Study Companion: S05 — Couples and Force–Couple Systems

Two equal and opposite forces that do not share a line add to zero force, yet they still turn a body. That pair is a couple, and its turning effect is the same about every point. It is pure turning, with no push at all. The couple is also the missing piece from S03 and S04: a force can be moved sideways to any point, provided a couple is added to make up the moment. That replacement, the force–couple system, is how every later lesson simplifies a set of forces before solving anything.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/5 ("Couple"), with Sample Problems 2/7 and 2/8. Marghitu, Dupac and Madsen, Statics with MATLAB, section 2.2 ("Couples"), the transfer formula of 2.1, and Example 2.4. Harper has no couple problem. Notation follows Meriam: M for a couple's moment (Marghitu also calls it the torque, as Meriam does for any 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 twin-screw ferry manoeuvres in harbour. Its starboard propeller pushes 200\,\mathrm{kN} ahead, and its port propeller, 10\,\mathrm m away, pushes 200\,\mathrm{kN} astern.

The two pushes add to zero, so the ship goes nowhere. Yet it spins. How much turning is there? And about which point? The centre of the ship, the stern, some point on the quay? The surprising answer is that the turning is 2000\,\mathrm{kN\cdot m} about every point. It comes in Nugget 01.

Nugget 01: Equal, Opposite, Parallel: A Couple Turns but Does Not Push

Claim. Two equal, opposite forces on parallel lines a distance d apart have zero resultant. Their combined moment is Fd, the same about every point in their plane.

The picture. Take the propeller thrusts: F = 200\,\mathrm{kN} and -F, on lines d = 10\,\mathrm m apart. Put a point O anywhere, say a = 25\,\mathrm m from the line of -F on the far side from F.

The far force has arm a + d and turns counter-clockwise about O. The near force has arm a and turns clockwise. Their difference is F(a + d) - Fa = Fd, and a has cancelled. Move O anywhere and the same cancellation happens.

Check with numbers. About this O: 200 \times 35 - 200 \times 25 = 7000 - 5000 = 2000\,\mathrm{kN\cdot m}. Now place the ship's axis along x with the bow at +x. The ahead thrust +200\mathbf i\ \mathrm{kN} acts at (0, -5)\,\mathrm m and the astern thrust -200\mathbf i\ \mathrm{kN} at (0, 5)\,\mathrm m. Using M = xF_y - yF_x:

about the centre (0, 0): 1000 + 1000 = 2000\,\mathrm{kN\cdot m};
about the port screw (0, 5): 200 \times 10 + 0 = 2000\,\mathrm{kN\cdot m};
about the bow (120, 0): 2000\,\mathrm{kN\cdot m} (the x offset does not matter for forces along x);
about Q, 30\,\mathrm m to port at (0, 30): 7000 - 5000 = 2000\,\mathrm{kN\cdot m}.

Every answer is 200 \times 10 = 2000\,\mathrm{kN\cdot m}, counter-clockwise: the bow swings to port. That answers the puzzle.

The idea. Meriam (2/5) calls the moment of two equal, opposite and non-collinear forces a couple. The two forces cannot be combined into a single force, because their sum is zero in every direction. Their only effect is a tendency to rotate. The moment centre's distance a drops out, so a couple has the same moment about every point. Marghitu (2.2) defines a couple by exactly these two properties: zero resultant and non-zero moment. A pair of forces that has them is a simple couple.

The equation (definition of the couple's moment; its independence of the point is a mathematical identity, shown above):

M = F(a + d) - Fa = Fd

Assumption: the two forces are equal in size, opposite in sense and on parallel, distinct lines. If the lines coincide, d = 0 and there is no couple.

Worked number. A driver turns a steering wheel of 0.38\,\mathrm m diameter with 30\,\mathrm N from each hand, one pushing up and the other pulling down, on opposite sides. M = 30 \times 0.38 = 11.4\,\mathrm{N\cdot m}. The column feels no sideways push at all.

Real example (one you can see). Watch a ferry or tug turn in its own length in a narrow harbour. It runs one screw ahead and one astern, and spins almost on the spot. A single rudder needs forward speed to turn the ship. A couple does not.

Plausible wrong answer. "The ship turns about its centre, so the moment is about the centre." The moment of a couple belongs to no particular point: it is 2000\,\mathrm{kN\cdot m} about Q, about the bow and about the quay. Where the ship actually rotates is decided by its mass distribution and the water. That is dynamics, not the couple.

