Study Companion: S30 — Work and the Principle of Virtual Work

Every earlier lesson found the forces on a body whose position was known: isolate it, draw every force, and write \Sigma\mathbf F = \mathbf 0 and \Sigma\mathbf M = \mathbf 0. A mechanism is a different kind of problem. Its links move relative to each other, many positions are possible, and the question is often which position the loads will hold, or what force holds a given one. Taking it apart works, but drags in every pin force only to cancel them again. This lesson builds a shorter route from one idea, the work done by a force during a small imagined motion. For a system in equilibrium that work is zero, and the forces that would have cluttered the free-body diagrams never appear.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 7, articles 7/1 ("Introduction"), 7/2 ("Work") and 7/3 ("Equilibrium"), with Sample Problems 7/1–7/3; Marghitu, Dupac and Madsen, Statics with MATLAB, section 6.1 ("Virtual Displacement and Virtual Work"); Harper, Solving Statics Problems in MATLAB, "Virtual Work, Equilibrium" (its Problem 7/27, the same number in Meriam's 9th edition). Units SI throughout; each coordinate's positive sense is stated on its figure, and every virtual displacement takes the sign its coordinate gives it.

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 scissor jack is four equal links pinned in a rhombus. The bottom joint sits on the ground, the top joint lifts the car's sill, and a horizontal screw joins the two side joints: turning the handle shortens the screw, the side joints are pulled together and the top rises. Under one side of a car the jack carries W = 5\,\mathrm{kN}. Just after it touches the sill it is nearly flat, its links at \theta = 15^\circ to the horizontal.

What tension T must the screw carry? The plausible guess is "not much more than the load, and probably less: the screw is horizontal and the load vertical, so the screw only steadies the joints". Anyone who has used one knows the first turns of the handle are the hardest, and the answer, 18.7\,\mathrm{kN}, is in Nugget 04.

Nugget 01: Work Is the Force Component Along the Displacement

Claim. A force does work only through the part of it that lies along the motion of its point: U = (F\cos\alpha)s, which is the same number as F(s\cos\alpha). Work is a scalar; it is positive when that component points with the motion and negative when against it.

The picture. Meriam's Fig. 7/1: the point A of a body moves through s while a force F acts at angle \alpha to the motion.

In (a) the force is resolved along the displacement: only F\cos\alpha pushes the body along s, and the perpendicular part neither helps nor hinders. In (b) the displacement is resolved along the force: the point advanced s\cos\alpha in the direction F points. Both products are the same, so work does not depend on which vector is resolved, and it has no direction of its own.

Check with numbers. A rope pulls a crate with F = 200\,\mathrm N at \alpha = 30^\circ above the floor while the crate slides s = 5\,\mathrm m. (a) F\cos\alpha = 173.2\,\mathrm N, times 5\,\mathrm m, is 866.0\,\mathrm J. (b) s\cos\alpha = 4.330\,\mathrm m, times 200\,\mathrm N, is again 866.0\,\mathrm J. A 150\,\mathrm N drag opposing the slide does -150 \times 5 = -750\,\mathrm J.

The idea. Meriam (7/2) defines the work of \mathbf F during an infinitesimal displacement d\mathbf r of its point as the dot product, and adds it up along the path for a finite motion:

To integrate, the force must be known as a function of position. For several forces at one point the work of their resultant equals the sum of their works, because the component of the resultant along the motion is the sum of the components. The SI unit is the joule, 1\,\mathrm J = 1\,\mathrm{N\cdot m} of force times distance along the force.

The equation (definition: work of a force):

Worked number. Two forces act at one ring, \mathbf F_1 = 300\mathbf i\,\mathrm N and \mathbf F_2 = 400\mathbf j\,\mathrm N, while it moves \mathbf s = 0.2\mathbf i + 0.1\mathbf j\,\mathrm m: U_1 = 300(0.2) = 60\,\mathrm J, U_2 = 400(0.1) = 40\,\mathrm J. Their resultant 300\mathbf i + 400\mathbf j\,\mathrm N does 60 + 40 = 100\,\mathrm J, the sum. A force that varies needs the integral: a spring force F_x = -kx with k = 2\,\mathrm{kN/m}, as the block it holds moves from x = 20 to 50\,\mathrm{mm}, does U = \int_{0.02}^{0.05}-2000x\,dx = -\frac12(2000)(0.05^2 - 0.02^2) = -2.1\,\mathrm J.

Real example. The scissor jack of the puzzle, opened from 15^\circ to 60^\circ with links of a = 0.2\,\mathrm m, raises its top joint from 2a\sin15^\circ = 103.5\,\mathrm{mm} to 2a\sin60^\circ = 346.4\,\mathrm{mm}. It does 5000 \times 0.2429 = 1214\,\mathrm J on the car, and the car's weight does -1214\,\mathrm J on the jack.

