Study Companion: S04 — Moment of a Force

A force pushes a body along its line, and it can also turn the body about an axis. That turning effect is the moment of the force. The whole lesson rests on one fact: the moment depends only on the size of the force and on how far its line of action passes from the axis. It does not depend on how long the lever is, or where along the line the force is applied. Everything else here (the cross product, Varignon's theorem, moments that change as a mechanism moves, zero moment as a design rule) is that fact put to work.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 2/4 ("Moment"), with Sample Problems 2/5 and 2/6. Marghitu, Dupac and Madsen, Statics with MATLAB, section 2.1 ("Moment of a Vector About a Point") and Example 2.1. Harper, Solving Statics Problems in MATLAB, Problems 2/49 and 2/168 ("2D Moment"). Notation follows Meriam: unit vectors \mathbf i, \mathbf j, \mathbf k (Marghitu prints them dotless) and d for the moment arm. Sign convention throughout: counter-clockwise moments are positive, which is +\mathbf k out of the page.

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 car's wheel nuts are tightened to a moment of 110\,\mathrm{N\cdot m} (a typical figure). You have a wheel brace 0.40\,\mathrm m long.

Everyone knows to hold the brace at the end. But how much better is the end, exactly? Why does pushing at an angle feel weaker? And why does a push straight along the handle do nothing at all, however hard you push? The answers (275\,\mathrm N at the end, 550\,\mathrm N at the middle, and no force is enough along the handle) come from Nugget 01 and Nugget 02.

Nugget 01: Turning Grows With the Force and With the Arm

Claim. The turning effect of a force about an axis is the force times the perpendicular distance from the axis to the force's line of action.

The picture. A door seen from above, hinged at O and 0.85\,\mathrm m wide. A door closer resists with a moment of 12\,\mathrm{N\cdot m}. Push the door square to its face at the handle, 0.78\,\mathrm m from the hinge, and then at 0.10\,\mathrm m from the hinge.

Both pushes produce the same turning, 12\,\mathrm{N\cdot m} counter-clockwise. The one near the hinge needs a much larger force, in the same ratio as the distances. Halve the distance and the force must double.

Check with numbers. F = M/d. At the handle, F = 12/0.78 = 15.4\,\mathrm N. At 0.10\,\mathrm m, F = 12/0.10 = 120\,\mathrm N. The distances are in the ratio 0.78/0.10 = 7.8, and so are the forces. For the wheel nut in the puzzle, F = 110/0.40 = 275\,\mathrm N at the end of the brace and 110/0.20 = 550\,\mathrm N at its middle.

The idea. Meriam (2/4) defines the moment M of a force about an axis as its tendency to rotate the body about that axis. The axis may be any line that neither crosses nor runs parallel to the force's line of action. Another word for moment is torque. Experience with a pipe wrench shows that the tendency grows with the force F and with the moment arm d, the perpendicular distance from the axis to the line of action.

The moment is a vector, \mathbf M, along the axis. Its sense comes from the right-hand rule: curl the fingers in the direction the force tends to turn the body, and the thumb points along \mathbf M. It obeys the rules of vector addition and is a sliding vector along the moment axis.

When every force lies in one plane, we speak of the moment about a point. This means the moment about the axis through that point, normal to the plane. Marghitu calls the same quantity the moment about a line. All such moments point along \pm\mathbf k, so a sign is enough to record them. Choose a sign convention, state it, and keep it for the whole problem. This book uses counter-clockwise positive.

The equation (definition):

Units: newton metres, \mathrm{N\cdot m}. A moment is never written in joules, even though 1\,\mathrm J = 1\,\mathrm{N\cdot m}. The joule is kept for work, which is a force times a distance along the force, not across it.

Worked number. A 250\,\mathrm N pull at right angles to the end of a 0.30\,\mathrm m spanner gives M = 250 \times 0.30 = 75\,\mathrm{N\cdot m}. To get the same moment with a 0.20\,\mathrm m spanner you need 375\,\mathrm N.

Real example. Door handles sit at the edge away from the hinges. Push near the hinge and the door barely moves, though the push is just as hard.

Plausible wrong answer. "Moment is force, so a harder push is all that matters." The distance counts equally. The 15.4\,\mathrm N push at the handle beats a 100\,\mathrm N push at 0.10\,\mathrm m, which gives only 10\,\mathrm{N\cdot m}.

Check. A nut needs 45\,\mathrm{N\cdot m} and you can push with 180\,\mathrm N. How long must the spanner be at least? (0.25\,\mathrm m.)

Narration spine. A door from above. A small arrow at the handle swings it open. The same arrow near the hinge cannot move it. Slide the push towards the hinge while a force readout climbs and the product F \times d stays at 12\,\mathrm{N\cdot m}. A curled right hand appears at the hinge, its thumb pointing out of the screen.

Bridge. In the door and the brace, the push is square to the lever, so d is just the lever length. Real pushes are rarely square. We need to know which distance d really is.

Nugget 02: The Arm Is the Perpendicular Distance to the Line of Action

Claim. The moment arm is the perpendicular distance from the axis to the force's line, not the length of the lever. The same force at the same point can give its full moment, part of it, or none, depending on its direction.

The picture. A 300\,\mathrm N force acts at the end of a 0.40\,\mathrm m spanner, in four directions: \alpha = 90^\circ, 60^\circ, 30^\circ and 0^\circ to the handle.