Check. Two 50\,\mathrm N forces, equal and opposite on the same line. What is their moment about a point 1\,\mathrm m away? (Zero: d = 0, so they form no couple. The arms are equal and the moments cancel.)

Narration spine. Two opposite arrows on parallel dashed lines. A point O appears with its two arms drawn. The two moment arcs, one large counter-clockwise and one smaller clockwise, subtract to a readout of 2000. Drag O around the screen, even across between the lines. The arms grow and shrink, but the readout never moves.

Bridge. If the couple's moment is the same everywhere, then the couple itself has no location. We need a way to write it that has no point in it.

Nugget 02: A Couple Is a Free Vector

Claim. A couple's moment is \mathbf M = \mathbf r\times\mathbf F, where \mathbf r runs from any point on -\mathbf F's line to any point on \mathbf F's. No moment centre appears, so \mathbf M is a free vector.

The picture. A couple of 150\,\mathrm N forces: \mathbf F upward at A(0.4, 0.1)\,\mathrm m and -\mathbf F downward at B(-0.2, 0.3)\,\mathrm m.

About O, the two moments are \mathbf r_A\times\mathbf F and \mathbf r_B\times(-\mathbf F). They share \mathbf F, so they combine into (\mathbf r_A - \mathbf r_B)\times\mathbf F. The difference \mathbf r_A - \mathbf r_B is the green arrow from B to A. It does not depend on where O is.

Check with numbers. \mathbf r = (0.4 - (-0.2))\mathbf i + (0.1 - 0.3)\mathbf j = 0.6\mathbf i - 0.2\mathbf j\ \mathrm m and \mathbf F = 150\mathbf j\ \mathrm N, so M = 0.6 \times 150 - (-0.2)(0) = 90\,\mathrm{N\cdot m}, counter-clockwise. Term by term about O: 0.4 \times 150 = 60 and (-0.2)(-150) = 30, total 90. About P(2, -1)\,\mathrm m: (-1.6)(150) = -240 and (-2.2)(-150) = 330, total 90 again. The lines are vertical and 0.6\,\mathrm m apart, and Fd = 150 \times 0.6 = 90\,\mathrm{N\cdot m}.

The idea. Meriam (2/5) writes \mathbf M = \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. Because no moment centre appears, a couple is a free vector. It can be drawn anywhere, normal to the plane of the forces, with its sense by the right-hand rule. In plane problems Meriam draws it as a curved arrow (or a zigzag of two half-arrows), clockwise or counter-clockwise. Marghitu (2.2) proves the same thing about any point A: \mathbf M = \mathbf r\times\mathbf v + \mathbf 0. He adds that the moment of a couple about a line is the component of \mathbf M along that line. For a plane couple that is M about any axis normal to the plane, and zero about any axis lying in the plane.

The equation (definition, equivalent to M = Fd by the cross product's geometry):

\mathbf M = \mathbf r\times\mathbf F, \qquad \mathbf r = \mathbf r_A - \mathbf r_B \text{ from the line of } -\mathbf F \text{ to the line of } \mathbf F

Worked number. A tap wrench is turned by 40\,\mathrm N pushed forward on one handle and 40\,\mathrm N pulled back on the other, the hands 0.30\,\mathrm m apart. M = 40 \times 0.30 = 12\,\mathrm{N\cdot m}, the same whether the wrench is held over the hole or the hole is off to one side.

Real example. An electric motor's shaft delivers a couple, its torque. The motor's mounting bolts feel that couple wherever the motor sits on its frame.

Plausible wrong answer. "A couple acts at the point midway between the forces." It acts nowhere in particular. Only its moment, sense and plane matter.

Check. Swap the labels so that \mathbf F is the downward force at B. What is \mathbf r, and does \mathbf M change? (\mathbf r now runs from A to B, -0.6\mathbf i + 0.2\mathbf j, and is crossed with -150\mathbf j, giving +90\mathbf k again.)

Narration spine. The two forces at A and B, and position arrows from O. Fade the moment centre away: the two position arrows merge into the single green arrow from B to A. Drag the whole couple across the plane, and a curled-arrow symbol labelled 90\,\mathrm{N\cdot m} rides along unchanged.

Bridge. If only the moment matters, then many different force pairs make the same couple. Which of them are interchangeable?

