointing down and to the left at \tan^{-1}(7/6) = 49.40^\circ below -x.

2. (a) The four forces cancel: \mathbf R = \mathbf 0. Each has arm 0.3\,\mathrm m about O, so M_O = 4(200)(0.3) = 240\,\mathrm{N\cdot m} counter-clockwise. The resultant is this couple; there is no single force. (b) \mathbf R = (200, 200)\,\mathrm N, 282.8\,\mathrm N at 45^\circ. Bottom edge, y = -0.3: -yF_x = 0.3(200) = 60. Right edge, x = 0.3: xF_y = 0.3(200) = 60. M_O = 120\,\mathrm{N\cdot m} counter-clockwise. d = 120/282.8 = 0.4243\,\mathrm m. The line 200x - 200y = 120, that is x - y = 0.6, passes through the corner (0.3, -0.3), where both lines of action meet, as Nugget 06 says it must.

3. Parallel forces along x: R = 400 + 250 - 150 = 500\,\mathrm N to the right. About the base, M = -\Sigma yF_x = -(0.5 \times 400 + 1.5 \times 250 - 2.0 \times 150) = -275\,\mathrm{N\cdot m}. The resultant at height h gives -500h, so h = 275/500 = 0.55\,\mathrm m.

4. Through O means M_O = 0. Taking P downward: (-0.3)(-400) + (0.5)(-P) - (0.12)(250) = 120 - 0.5P - 30 = 0, so P = 180\,\mathrm N downward.

5. R = 110 + 66 = 176\,\mathrm{kN} forward. With y to port, M_O = -(6)(110) - (-6)(66) = -660 + 396 = -264\,\mathrm{kN\cdot m}, clockwise from above (towards the weak engine). The line is \bar y = 264/176 = 1.5\,\mathrm m to port.

6. R_x = 100 + 200\cos30^\circ = 273.2\,\mathrm N, R_y = -150 + 200\sin30^\circ = -50\,\mathrm N. R = 277.7\,\mathrm N at -10.37^\circ (below +x). Moments: 100\,\mathrm N at (0, 3): -3(100) = -300; 150\,\mathrm N at (4, 3): 4(-150) = -600; 200\,\mathrm N at (4, 0): 4(100) = 400; couple -80. M_O = -580\,\mathrm{N\cdot m}. d = 580/277.7 = 2.088\,\mathrm m. Line: xR_y - yR_x = M_O gives -50x - 273.2y = -580, that is 50x + 273.2y = 580. Intercepts: x = 11.6\,\mathrm m, y = 2.123\,\mathrm m.

7. R = 180\,\mathrm N down. Moment about A: 40(0.2) + 80(0.5) + 60(0.9) = 102\,\mathrm{N\cdot m} (clockwise), so the single force is at \bar x = 102/180 = 0.567\,\mathrm m from A. For the pair: moments about A, 1.2F_B = 102, F_B = 85\,\mathrm N; forces, F_A = 180 - 85 = 95\,\mathrm N, both down.

8. Head to tail around a closed triangle: \mathbf R = \mathbf 0. Each force is at the inradius r = 0.5/(2\sqrt3) = 0.1443\,\mathrm m from the centre, all turning the same way, so M = 3(60)(0.1443) = 25.98\,\mathrm{N\cdot m}. The resultant is a couple of 25.98\,\mathrm{N\cdot m} in the sense of circulation.

9. Set (3\cos\theta - 3 - 10\sin\theta)/\sin\theta = -10.5, so 3\cos\theta - 3 + 0.5\sin\theta = 0. Solving numerically, \theta = 18.92^\circ. Check: 3(0.9460) - 3 + 0.5(0.3242) = 0.0000. Then R/T = 2\cos(9.46^\circ) = 1.973.

10. Zero force: 40 + F_B + F_C - 100 - 50 = 0, so F_B + F_C = 110\,\mathrm N. Zero moment about O (M = xF_y): 0.2(40) + 1.0F_B + 1.8F_C - 0.6(100) - 1.5(50) = 0, so F_B + 1.8F_C = 127. Subtracting, 0.8F_C = 17: F_C = 21.25\,\mathrm N and F_B = 88.75\,\mathrm N, both up.

11. xR_y - yR_x = M_O: 400x + 300y = 600. Intercepts: x = 1.5\,\mathrm m, y = 2\,\mathrm m. R = 500\,\mathrm N, so d = 600/500 = 1.2\,\mathrm m. Check: the distance from O to the line 400x + 300y = 600 is 600/\sqrt{400^2 + 300^2} = 1.2\,\mathrm m.