Extend the 60^\circ force's line of action backwards (dashed) and drop a perpendicular to it from O. That perpendicular is the moment arm. It is shorter than the handle. As \alpha shrinks the line swings towards O and the arm shrinks with it. At \alpha = 0^\circ the line passes straight through O, and the arm is zero.

Check with numbers. In the right triangle formed by O, the end of the handle and the foot of the perpendicular, the handle is the hypotenuse, so d = 0.40\sin\alpha.

That answers the rest of the puzzle. A push at an angle wastes part of itself, and a push along the handle (\alpha = 0^\circ) has no moment about the nut at any size.

The idea. Meriam defines d as "the perpendicular distance from the axis to the line of action of the force". Two facts follow, and both matter.

• Sliding the force along its line does not change d. This agrees with transmissibility (S03): the moment is an external effect.
• The moment is zero exactly when the line of action passes through the point, or when the force itself is zero (Marghitu 2.1).

Meriam's pipe wrench says the same thing in words: "a pull which is not perpendicular to the wrench handle is less effective than the right-angle pull".

The equation (geometric relationship, with r the distance from the axis to any point on the line and \alpha the angle between that radius and the force):

Worked number. A cyclist pushes down on a pedal with 400\,\mathrm N. The crank is 0.17\,\mathrm m long. With the crank horizontal (\alpha = 90^\circ), M = 400 \times 0.17 = 68\,\mathrm{N\cdot m}. With the crank 30^\circ past top dead centre, a vertical push makes \alpha = 30^\circ and M = 400 \times 0.17 \times 0.5 = 34\,\mathrm{N\cdot m}. At top dead centre the push runs through the axle and M = 0.

Real example (one you can see). Watch someone pedal slowly up a hill. They stall with the cranks vertical, because there the push points straight at the axle.

Plausible wrong answer. "Moment = force × lever length." That gives 120\,\mathrm{N\cdot m} for every row of the table. It is right only for a push at 90^\circ to the lever.

Check. 250\,\mathrm N at 45^\circ to a 0.30\,\mathrm m handle, at its end. What is M? (250 \times 0.30 \times \sin45^\circ = 53.0\,\mathrm{N\cdot m}.)

Narration spine. The force arrow sits at the spanner's end at 90^\circ. A dashed line of action and the perpendicular from O appear; the perpendicular equals the handle. Rotate the arrow slowly. The dashed line swings, the perpendicular shortens, and a bar labelled M shrinks with it. At 0^\circ the line passes through O, the perpendicular vanishes and the bar hits zero.

Bridge. Finding d needed a right triangle. When a force's direction and point are given as coordinates, the triangle can be awkward to find. A vector formula finds d for us.

Nugget 03: The Cross Product Does the Geometry for You

Claim. \mathbf M = \mathbf r\times\mathbf F, with \mathbf r drawn from the moment centre to any point on the line of action, gives the moment's size and sense in one step.

The picture. The 60^\circ force again, now with two position vectors. One goes to the point of application E. The other goes to D, the foot of the perpendicular from O.

The size of a cross product is rF\sin\alpha, and r\sin\alpha is exactly the perpendicular distance d of Nugget 02. So the formula builds the right triangle by itself. Any point on the line works, because moving the tip of \mathbf r along the line adds a piece parallel to \mathbf F, and a vector crossed with a parallel vector gives zero.

Check with numbers. \mathbf F = 300(\cos60^\circ\,\mathbf i + \sin60^\circ\,\mathbf j) = 150\mathbf i + 259.8\mathbf j\ \mathrm N. In a plane, \mathbf r\times\mathbf F = (xF_y - yF_x)\mathbf k.

• With \mathbf r = 0.40\mathbf i\ \mathrm m: M = 0.40 \times 259.8 - 0 \times 150 = 103.9\,\mathrm{N\cdot m}.
• With \mathbf r' = \overrightarrow{OD} = 0.300\mathbf i - 0.1732\mathbf j\ \mathrm m: M = 0.300 \times 259.8 - (-0.1732)(150) = 77.94 + 25.98 = 103.9\,\mathrm{N\cdot m}.

Both give +103.9\mathbf k, counter-clockwise. \mathbf r' is perpendicular to \mathbf F, so its length \sqrt{0.300^2 + 0.1732^2} = 0.3464\,\mathrm m is d itself.

The idea. Meriam (2/4) writes \mathbf M = \mathbf r\times\mathbf F, with \mathbf r running from the moment reference point to any point on the line of action. The right-hand rule applied to the order \mathbf r\times\mathbf F gives the sense. The order matters: \mathbf F\times\mathbf r = -\mathbf r\times\mathbf F gives -103.9\,\mathrm{N\cdot m}, the wrong sense. When should you use which? If d is easy to see, the scalar Fd is usually simpler. If the force and the point are given in coordinates and are not perpendicular, the cross product is usually simpler. In three dimensions (S08) the cross product is almost always the better choice.

Marghitu (2.1) defines the moment of any vector \mathbf v about a point A as \mathbf M_A = \mathbf r_{AB}\times\mathbf v, for B anywhere on its line. He proves the independence of B directly. If B' is another point on the line, then \mathbf r_{AB} = \mathbf r_{AB'} + \mathbf r_{B'B}, and \mathbf r_{B'B}\times\mathbf v = \mathbf 0 because the two are parallel. The moment is perpendicular to both \mathbf r and \mathbf v.

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

Here (x, y) is any point on the line of action, measured from O.