Nugget 03: Only the Product Fd Matters

Claim. Two couples are equivalent if they have the same moment Fd and the same sense in the same or a parallel plane. The individual F, d, position and orientation are free.

The picture. On the left, two 120\,\mathrm N forces 80\,\mathrm{mm} apart. On the right, two 300\,\mathrm N forces at the two sides of a 50\,\mathrm{mm} post, each tilted by an angle \theta from the vertical.

The larger forces are tilted until their lines come close enough together. The points they act at are still 50\,\mathrm{mm} apart, but the perpendicular distance between the lines is 50\cos\theta. When 300 \times 0.050\cos\theta equals 9.6\,\mathrm{N\cdot m}, the two couples are interchangeable. Both are clockwise, so the senses match.

Check with numbers. Left: M = 120 \times 0.080 = 9.6\,\mathrm{N\cdot m}. Right: 300 \times 0.050\cos\theta = 15\cos\theta = 9.6, so \cos\theta = 0.64 and \theta = 50.21^\circ. The lines are then 0.050 \times 0.64 = 0.032\,\mathrm m apart, and 300 \times 0.032 = 9.6\,\mathrm{N\cdot m}. Meriam's Sample Problem 2/7 does the same: a 10\,\mathrm{N\cdot m} couple made from 400\,\mathrm N forces at points 40\,\mathrm{mm} apart needs \cos\theta = 10/16, so \theta = 51.3^\circ.

The idea. Meriam (2/5): changing F and d does not change a couple as long as Fd stays the same. Neither does moving its forces to a different but parallel plane. His Fig. 2/11 shows four blocks carrying the same couple: forces on the top face or on a side face, at distance d or 2F at d/2. All are described by one free vector. As Meriam's hint puts it, the only dimensions that count are those giving the perpendicular distances between the forces.

The equation (mathematical identity: equal free vectors describe the same couple):

F_1d_1 = F_2d_2 \text{ (same sense, parallel planes)} \;\Rightarrow\; \text{equivalent couples}

Worked number. A cross-shaped lug wrench takes 200\,\mathrm N from each hand, the hands 0.70\,\mathrm m apart: M = 200 \times 0.70 = 140\,\mathrm{N\cdot m}. The square nut, 25\,\mathrm{mm} across the flats, receives that couple as four equal forces F, one on each flat, in two opposite pairs 25\,\mathrm{mm} apart. That makes two couples of 0.025F each. Equivalence gives 2 \times 0.025F = 140, so F = 2800\,\mathrm N. The modest hand forces become large contact forces, which is why a worn nut rounds off. (Meriam's Problem 2/60 is the same, in US units.)

Real example. Tugs that hold the ferry of the puzzle against its own propellers push sideways, one at the bow and one at the stern, 80\,\mathrm m apart and in opposite directions. They need a 2000\,\mathrm{kN\cdot m} couple of the opposite sense, so each pushes with 2000/80 = 25\,\mathrm{kN}. That is far less than the 200\,\mathrm{kN} thrusts, because their arm is eight times longer.

Plausible wrong answer. "To balance 200\,\mathrm{kN} thrusts the tugs must push 200\,\mathrm{kN}." Only the couple matters: 25\,\mathrm{kN} at 80\,\mathrm m equals 200\,\mathrm{kN} at 10\,\mathrm m.

Check. Is a couple of two 50\,\mathrm N forces 0.2\,\mathrm m apart equivalent to two 20\,\mathrm N forces 0.5\,\mathrm m apart in the same sense? (Yes, both are 10\,\mathrm{N\cdot m}.)

Narration spine. The 120\,\mathrm N couple sits on the left. On the right, two 300\,\mathrm N arrows start square to the post, and a moment readout shows 15\,\mathrm{N\cdot m}, too much. They tilt together. Their lines close in, and the readout falls until it matches 9.6 at 50.2^\circ. Finally the whole right-hand couple slides up the post and onto a parallel face, and the readout stays.

Bridge. Couples can be built and traded. The most useful trade of all puts one in deliberately: moving a force sideways to a convenient point leaves its moment behind, and a couple carries it.

Nugget 04: Moving a Force Sideways Costs a Couple

Claim. A force \mathbf F at A is equivalent to the same force at any other point B plus a couple M = Fd, where d is the perpendicular distance between the two lines. The sense of the couple is that of the original force's moment about B.