Plausible wrong answer. "Work is force times distance." Only the component along the motion counts: a force at right angles to the motion does no work at all, however large.

Check. A 100\,\mathrm N force acts at right angles to a 3\,\mathrm m motion. What work does it do? (None: \cos90^\circ = 0.)

Narration spine. A crate slides from A to A' under a slanting rope. The force splits into a green part along the floor and a grey part across it; the green part times s fills in as 866\,\mathrm J. The picture flips: the displacement is projected onto the rope instead, and the same 866\,\mathrm J appears.

Bridge. Mechanisms are also driven by couples: a torque on a crank, a motor shaft. What work does a couple do?

Nugget 02: A Couple Does Work Only by Turning

Claim. A couple M does no work while the body it acts on translates, and work M\,d\theta while the body turns through d\theta in the couple's plane.

The picture. Meriam's Fig. 7/4: the couple as two forces F and -F at points A and B, a distance b apart, with M = Fb. Move AB to its new place in two steps.

First step (left): both points move by the same d\mathbf r, a translation. F does \mathbf F\cdot d\mathbf r and -F does -\mathbf F\cdot d\mathbf r: they cancel. Second step (right): A turns about B through d\theta, moving b\,d\theta along F, while B stays put. Only F works, F(b\,d\theta) = M\,d\theta.

Check with numbers. F = 100\,\mathrm N at b = 0.4\,\mathrm m is M = 40\,\mathrm{N\cdot m}. Translate by 10\,\mathrm{mm}: 100(0.01) - 100(0.01) = 0. Turn by 0.01\,\mathrm{rad}: A moves 0.4 \times 0.01 = 4\,\mathrm{mm}, and 100 \times 0.004 = 0.4\,\mathrm J = 40 \times 0.01.

The idea. Meriam (7/2), work of a couple: the work is positive when M has the same sense as the rotation, negative when opposite, and for a finite rotation it is integrated:

with \theta in radians. Work and moment have the same dimensions, force times length, but are different things: work is a scalar from a dot product, force and distance measured along the same line; moment is a vector from a cross product, with the distance at right angles to the force. So the SI writes the joule (J) for work and keeps the newton metre (N·m) for moment.

The equation (definition: work of a couple):

Worked number. A wheel-nut wrench holds a steady 40\,\mathrm{N\cdot m} while the nut turns 60^\circ = \pi/3\,\mathrm{rad}: U = 40\pi/3 = 41.89\,\mathrm J. The wrench reads 40\,\mathrm{N\cdot m}; the work is 41.89\,\mathrm J.

Real example. A motor delivering 1\,\mathrm{kW} at 1500\,\mathrm{rpm} (157.08\,\mathrm{rad/s}) gives a torque of 1000/157.08 = 6.366\,\mathrm{N\cdot m} and does 6.366 \times 2\pi = 40.0\,\mathrm J every revolution (our example).

Plausible wrong answer. "A couple of 40\,\mathrm{N\cdot m} turning one radian does 40\,\mathrm{N\cdot m} of work." The number is right but the unit names the wrong quantity: 40\,\mathrm J.

Check. A couple acts on a body that slides without turning. What work does it do? (None.)

Narration spine. Two antiparallel forces on a rod. The rod slides sideways and two work counters tick up and down and cancel. Then B is pinned and A swings through d\theta; the arc b\,d\theta lights up along F and the counter reads M\,d\theta.

Bridge. Real work needs a real motion. A body in equilibrium is not moving at all, so to use work in statics the motion has to be imagined.

Nugget 03: A Virtual Displacement Is a Small Imagined Move the Supports Allow

Claim. A virtual displacement is an assumed, infinitesimal move away from the equilibrium position, one the supports and connections permit. The work the forces would do in it is the virtual work, computed with the forces held at their present values.

The picture. The force of a spring, F = kx, against its stretch x, with a small step \delta x from x.

The exact work over the step is the trapezoid under the line. Virtual work takes the force as constant through the step: the green rectangle, F\,\delta x. They differ by the orange sliver, a triangle of area \frac12k\,\delta x^2. Halve \delta x and the rectangle halves but the sliver quarters: as the step shrinks the sliver becomes a vanishing fraction of the rectangle. That is why the change in a force during the step can be ignored.

Check with numbers. A spring of k = 20\,\mathrm{kN/m} stretched x = 25\,\mathrm{mm} pulls with 500\,\mathrm N. For \delta x = 1\,\mathrm{mm}: F\,\delta x = 0.500\,\mathrm J, the exact work \frac12(20\,000)(0.026^2 - 0.025^2) = 0.510\,\mathrm J, sliver 0.010\,\mathrm J, 2\,\% of the rectangle. For \delta x = 0.1\,\mathrm{mm}: 0.0500 against 0.0501\,\mathrm J, sliver 0.2\,\%. Ten times smaller step, ten times smaller fraction.