Worked number. Marghitu's demonstration takes B(1, 2, 0) and C(3, 3, 0), with \mathbf v = \mathbf r_C - \mathbf r_B = 2\mathbf i + \mathbf j along the line BC. About the origin, \mathbf r_B\times\mathbf v = (1 \times 1 - 2 \times 2)\mathbf k = -3\mathbf k. The point B' = B + 0.75(C - B) = (2.5, 2.75, 0) gives (2.5 \times 1 - 2.75 \times 2)\mathbf k = -3\mathbf k, and C gives (3 \times 1 - 3 \times 2)\mathbf k = -3\mathbf k. All three agree, as they must.

Real example. A torque-angle gauge on a power tool computes \mathbf r\times\mathbf F from measured force components and a known geometry. This is the same arithmetic as above, done thousands of times a second.

Plausible wrong answer. "\mathbf r must go to the point where the force is applied." Any point on the line works. Choosing the foot of the perpendicular, or a point on an axis (Nugget 04), often makes the arithmetic shorter.

Check. A force \mathbf F = 3\mathbf i - 2\mathbf j\ \mathrm{kN} acts through (1, 4)\,\mathrm m. What is its moment about (-1, 1)\,\mathrm m? (\mathbf r = 2\mathbf i + 3\mathbf j, so M = 2(-2) - 3(3) = -13\,\mathrm{kN\cdot m}, clockwise.)

Second view (Marghitu, MATLAB). A plane force is a 3-vector with a zero third entry, and cross returns the moment:

r  = [0.40 0 0];                   % m
F  = 300*[cosd(60) sind(60) 0];    % N
M  = cross(r, F)                   % [0 0 103.9]
rD = [0.300 -0.1732 0];
cross(rD, F)                       % [0 0 103.9], any point on the line

Marghitu's Example 2.1 is a 500\,\mathrm N force at 45^\circ, applied a = 1\,\mathrm m across and b = 5\,\mathrm m up from the base O of an L-shaped post. It gives M_O = (aF\cos\theta - bF\sin\theta)\mathbf k = (353.6 - 1767.8)\mathbf k = -1414\mathbf k\ \mathrm{N\cdot m}. The book's substitution line prints "5" for the force in the first term, and its program runs with F = 5, reporting -14.14\,\mathrm{kN\cdot m}. With the stated 500\,\mathrm N the moment is -1414\,\mathrm{N\cdot m}.

Narration spine. Show the spanner, the 60^\circ force and its dashed line. Draw \mathbf r from O to E and shade the parallelogram of \mathbf r and \mathbf F; its area reads 103.9. Now slide the tip of \mathbf r along the dashed line to D. The parallelogram shears but keeps its height and its area. At D, \mathbf r meets the line at a right angle, and its length is d.

Bridge. The cross product splits naturally into x and y terms: xF_y and -yF_x. Each looks like the moment of one component of the force. That is a theorem, and it is the most useful shortcut in plane statics.

Nugget 04: Varignon: Split the Force, Add the Moments

Claim. The moment of a force about a point equals the sum of the moments of its components about that point. Any split works, at any point on the line.

The picture. A 900\,\mathrm N cable pull acts at A, the end of a sign post's arm. A is 1.5\,\mathrm m across from the post's base O and 3.2\,\mathrm m above it. The pull points 35^\circ below the horizontal.

The force tends to turn the post clockwise about its base. The green components show why. The horizontal part pulls the top to the right, and the downward part pushes the arm down. Both turn the post clockwise, and their two moments add up to the moment of the whole force. The line of action also crosses the y-axis at B and the x-axis at C. Slid to either point, the force has one component that passes through O, so only the other one turns.

Check with numbers. F_x = 900\cos35^\circ = 737.2\,\mathrm N and F_y = -900\sin35^\circ = -516.2\,\mathrm N. Here are five ways to the same moment (Meriam's Sample Problem 2/5 does the same with its own numbers).

1. M = Fd. The perpendicular distance from O to the line is d = 3.2\cos35^\circ + 1.5\sin35^\circ = 2.621 + 0.860 = 3.482\,\mathrm m, so M = 900 \times 3.482 = 3133\,\mathrm{N\cdot m} clockwise.
2. Varignon at A. 737.2\,\mathrm N with arm 3.2\,\mathrm m gives 2359\,\mathrm{N\cdot m}; 516.2\,\mathrm N with arm 1.5\,\mathrm m gives 774.3\,\mathrm{N\cdot m}; both are clockwise, total 3133\,\mathrm{N\cdot m}.
3. Slide to B. B is at height 3.2 + 1.5\tan35^\circ = 3.2 + 1.050 = 4.250\,\mathrm m on the y-axis. There F_y passes through O, and 737.2 \times 4.250 = 3133\,\mathrm{N\cdot m}.
4. Slide to C. C is at 1.5 + 3.2/\tan35^\circ = 1.5 + 4.570 = 6.070\,\mathrm m on the x-axis. There F_x passes through O, and 516.2 \times 6.070 = 3133\,\mathrm{N\cdot m}.
5. Cross product. M = xF_y - yF_x = 1.5(-516.2) - 3.2(737.2) = -3133\,\mathrm{N\cdot m}. The minus sign means clockwise under our convention.