The picture. A lever 0.25\,\mathrm m long stands at 60^\circ on a pivot O. A 400\,\mathrm N horizontal force pulls its top A to the left.

In the middle panel, two forces are added at O: 400\,\mathrm N to the right (green) and 400\,\mathrm N to the left. They cancel, so nothing has changed. But now regroup. The original force at A and the green force at O are equal, opposite and on parallel lines: a couple. What is left over is a 400\,\mathrm N force to the left at O. So the right-hand panel, a force at O plus a couple, is the same loading as the left-hand one.

Check with numbers. A is 0.25\sin60^\circ = 0.2165\,\mathrm m above O, and that is the distance between the two horizontal lines. The couple is M = 400 \times 0.2165 = 86.60\,\mathrm{N\cdot m}. A leftward pull above O turns the lever counter-clockwise, so M = +86.60\,\mathrm{N\cdot m}. By the cross product, with \mathbf r_{OA} = (0.125, 0.2165)\,\mathrm m and \mathbf F = (-400, 0)\,\mathrm N: M_O = 0.125 \times 0 - 0.2165 \times (-400) = 86.60\,\mathrm{N\cdot m}. Meriam's Sample Problem 2/8 is the same lever in US units. An 80\,\mathrm{lb} (355.9\,\mathrm N) force at the end of a 9\,\mathrm{in.} (0.2286\,\mathrm m) lever at 60^\circ gives M = 80(9\sin60^\circ) = 624\,\mathrm{lb\text{-}in.} (70.45\,\mathrm{N\cdot m}), counter-clockwise, with the 80\,\mathrm{lb} force moved to O.

The idea. Meriam (2/5): a force has two effects, a push or pull along its line and a tendency to turn the body about any axis that misses that line. The force–couple system shows them separately. The force, moved to the point of interest, carries the push. The couple carries the turning that the move would otherwise lose. Adding \mathbf F and -\mathbf F at B changes no external effect, so the replacement is exact for a rigid body.

Marghitu (2.1) states the general rule for any system. Moments about two points A and P differ by the moment of the resultant: \mathbf M_A = \mathbf M_P + \mathbf r_{AP}\times\mathbf R. For one force moved from its own line (where its moment is zero) to a new point B, that gives the couple \mathbf M_B = \mathbf r_{BA}\times\mathbf F, the moment of the original force about B.

The equation (mathematical identity: adding \mathbf F - \mathbf F = \mathbf 0 and regrouping):

\mathbf F \text{ at } A \;\equiv\; \mathbf F \text{ at } B \;+\; \mathbf M = \mathbf r_{BA}\times\mathbf F, \qquad M = Fd

Assumption: rigid body. The replacement preserves external effects only, exactly as transmissibility did (S03).

Worked number (second view, Marghitu's Example 2.4). A 100\,\mathrm{lb} (444.8\,\mathrm N) force acts at the end A of a link OA = 1\,\mathrm{ft} (0.3048\,\mathrm m) inclined at \theta = 45^\circ. The force makes \alpha = 100^\circ with the link. At O it becomes \mathbf R = -F\cos(\alpha - \theta)\,\mathbf i + F\sin(\alpha - \theta)\,\mathbf j = -255.1\mathbf i + 364.4\mathbf j\ \mathrm N, plus the couple M = Fl\sin\alpha = 444.8 \times 0.3048 \times \sin100^\circ = 133.5\,\mathrm{N\cdot m}, counter-clockwise. The book's values are (-57.358, 81.915)\,\mathrm{lb} and 98.481\,\mathrm{lb\cdot ft}.

F = 444.8;  l = 0.3048;  theta = 45;  alpha = 100;   % N, m, deg
FA = [-F*cosd(alpha-theta)  F*sind(alpha-theta)  0];
rA = [l*cosd(theta)  l*sind(theta)  0];
R  = FA                       % [-255.1 364.4 0] N, now at O
M  = cross(rA, FA)            % [0 0 133.5] N m, the couple

Real example. A shelf bracket on a column carries 5\,\mathrm{kN} at 1.2\,\mathrm m from the column's centre line. For the column, that load is 5\,\mathrm{kN} down its centre plus a 6\,\mathrm{kN\cdot m} couple. The engineer designs the column for both: compression from the force, bending from the couple.

Plausible wrong answer. "Slide the force to O, as transmissibility allows." Transmissibility allows sliding along the line of action. Moving to O is a sideways move of 0.2165\,\mathrm m, and it drops 86.6\,\mathrm{N\cdot m} unless the couple is added.

Check. If the same 400\,\mathrm N pointed to the right at A, what would the force–couple system at O be? (400\,\mathrm N to the right at O and 86.6\,\mathrm{N\cdot m} clockwise.)

Narration spine. The lever with its force at A. Two ghost arrows pop into existence at O, equal and opposite, with a "+0" tag. The force at A and the green ghost light up together and fold into a curved couple arrow. The remaining arrow at O is the moved force. Run it backwards to show the trade is exact both ways.

Bridge. The trade also runs in reverse. A force and a couple acting together can be merged back into one force. The question is where that force goes.

Nugget 05: A Force and a Couple in Its Plane Are One Force, Shifted

Claim. A force \mathbf F at O together with a couple M in the same plane is equivalent to the force alone, moved sideways by d = M/F. It moves to the side where its moment about O has the couple's sense.

The picture. A 5\,\mathrm{kN} force along +x through O, and a 750\,\mathrm{N\cdot m} counter-clockwise couple.

Replace the couple by two 5\,\mathrm{kN} forces, 0.15\,\mathrm m apart, with one of them placed on the given force's line and pointing the opposite way. That one cancels the given force. What remains is a single 5\,\mathrm{kN} force on a parallel line 0.15\,\mathrm m away. To make the counter-clockwise turning, a force pointing along +x must pass below O.

Check with numbers. d = M/F = 750/5000 = 0.15\,\mathrm m. The force 5000\mathbf i\ \mathrm N on the line y = -0.15\,\mathrm m has M_O = xF_y - yF_x = 0 - (-0.15)(5000) = +750\,\mathrm{N\cdot m}. That matches the couple in size and sense, and the force is unchanged, so the two systems are equivalent.

The idea. Meriam (2/5): by reversing the force–couple construction, a couple and a force lying in the couple's plane (that is, normal to the couple vector) combine into a single equivalent force. In his Sample Problem 2/8 hint, the 624\,\mathrm{lb\text{-}in.} couple and the 80\,\mathrm{lb} force at O recombine into the original force on a line 624/80 = 7.79\,\mathrm{in.} (0.198\,\mathrm m) from O. That is 9\sin60^\circ, the original line. In a plane this always works, because the couple vector is always normal to the plane's forces. In three dimensions it fails when the couple has a part along the force. That leftover is the wrench of S09.

The equation (mathematical identity, the force–couple identity read backwards):

\mathbf F \text{ at } O + M \;\equiv\; \mathbf F \text{ on a parallel line at } d = \frac{M}{F}, \text{ on the side that gives } M \text{ its sense}

Worked number. Meriam's Problem 2/54: a small shaft at O carries a 6\,\mathrm{kN} force along +y and a 400\,\mathrm{N\cdot m} counter-clockwise couple. The single force is 6\,\mathrm{kN} along +y through x = 400/6000 = 0.0667\,\mathrm m (66.7\,\mathrm{mm}) on the positive side, since a force along +y to the right of O turns counter-clockwise.

Real example. A crane's slewing ring sees a vertical load and a couple from the boom. The engineer replaces them by one vertical force offset by M/F. If that offset stays inside the ring's bolt circle, the crane is in no danger of tipping. The same "where does the resultant fall" check underlies every retaining wall and footing.

Plausible wrong answer. "Shift the force by M/F in either direction." The side matters. Shifting the +x force above O gives -750\,\mathrm{N\cdot m}, the wrong sense, and the loading becomes the mirror image.

Check. A 2\,\mathrm{kN} force along -y through O and a 300\,\mathrm{N\cdot m} clockwise couple. Where is the single equivalent force? (2\,\mathrm{kN} along -y on the line x = 0.15\,\mathrm m: M_O = x(-2000) = -300\,\mathrm{N\cdot m}.)

Narration spine. A force and a curled couple at O. The couple unfolds into two parallel arrows, one landing on the force and annihilating it with a flash. The survivor is the same arrow, slid 0.15\,\mathrm m below O. Flip the couple's sense and the arrow hops to the other side.