The idea. Meriam (2/4) proves the theorem in one line. If \mathbf R = \mathbf P + \mathbf Q, then by the distributive law of the cross product (S02), \mathbf r\times\mathbf R = \mathbf r\times\mathbf P + \mathbf r\times\mathbf Q. The same \mathbf r serves all three because the components are concurrent with \mathbf R at the tip of \mathbf r. Nothing limits the split to two parts, or to perpendicular ones. Meriam notes that the theorem was first stated only for two concurrent components. Marghitu gives the same distributive law the name "Varignon" as early as his Chapter 1.

Meriam writes the theorem for a general split as M_O = Rd = -pP + qQ, where p and q are the arms of the two components. In that one equation he takes clockwise as positive. This lesson keeps counter-clockwise positive everywhere, so the same statement reads M_O = pP - qQ.

The equation (mathematical identity: the distributive law of the cross product):

Worked number. Meriam's Sample Problem 2/5 is a 600\,\mathrm N force at 40^\circ below the horizontal. It acts 2\,\mathrm m across and 4\,\mathrm m up from the base. Its arm is 4\cos40^\circ + 2\sin40^\circ = 4.35\,\mathrm m, so M = 2610\,\mathrm{N\cdot m} clockwise. The components 460\,\mathrm N and 386\,\mathrm N give 460(4) + 386(2) = 2610. Sliding the force to the axes gives arms 4 + 2\tan40^\circ = 5.68\,\mathrm m and 2 + 4\cot40^\circ = 6.77\,\mathrm m.

Real example. Engineers check the base of a road sign, a lamp post or a crane mast this way. The wind load and the cable pulls are split into horizontal and vertical parts, and each part's moment about the base is added. The total sets the size of the base bolts.

Plausible wrong answer. "Add the components' moments, but use the distance from O to A as the arm for both." Each component has its own perpendicular arm: 3.2\,\mathrm m for the horizontal part and 1.5\,\mathrm m for the vertical. Using OA = 3.534\,\mathrm m for both gives nonsense.

Check. In method 3 the force was slid to B, a point off the post. Is that allowed? (Yes. A moment is a calculation about a line of action, and the point need not be on the body. Meriam makes the same remark.)

Narration spine. The force at A and its dashed line. Split it into green components; each sweeps an arc about O labelled with its own moment, and the two arcs add to 3133. Then slide the force along its line up to B; the vertical component's arm collapses to zero. Slide it down to C; now the horizontal component's arm vanishes. Every route lands on the same number.

Bridge. So far every force and every point has stayed fixed. In a mechanism the force's line moves as the parts turn, and the moment arm changes with it. Sometimes it shrinks to nothing.

Nugget 05: Moment Arms Change as Things Move

Claim. In a moving mechanism the moment arm is a function of position. It peaks when the force is perpendicular to the crank, and it falls to zero when the line of action passes through the pivot.

The picture. A crank of radius 60\,\mathrm{mm} turns on a shaft O. A spring anchored at S, 240\,\mathrm{mm} to the left of O, pulls on the crank pin P. The crank angle \varphi is measured from +x.

The spring always pulls along the line PS. At \varphi = 60^\circ that line passes 45.4\,\mathrm{mm} from O, less than the 60\,\mathrm{mm} crank. At \varphi = 0^\circ the pin lies on the line OS, and the spring pulls straight through the shaft (dashed). The arm is zero, so the spring cannot turn the crank at all, however hard it pulls. This position is called dead centre.

Check with numbers. Place O at the origin, so that P = (0.06\cos\varphi, 0.06\sin\varphi)\,\mathrm m and S = (-0.24, 0)\,\mathrm m. The arm is the distance from O to the line PS. By the cross product,

At \varphi = 60^\circ, PS = 275.0\,\mathrm{mm} and the numerator is 0.01247\,\mathrm{m^2}, so d = 45.4\,\mathrm{mm}. Take a spring force of 500\,\mathrm N at each position (for this comparison only; a real spring's force changes with its length):

The arm reaches the full crank radius only at \varphi = 104.5^\circ. That is where PS is perpendicular to OP, given by \cos\varphi = -0.06/0.24.

The idea. Meriam's Sample Problem 2/6 makes this point with a trap door. The door is 0.5\,\mathrm m long, raised by a cable over a pulley 0.4\,\mathrm m above its hinge. The ratio M_O/T = 0.2\cos\theta/\sqrt{0.41 - 0.4\sin\theta}\ \mathrm m is the moment arm. It is largest, 0.4\,\mathrm m, at \theta = 53.1^\circ, where the cable is horizontal. It falls to zero at \theta = 90^\circ, where the cable runs through the hinge. Meriam notes that the ratio holds even if the tension varies. The moment arm depends on geometry alone.

The equation (geometric relationship: the perpendicular distance from a point to a line, by the cross product):

Worked number (second view, Harper's Problem 2/168). A 75\,\mathrm N erecting cable raises a flagpole through a light triangular frame with 3\,\mathrm m sides. The cable runs from the frame's tip A to a winch D, 6\,\mathrm m from the pivot O. Harper slides the tension along its line to D. There only its vertical part has an arm about O, and M_O = 6T\sin\alpha, where \alpha is the cable's angle at D. The frame geometry gives \alpha through the laws of cosines and sines. With elevation \theta, the angle at O is \gamma = 120^\circ - \theta, and

This gives 147.3\,\mathrm{N\cdot m} at \theta = 0, 225\,\mathrm{N\cdot m} at \theta = 60^\circ and 181.6\,\mathrm{N\cdot m} at \theta = 90^\circ. Harper finds the maximum three ways. By geometry: it occurs when \mathbf T is perpendicular to OA, so \cos\gamma = 3/6 gives \gamma = 60^\circ, \theta = 60^\circ, and M = 3 \times 75 = 225\,\mathrm{N\cdot m}. By setting dM_O/d\theta = 0 symbolically. And by MATLAB's max on a grid, which is approximate:

theta = 0.9:0.001:1.1;                  % rad
gamma = 2*pi/3 - theta;
AD = sqrt(6^2 + 3^2 - 2*6*3*cos(gamma));
alpha = asin(3*sin(gamma)./AD);
Mo = 6*75*sin(alpha);
[y, i] = max(Mo)                        % 225.0000 at i = 148
theta(i)*180/pi                         % 59.99 deg, not exactly 60

Real example (one you can see). The owner's switchgear contact is this crank and spring. A crank turns the shaft, the spring's line swings towards the shaft centre, and its moment arm shrinks to zero at dead centre. Just past that point the arm reappears on the other side, and the spring snaps the contact shut. The same thing happens with a car's boot strut, which holds the lid up once its line passes the hinge.

Plausible wrong answer. "The spring's torque is force × crank length, 500 \times 0.06 = 30\,\mathrm{N\cdot m}, at every angle." That is right at one angle only, 104.5^\circ. At 30^\circ the true value is 12.3\,\mathrm{N\cdot m}, and at dead centre it is zero.

Check. At which crank angles is the spring's moment zero, and why? (0^\circ and 180^\circ, where P lies on the line OS, so the spring's line passes through O.)

Narration spine. The crank turns slowly. The spring's line sweeps around, the perpendicular from O to it is drawn live, and a graph of d against \varphi traces itself beside the mechanism. At \varphi = 0^\circ the spring's line passes through O and the trace touches zero. A ghost arrow labelled "F \times crank" stays flat at 30\,\mathrm{N\cdot m} above the true curve, the wrong answer made visible.

Bridge. A single force's moment can now be computed and tracked. Real fittings carry several forces, and often the design requirement is that their moments cancel.

Nugget 06: Moments Add, and Zero Total Moment Is a Design Condition

Claim. The moment of several forces about a point is the sum of their moments. Setting that sum to zero is a design condition that fixes an unknown force.

The picture. A masthead fitting is clamped around a mast whose centre is O. A 4\,\mathrm{kN} load hangs from pin A (100\,\mathrm{mm} left of O, 50\,\mathrm{mm} up), at an angle \theta from the vertical. A stay at pin B (130\,\mathrm{mm} right, 50\,\mathrm{mm} up) pulls down and to the right on a slope of 1 across to 3 down. Choose the stay tension T so that the mast is not bent at O: zero total moment about O.

The load at A tends to turn the fitting counter-clockwise about O, and the stay at B turns it clockwise. For zero total, the two must match. As the load swings outward (\theta grows), its line moves closer to O, its moment falls, and less stay tension is needed.

Check with numbers. Use M = xF_y - yF_x in metres and kilonewtons. The load is \mathbf L = 4(-\sin\theta\,\mathbf i - \cos\theta\,\mathbf j) at (-0.100, 0.050):

The stay is \mathbf T = T(\mathbf i - 3\mathbf j)/\sqrt{10} at (0.130, 0.050):

Setting M_L + M_T = 0 gives T = (0.4\cos\theta + 0.2\sin\theta)/0.1391. At \theta = 30^\circ: M_L = 0.4464\,\mathrm{kN\cdot m} and T = 3.208\,\mathrm{kN}.

The idea. Marghitu (2.1) defines the moment of a system of vectors about a point as the sum of their moments, \mathbf M_A = \sum\mathbf r_i\times\mathbf v_i. In the plane that is a signed sum of F d terms. The largest T comes where 0.4\cos\theta + 0.2\sin\theta is largest, at \tan\theta = 0.5, or \theta = 26.57^\circ. That is where the load's line is perpendicular to OA, so its arm equals the full distance OA. T falls to zero at \tan\theta = -2, \theta = 116.6^\circ, where the load's line passes through O. Beyond that, the formula gives a negative T (for example -1.771\,\mathrm{kN} at 150^\circ). A stay can pull but cannot push, so no stay tension can keep the moment zero there. The design needs a strut, or a different geometry.

The equation (definition of the system's moment; the zero condition is a design requirement imposed on it):

Worked number (second view, Harper's Problem 2/49). Harper's fitting has a 5\,\mathrm{kN} load on a pin 90\,\mathrm{mm} left of and 60\,\mathrm{mm} above O, and the stay on a 2:5 slope at 120\,\mathrm{mm} right, 60\,\mathrm{mm} up. Zero moment gives T = \frac{5\sqrt{29}}{24}(3\cos\theta + 2\sin\theta), which is 4.037\,\mathrm{kN} at the drawn \theta = 30^\circ. He then has MATLAB plot T and the resultant R of the two forces for 0^\circ \le \theta \le 180^\circ:

theta = 0:0.01:pi;
T  = 5*sqrt(29)/24*(3*cos(theta) + 2*sin(theta));
Rx = 2/sqrt(29)*T - 5*sin(theta);
Ry = 5/sqrt(29)*T + 5*cos(theta);
plot(theta*180/pi, T, theta*180/pi, sqrt(Rx.^2 + Ry.^2))

His plot shows T going negative beyond about 123.7^\circ, where \tan\theta = -1.5. As with our fitting, those angles cannot be reached with a cable. Harper's plot does not comment on this.