Synthesis

The ferry puzzle is a couple: zero force and 2000\,\mathrm{kN\cdot m} about every point. Holding it still takes another couple, not another force, so two 25\,\mathrm{kN} tugs 80\,\mathrm m apart are enough. The deeper result is the two-way trade of Nuggets 04 and 05. Any force may be moved to any point, if the couple equal to its moment about that point goes with it. And any force plus a couple in its plane is one force on a shifted line. That makes every force in a plane interchangeable with a standard form: the force at a chosen point, plus one couple. Adding several such forms is just adding forces and adding couples, which is how S06 reduces a whole system to its resultant.

Central idea: a couple is pure turning, a free vector of magnitude Fd with the same moment about every point. It lets a force be moved off its line without changing its external effect.
Key relationship: M = Fd and \mathbf M = \mathbf r\times\mathbf F for a couple. \mathbf F at A \equiv \mathbf F at B plus \mathbf r_{BA}\times\mathbf F, and conversely d = M/F.
Most important assumption: rigid body. The force–couple replacement, like transmissibility, preserves external effects only. The column of Nugget 04 feels compression and bending.
Open question: several forces and couples all reduce to one force and one couple at a point. When does that become a single force, where does it act, and when is it only a couple? (S06.)

Computed views

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

Python (numpy). S05_couples.py takes the steering-wheel couple's moment about three different points, moves Marghitu's Example 2.4 force to O with its couple, and goes back from a force and a couple to one shifted force.

Part 1: a couple is a free vector
  steering wheel about (0, 0): M = 11.40 N m
  steering wheel about (0.5, 0.2): M = 11.40 N m
  steering wheel about (-3, 7): M = 11.40 N m
  lug wrench: F = 140 / (2 x 0.025) = 2800 N
  tugs 80 m apart for 2000 kN m: 25 kN each
Part 2: Marghitu Example 2.4, force at A moved to O
  R at O = [-255.1  364.4] N, couple = 133.5 N m
  F l sin(alpha) = 133.5 N m
Part 3: a force and a couple become one force
  P 2/54: 6 kN up through x = M / F = 66.7 mm
  bracket: 5 kN at 1.2 m is 5 kN at the axis plus 6.0 kN m

MATLAB (Octave-compatible). S05_couples.m does the same with cross.

steering wheel about (0.0, 0.0): M = 11.40 N m
steering wheel about (0.5, 0.2): M = 11.40 N m
steering wheel about (-3.0, 7.0): M = 11.40 N m
Ex 2.4: R at O = [-255.1 364.4] N, couple = 133.5 N m
P 2/54: x = M/F = 66.7 mm

Traps

A couple about "its" point. The moment of a couple is the same about every point: 2000\,\mathrm{kN\cdot m} about the ship's centre, its bow and the quay. (From Nugget 01.)
Collinear opposite forces as a couple. With d = 0 the moments cancel and there is no couple. (From Nugget 01.)
Giving a couple a location. A couple is a free vector. Drawing it "at the midpoint" adds nothing and suggests it would change if moved. (From Nugget 02.)
Equating forces instead of moments. To cancel 200\,\mathrm{kN} thrusts 10\,\mathrm m apart, tugs 80\,\mathrm m apart need 25\,\mathrm{kN}, not 200\,\mathrm{kN}. (From Nugget 03.)
Using the distance between the points, not the lines. Tilted 300\,\mathrm N forces at points 50\,\mathrm{mm} apart have lines only 32\,\mathrm{mm} apart. (From Nugget 03.)
Moving a force sideways without its couple. Moving the lever's 400\,\mathrm N to O drops 86.6\,\mathrm{N\cdot m} unless the couple goes with it. (From Nugget 04.)
The wrong sense for the transfer couple. The couple is the original force's moment about the new point, \mathbf r_{BA}\times\mathbf F, with its sign. A leftward force above O gives a counter-clockwise couple. (From Nugget 04.)
Adding a couple to a force. "R = 5\,\mathrm{kN} + 750\,\mathrm{N\cdot m}" mixes units. Forces add to forces and couples to couples. (From Nugget 05.)
Shifting to the wrong side. The single force goes on the side where its moment about O has the couple's sense. A +x force with a counter-clockwise couple goes below O. (From Nugget 05.)