Real example. The forestay and backstay of a sailing boat's mast are set so that the sail loads and stay tensions put no bending moment on the masthead. The rigger tensions them to exactly this kind of moment balance.

Plausible wrong answer. "Zero moment means the forces are equal." Here the load is 4\,\mathrm{kN} and the stay needs 3.208\,\mathrm{kN}. It is the moments that are equal, and different arms need different forces.

Check. At \theta = 0, what are the stay tension and the size of the resultant of the two forces on the fitting? (See self-check 10: 2.875\,\mathrm{kN} and 6.788\,\mathrm{kN}.)

Narration spine. Show the fitting with the load and stay arrows, and arcs for their moments about O (orange counter-clockwise, blue clockwise). Swing the load outward. Its arc shrinks, the stay arrow shrinks to keep the arcs equal, and a T against \theta curve traces out. At 116.6^\circ the load's line crosses O and the stay arrow vanishes. Past it, the arrow flips to a push and turns red.

Synthesis

The wheel brace needed only Nuggets 01 and 02: 275\,\mathrm N at the end, 550\,\mathrm N at the middle, and nothing at all along the handle. Every later nugget used the same fact. The cross product is Fd computed from coordinates. Varignon's theorem is the cross product distributed over components. The spring on the crank lost its moment when its line reached the shaft. The masthead stay was set by matching two Fd products. A force turns a body about a point by its size times the perpendicular distance of its line from that point, and by nothing else. Where the force is applied along that line, how long the lever is, and how the force is split into components all drop out.

• Central idea: the moment of a force about a point depends only on F and on the perpendicular distance d from the point to its line of action.
• Key relationship: M = Fd = Fr\sin\alpha, and as a vector \mathbf M = \mathbf r\times\mathbf F (M = xF_y - yF_x in the plane), with Varignon's \mathbf r\times(\mathbf P + \mathbf Q) = \mathbf r\times\mathbf P + \mathbf r\times\mathbf Q.
• Most important assumption: the body is rigid, so a force may slide along its line, and a stated sign convention is kept consistently. Here that is counter-clockwise positive, +\mathbf k.
• Open question: two equal and opposite forces on parallel lines add to zero force, yet together they still turn a body. What is that pure turning, and does it depend on the point we take moments about? (S05.)

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). S04_moment.py takes xF_y - yF_x for points along one line of action, checks Varignon on Sample Problem 2/5, sweeps Harper's flagpole on his grid and at exactly 60^\circ, and tabulates the stay tension of Harper's fitting.

Part 1: r x F
  spanner, r to E: M = 103.9 N m
  spanner, r to D: M = 103.9 N m
  B + 0(C - B): r x v = -3.0
  B + 0.75(C - B): r x v = -3.0
  B + 1(C - B): r x v = -3.0
  Marghitu Ex 2.1: M_O = -1414.2 N m
Part 2: Varignon, Sample Problem 2/5
  components [ 460. -386.] N, r x F = -2610 N m
  arm 4.35 m, 600 x arm = 2610 N m clockwise
Part 3: Harper's flagpole, T = 75 N
  grid max 225.0000 N m at 59.99 deg
  at exactly 60 deg: 225.0000 N m
Part 4: Harper's fitting, zero moment about O
  theta  0 deg: T = 3.366 kN, R = 8.221 kN
  theta 30 deg: T = 4.037 kN, R = 8.140 kN
  theta 60 deg: T = 3.626 kN, R = 6.582 kN
  theta 90 deg: T = 2.244 kN, R = 4.658 kN

The grid maximum lands at 59.99^\circ only because 60^\circ is not a grid point; the moment there is the same 225\,\mathrm{N\cdot m}.

MATLAB (Octave-compatible). S04_moment.m uses cross on 3-vectors with a zero third entry, and max on the flagpole sweep.

spanner: M from E 103.9 N m, from D 103.9 N m
Marghitu Ex 2.1: M_O = -1414.2 N m
SP 2/5: components [460 -386] N, M = -2610 N m
flagpole: grid max 225.0000 N m at 59.99 deg
theta  0 deg: T = 3.366 kN, R = 8.221 kN
theta 30 deg: T = 4.037 kN, R = 8.140 kN
theta 60 deg: T = 3.626 kN, R = 6.582 kN
theta 90 deg: T = 2.244 kN, R = 4.658 kN

Traps

• Lever length for moment arm. The spring on the crank gives 500 \times 0.06 = 30\,\mathrm{N\cdot m} at one angle only; at 30^\circ it gives 12.3, and at dead centre 0. Use the perpendicular distance to the line of action. (From Nugget 02 and Nugget 05.)
• A big force has a big moment. A push of any size along the handle, through O, has zero moment. (From Nugget 02.)
• \mathbf F\times\mathbf r. Reversing the order flips the sign: \mathbf F\times\mathbf r = -103.9\mathbf k for a counter-clockwise 103.9\,\mathrm{N\cdot m}. Write \mathbf r\times\mathbf F, with \mathbf r from the point to the line. (From Nugget 03.)
• Mixed sign conventions. Meriam's M_O = -pP + qQ takes clockwise positive. Pick one convention, write it on the diagram, and keep it. (From Nugget 04.)
• One arm for both components. Each component has its own perpendicular arm: 3.2\,\mathrm m for F_x and 1.5\,\mathrm m for F_y, not OA for both. (From Nugget 04.)
• Maximum moment at the end of travel. The flagpole's moment peaks at \theta = 60^\circ, where the cable is perpendicular to OA, not at either end. (From Nugget 05.)
• Trusting a negative cable tension. T = -1.771\,\mathrm{kN} means the required force is a push, which a cable cannot give. (From Nugget 06.)
• Moment in joules. A moment is in \mathrm{N\cdot m}; the joule is reserved for work. (From Nugget 01.)
• Copying a mixed-up example. Marghitu's Example 2.1 states F = 500\,\mathrm N but computes with 5. The moment is -1414\,\mathrm{N\cdot m}. Check that every number in a substitution is the stated one, with its unit. (From Nugget 03.)