Self-check

A 120\,\mathrm N force acts upward through (-0.25, 0)\,\mathrm m, and a 120\,\mathrm N force acts downward through (0.10, 0)\,\mathrm m. Find their combined moment about O and about A(-0.15, 0)\,\mathrm m.
A caster carries two 350\,\mathrm N vertical forces, one up and one down, whose lines are 36\,\mathrm{mm} apart. What couple acts on the caster?
A 300\,\mathrm N force acts at A(-3, 4)\,\mathrm m, pointing downward and 20^\circ to the right of the vertical. An equal and opposite force acts at B(6, -2)\,\mathrm m. Find their combined moment about O, about C(-5, -5) and about D(8, 0), and the distance between their lines.
A 10\,\mathrm{kN} force at 30^\circ above +x acts at A(-3, 0)\,\mathrm m. Replace it by a force–couple system (a) at O and (b) at B(0, -4)\,\mathrm m.
A revolving door's two wings are pushed by two people with equal forces F, each 0.9\,\mathrm m from the axis on opposite wings and 20^\circ off the perpendicular to the wing, so that the pair forms a couple. If the couple is 30\,\mathrm{N\cdot m}, find F.
During an engine test an aircraft's two engines, 4.4\,\mathrm m apart, give 2.2\,\mathrm{kN} of thrust forward and backward respectively. The braked main wheels are 2.6\,\mathrm m apart. What equal and opposite force must the ground exert on each wheel to stop the aircraft turning?
A cantilever beam is fixed at O. A 6\,\mathrm{kN} force acts at the end, 2.5\,\mathrm m along the beam and 0.4\,\mathrm m below its centre line, pointing downward and 30^\circ to the right of the vertical. Find the equivalent force–couple system at O.
A lug wrench takes 180\,\mathrm N from each hand, the hands 0.60\,\mathrm m apart. The square nut is 22\,\mathrm{mm} across its flats. Find the four equal contact forces on the flats.
A cable from A(0.30, 0.20)\,\mathrm m towards C(0.90, -0.25)\,\mathrm m pulls on A with 800\,\mathrm N. Find the equivalent force–couple systems at O and at B(0.5, 0)\,\mathrm m in vector form, and check one from the other.
A tape passes over two pulleys. The upper strand pulls with 150\,\mathrm N to the left on the line y = 0.12\,\mathrm m, and the lower strand pulls with 150\,\mathrm N to the right on y = 0.04\,\mathrm m. Find the equivalent force–couple system at O and at (1, -2)\,\mathrm m.
A trolley at O carries a 9\,\mathrm{kN} downward force and a 0.5\,\mathrm{kN\cdot m} counter-clockwise couple. Replace them by two vertical forces at the wheel connections A and B, 0.25\,\mathrm m to the left and right of O.
A 500\,\mathrm N force acts at the end A of a link OA = 0.4\,\mathrm m inclined at 30^\circ, making 110^\circ with the link (Marghitu's angle convention: the force's direction is \alpha - \theta = 80^\circ measured clockwise from the negative x-axis). Find the force–couple system at O, and where a single force could act instead.

Fully worked solutions

1. About O, using M = xF_y for vertical forces: (-0.25)(120) + (0.10)(-120) = -30 - 12 = -42\,\mathrm{N\cdot m}, that is 42\,\mathrm{N\cdot m} clockwise. About A: (-0.10)(120) + (0.25)(-120) = -12 - 30 = -42\,\mathrm{N\cdot m}. The same, as it must be for a couple: Fd = 120 \times 0.35 = 42\,\mathrm{N\cdot m}.

2. M = Fd = 350 \times 0.036 = 12.6\,\mathrm{N\cdot m}. The sense is set by which force is on which side; the caster feels no net force.

3. \mathbf F = 300(\sin20^\circ\,\mathbf i - \cos20^\circ\,\mathbf j) = 102.6\mathbf i - 281.9\mathbf j\ \mathrm N at A, and -\mathbf F at B. \mathbf r = \mathbf r_A - \mathbf r_B = -9\mathbf i + 6\mathbf j\ \mathrm m. M = (-9)(-281.9) - 6(102.6) = 2537.2 - 615.6 = 1921.5\,\mathrm{N\cdot m}, counter-clockwise, about O, C and D alike (a couple). The distance between the lines is d = M/F = 1921.5/300 = 6.405\,\mathrm m.

4. \mathbf F = 10(\cos30^\circ\,\mathbf i + \sin30^\circ\,\mathbf j) = 8.660\mathbf i + 5\mathbf j\ \mathrm{kN}. (a) At O: \mathbf F plus M_O = (-3)(5) - 0 = -15\,\mathrm{kN\cdot m}, which is 15\,\mathrm{kN\cdot m} clockwise. (b) At B: \mathbf r_{BA} = (-3, 4)\,\mathrm m, so M_B = (-3)(5) - 4(8.660) = -15 - 34.64 = -49.64\,\mathrm{kN\cdot m}, clockwise, plus \mathbf F at B. Check by transfer: M_B = M_O + \mathbf r_{BO}\times\mathbf F, with \mathbf r_{BO} = (0, 4), gives -15 - 4(8.660) = -49.64.

5. Each force's arm is 0.9\cos20^\circ, so the couple is 2F \times 0.9\cos20^\circ = 1.691F = 30, and F = 30/1.691 = 17.74\,\mathrm N.

6. The engines make a couple of 2.2 \times 4.4 = 9.68\,\mathrm{kN\cdot m}. The wheel forces must form an equal and opposite couple: F \times 2.6 = 9.68, so F = 3.723\,\mathrm{kN} on each wheel, in opposite directions.

7. \mathbf F = 6(\sin30^\circ\,\mathbf i - \cos30^\circ\,\mathbf j) = 3\mathbf i - 5.196\mathbf j\ \mathrm{kN} at (2.5, -0.4)\,\mathrm m. At O: the same \mathbf F plus M_O = 2.5(-5.196) - (-0.4)(3) = -12.99 + 1.2 = -11.79\,\mathrm{kN\cdot m}, which is 11.79\,\mathrm{kN\cdot m} clockwise. The offset below the centre line reduces the couple, because the force's horizontal part pulls below O.

8. The couple is 180 \times 0.60 = 108\,\mathrm{N\cdot m}. Four forces F in two pairs 0.022\,\mathrm m apart give 2 \times 0.022F = 108, so F = 108/0.044 = 2455\,\mathrm N.

9. \overrightarrow{AC} = (0.60, -0.45)\,\mathrm m, of length 0.75\,\mathrm m, so \mathbf T = 800(0.8\mathbf i - 0.6\mathbf j) = 640\mathbf i - 480\mathbf j\ \mathrm N. At O: \mathbf T plus \mathbf M_O = [0.30(-480) - 0.20(640)]\mathbf k = (-144 - 128)\mathbf k = -272\mathbf k\ \mathrm{N\cdot m}. At B: \mathbf r_{BA} = (-0.20, 0.20)\,\mathrm m, so \mathbf M_B = [(-0.20)(-480) - 0.20(640)]\mathbf k = (96 - 128)\mathbf k = -32\mathbf k\ \mathrm{N\cdot m}. Check: \mathbf M_B = \mathbf M_O + \mathbf r_{BO}\times\mathbf T with \mathbf r_{BO} = (-0.5, 0): -272 + (-0.5)(-480) = -272 + 240 = -32.

10. The two strands are equal, opposite and parallel, so they form a couple and the resultant force is zero. M = -yF_x summed: -(0.12)(-150) - (0.04)(150) = 18 - 6 = 12\,\mathrm{N\cdot m}, counter-clockwise. The system at O is no force and a 12\,\mathrm{N\cdot m} couple, and at (1, -2) it is exactly the same: a couple needs no transfer term, because its resultant is zero.

11. Downward forces F_A at x = -0.25 and F_B at x = 0.25. Force: F_A + F_B = 9. Moment about O: (-0.25)(-F_A) + (0.25)(-F_B) = 0.25(F_A - F_B) = 0.5, so F_A - F_B = 2. Then F_A = 5.5\,\mathrm{kN} and F_B = 3.5\,\mathrm{kN}, both downward. The couple loads the left wheel more.

12. \mathbf F = -500\cos80^\circ\,\mathbf i + 500\sin80^\circ\,\mathbf j = -86.82\mathbf i + 492.40\mathbf j\ \mathrm N, and \mathbf r_{OA} = 0.4(\cos30^\circ, \sin30^\circ) = (0.3464, 0.2)\,\mathrm m. At O: \mathbf F plus M = 0.3464(492.40) - 0.2(-86.82) = 170.57 + 17.36 = 187.9\,\mathrm{N\cdot m}, counter-clockwise. This agrees with Fl\sin\alpha = 500 \times 0.4 \times \sin110^\circ = 187.9\,\mathrm{N\cdot m}. A single force needs d = M/F = 187.9/500 = 0.3759\,\mathrm m, which is just the original line through A. The force–couple system and the original force are the same loading.