Self-check

1. A 5\,\mathrm{kN} force acts at A(2.0, 0.6)\,\mathrm m. It points down and to the left, 4 units left for every 3 down. Find its moment about O as a scalar and as a vector, and the points on the x- and y-axes about which its moment is zero.
2. A force F = 250\,\mathrm N acts along the sloping edge of a right-triangular plate. The edge runs from the top of the vertical side (h = 0.30\,\mathrm m above the corner O) down to the end of the base (b = 0.40\,\mathrm m from O). Find the moment about O.
3. A wrench handle is gripped at A, 0.22\,\mathrm m to the right of the bolt centre O and 0.035\,\mathrm m above it. The hand pushes with 320\,\mathrm N, downward and 20^\circ to the left of the vertical. Find the moment about O.
4. A 80\,\mathrm N force acts on a gear tooth at the top of a pitch circle of radius 0.12\,\mathrm m, pointing to the right and 25^\circ below the horizontal. Find its moment about the gear centre.
5. A drum rests on the floor at P. Its outer radius is 0.18\,\mathrm m. A cord wound on its hub (radius 0.10\,\mathrm m) pulls with 240\,\mathrm N up and to the left, at \theta above the horizontal, leaving the hub on its lower left side. Find the moment about the centre C, and the angle \theta for which the moment about P is zero.
6. A forearm is horizontal. Its weight, 22\,\mathrm N, acts 0.15\,\mathrm m from the elbow O. A 36\,\mathrm N ball is held 0.33\,\mathrm m from O. The biceps pulls vertically upward 0.05\,\mathrm m from O. Find the moment of the two weights about O, and the biceps tension that makes the total moment zero.
7. A wheelbarrow and its load weigh 800\,\mathrm N, acting 0.25\,\mathrm m horizontally from the wheel's ground contact B. The handles are 1.20\,\mathrm m horizontally from B. What vertical lift F makes the total moment about B zero?
8. A 350\,\mathrm N force acts at the corner B of a crate, 0.5\,\mathrm m right of and 0.3\,\mathrm m above the lower corner A. It points to the left and 20^\circ below the horizontal. Find M_A by components, find the moment arm, and check by sliding the force to the vertical line through A.
9. A trap door 0.6\,\mathrm m long is hinged at O. It is raised by a cable from its free end to a pulley 0.45\,\mathrm m directly above O, with tension 200\,\mathrm N. Find the moment about O at a door elevation of 30^\circ, the largest possible moment arm and the angle at which it occurs.
10. For the masthead fitting of Nugget 06 with \theta = 0, find the stay tension for zero moment about O and the magnitude of the resultant of the load and the stay.
11. A crank of radius 50\,\mathrm{mm} carries a spring anchored 200\,\mathrm{mm} to the left of the shaft, pulling with 400\,\mathrm N. At a crank angle of 60^\circ, find the moment arm and the moment. At what angle is the arm largest, and what is the "force × crank length" answer?
12. A force \mathbf F = 3\mathbf i - 2\mathbf j\ \mathrm{kN} acts at A(1, 4)\,\mathrm m. Find its moment about P(-1, 1)\,\mathrm m by the cross product and check it by Varignon's theorem.

Fully worked solutions

1. \mathbf F = 5 \times \frac{1}{5}(-4\mathbf i - 3\mathbf j) = -4\mathbf i - 3\mathbf j\ \mathrm{kN}. M_O = xF_y - yF_x = 2.0(-3) - 0.6(-4) = -6 + 2.4 = -3.6\,\mathrm{kN\cdot m}, that is 3.6\,\mathrm{kN\cdot m} clockwise, or \mathbf M_O = -3.6\mathbf k\ \mathrm{kN\cdot m}. The moment is zero about any point on the line of action. Along the line, (x, y) = (2.0 - 4s, 0.6 - 3s). Setting y = 0 gives s = 0.2 and x = 1.2, so the point is (1.2, 0)\,\mathrm m. Setting x = 0 gives s = 0.5 and y = -0.9, so the point is (0, -0.9)\,\mathrm m.

2. The edge runs from (0, 0.30) towards (0.40, 0), a length of 0.50\,\mathrm m, so \mathbf F = 250(0.40\mathbf i - 0.30\mathbf j)/0.50. Take \mathbf r = 0.30\mathbf j: M = 0 \times F_y - 0.30 \times 250 \times 0.80 = -60\,\mathrm{N\cdot m}, 60\,\mathrm{N\cdot m} clockwise. In general M = Fbh/\sqrt{b^2 + h^2}, since the arm is the altitude of the triangle onto its hypotenuse.

3. \mathbf F = 320(-\sin20^\circ\,\mathbf i - \cos20^\circ\,\mathbf j) = -109.4\mathbf i - 300.7\mathbf j\ \mathrm N. M_O = 0.22(-300.7) - 0.035(-109.4) = -66.15 + 3.83 = -62.32\,\mathrm{N\cdot m}, 62.3\,\mathrm{N\cdot m} clockwise. The small offset of the grip changes the answer by about 6 %.

4. \mathbf r = 0.12\mathbf j and \mathbf F = 80\cos25^\circ\,\mathbf i - 80\sin25^\circ\,\mathbf j = 72.50\mathbf i - 33.81\mathbf j\ \mathrm N. M = 0 - 0.12 \times 72.50 = -8.70\,\mathrm{N\cdot m}, clockwise. Only the tangential component, 72.50\,\mathrm N, has an arm; the radial part passes through the centre.

5. The cord's line is tangent to the hub, so its arm about C is the hub radius. It leaves the hub below the centre and pulls up and to the left, which turns the drum clockwise: M_C = -240 \times 0.10 = -24\,\mathrm{N\cdot m}, for any \theta. About P, which is 0.18\,\mathrm m below C, the transfer adds the moment of \mathbf T at the arm 0.18 from its horizontal part: M_P = -24 + 0.18 \times 240\cos\theta = 240(0.18\cos\theta - 0.10). This is zero when \cos\theta = 0.10/0.18 = 0.5556, so \theta = 56.25^\circ. Then the cord's line passes through P.

6. Both weights turn the forearm clockwise: M = -(22 \times 0.15 + 36 \times 0.33) = -(3.30 + 11.88) = -15.18\,\mathrm{N\cdot m}. The biceps turns it counter-clockwise with 0.05\,T. Zero total: T = 15.18/0.05 = 303.6\,\mathrm N, nearly six times the weight in the hand.

7. For vertical forces the arms are horizontal distances. About B: 1.20F - 0.25 \times 800 = 0, so F = 166.7\,\mathrm N.

8. \mathbf F = -350\cos20^\circ\,\mathbf i - 350\sin20^\circ\,\mathbf j = -328.9\mathbf i - 119.7\mathbf j\ \mathrm N. By components, M_A = 0.5(-119.7) - 0.3(-328.9) = -59.85 + 98.67 = 38.81\,\mathrm{N\cdot m}, counter-clockwise. The arm is d = 38.81/350 = 0.1109\,\mathrm m. Now slide the force along its line to x = 0. The line falls 0.5\tan20^\circ = 0.1820\,\mathrm m over that run, so it crosses at y = 0.3 - 0.1820 = 0.1180\,\mathrm m. There only F_x has an arm: M_A = 0.1180 \times 328.9 = 38.81\,\mathrm{N\cdot m}, counter-clockwise.

9. With a = 0.6, b = 0.45 and the pulley at (0, b), Sample Problem 2/6's method gives the arm ab\cos\theta/\sqrt{a^2 + b^2 - 2ab\sin\theta}. At 30^\circ: 0.27 \times 0.8660/\sqrt{0.36 + 0.2025 - 0.54 \times 0.5} = 0.2338/0.5408 = 0.4323\,\mathrm m, so M_O = 200 \times 0.4323 = 86.47\,\mathrm{N\cdot m}. The arm is largest when the cable is horizontal, at 0.6\sin\theta = 0.45, which gives \theta = 48.59^\circ and an arm of 0.45\,\mathrm m, the pulley height. At \theta = 90^\circ the cable runs through the hinge and the moment is zero.

10. At \theta = 0, M_L = 0.4\,\mathrm{kN\cdot m}, so T = 0.4/0.1391 = 2.875\,\mathrm{kN}. The stay is (T/\sqrt{10})(\mathbf i - 3\mathbf j) = 0.909\mathbf i - 2.727\mathbf j\ \mathrm{kN}, and the load is -4\mathbf j\ \mathrm{kN}. So \mathbf R = 0.909\mathbf i - 6.727\mathbf j\ \mathrm{kN} and R = 6.788\,\mathrm{kN}.

11. P = (0.025, 0.0433)\,\mathrm m and S = (-0.2, 0)\,\mathrm m, so PS = \sqrt{0.225^2 + 0.0433^2} = 229.1\,\mathrm{mm}. The arm is d = 0.2 \times 0.05\sin60^\circ/0.2291 = 0.008660/0.2291 = 37.80\,\mathrm{mm}, and M = 400 \times 0.03780 = 15.12\,\mathrm{N\cdot m}. The arm is largest, 50\,\mathrm{mm}, when PS \perp OP: \cos\varphi = -0.05/0.2, so \varphi = 104.48^\circ. "Force × crank length" gives 400 \times 0.05 = 20\,\mathrm{N\cdot m}, which is true at that one angle only.

12. \mathbf r_{PA} = (1 - (-1))\mathbf i + (4 - 1)\mathbf j = 2\mathbf i + 3\mathbf j\ \mathrm m. M_P = 2(-2) - 3(3) = -13\,\mathrm{kN\cdot m}, 13\,\mathrm{kN\cdot m} clockwise. Varignon: F_x = 3\,\mathrm{kN} acts 3\,\mathrm m above P and pushes right, giving -9\,\mathrm{kN\cdot m}. F_y = -2\,\mathrm{kN} acts 2\,\mathrm m right of P and pushes down, giving -4\,\mathrm{kN\cdot m}. The total is -13\,\mathrm{kN\cdot m}.