The idea. Meriam (7/2) and Marghitu (6.1) define a virtual displacement \delta\mathbf r of a particle in equilibrium as any assumed, arbitrary small displacement away from its natural position that is consistent with the constraints. "Virtual" means it does not really happen: it is assumed so that possible equilibrium positions can be compared. The virtual work of a force or a couple is

d\mathbf r is an actual change and can be integrated; \delta\mathbf r is assumed and is not integrated. Both are first-order differentials and obey the same rules of differentiation. A force or couple is taken as constant during the step: any change in it adds only higher-order terms, which vanish in the limit, the same reasoning that drops dx\,dy when writing dA = y\,dx under a curve (Meriam).

The equation (definition: virtual work):

Assumptions. The step is infinitesimal (first order); it is admissible: a pinned point does not move, a link keeps its length, a roller stays on its track.

Worked number. Give the puzzle's jack a virtual opening \delta\theta = 0.001\,\mathrm{rad} at 15^\circ with a = 0.2\,\mathrm m: the top rises \delta h = 2a\cos\theta\,\delta\theta = 0.386\,\mathrm{mm}, and the load does \delta U = -W\,\delta h = -1.93\,\mathrm J.

Real example. Engineers check the mechanical advantage of a linkage by exactly this thought experiment: "if the input moves a little, how far does the output move?"

Plausible wrong answer. "A virtual displacement can be any motion, say lifting the jack off the ground." It must respect the constraints: the bottom joint is held, so lifting it is not admissible.

Check. Why may the spring's force be held at 500\,\mathrm N through a virtual \delta x? (The change adds only \frac12k\,\delta x^2, second order, which vanishes against F\,\delta x.)

Narration spine. A straight line F = kx. A thin step \delta x is shaded green as a rectangle, an orange triangle sits on top. The step shrinks and the triangle disappears faster than the rectangle; the words "forces held constant" appear.

Bridge. Now apply the imagined move to bodies in equilibrium, and ask which forces do virtual work.

Nugget 04: Only Active Forces Do Virtual Work

Claim. For any system in equilibrium the virtual work is zero for every admissible virtual displacement. In an ideal system of connected rigid bodies only the external active forces do any, so \delta U = 0 relates the loads directly, without reactions and without taking the system apart.

The picture. The jack carries Meriam's three kinds of force (his Fig. 7/7).

(a) Active forces, the load W and the screw's pull T at the side joints, move when the jack moves, so they work. (b) The ground's force R acts on the bottom joint, which cannot move: \delta\mathbf r = \mathbf 0, so it does no work. (c) At every pin the two links push on each other with equal and opposite forces at the same point: whatever that point does, the two works cancel.

Check with numbers. First a particle: three forces 300\mathbf i, -100\mathbf i + 250\mathbf j and -200\mathbf i - 250\mathbf j\,\mathrm N sum to zero, and for \delta\mathbf r = 3\mathbf i - 2\mathbf j\,\mathrm{mm}, \delta U = 0(0.003) + 0(-0.002) = 0. Were the first force 310\,\mathrm N, the same move would give \delta U = 10 \times 0.003 = 0.03\,\mathrm J \ne 0. Then the jack (the puzzle): with the bottom joint fixed, the top joint is at height h = 2a\sin\theta and the side joints are x = 2a\cos\theta apart. Differentiating, \delta h = 2a\cos\theta\,\delta\theta (up, positive) and \delta x = -2a\sin\theta\,\delta\theta (negative: the side joints close as the jack opens). W points down, so it does -W\,\delta h. Each side joint is pulled towards the centre with T, so the pair does T(-\delta x). Then

At 15^\circ, T = 5 \times 3.732 = 18.66\,\mathrm{kN}: the screw carries 3.7 times the load. At 30^\circ, 8.66\,\mathrm{kN}; at 45^\circ, 5.00\,\mathrm{kN}; at 60^\circ, 2.89\,\mathrm{kN}. The link length a cancelled.

The idea. Meriam (7/3) builds the principle in three steps. Particle: \delta U = \Sigma\mathbf F\cdot\delta\mathbf r = \Sigma F_x\,\delta x + \Sigma F_y\,\delta y + \Sigma F_z\,\delta z. In equilibrium each sum is zero, so \delta U = 0; and taking \delta x, \delta y, \delta z one at a time, \delta U = 0 for all of them forces each sum to zero, so the condition is both necessary and sufficient. Rigid body: a body is particles rigidly joined, each with zero virtual work; the internal forces come in equal, opposite, collinear pairs and do no net work. For one body this gives nothing new: for Meriam's hinged plate (Fig. 7/6) a rotation \delta\theta about the hinge gives -Pa\,\delta\theta + Rb\,\delta\theta = 0, which is just moments about the hinge. Ideal system: rigid members joined by connections that absorb no energy, with negligible friction. Its forces are active (external forces that can do virtual work), reactive (at fixed supports, where no virtual displacement occurs along them) and internal (equal and opposite pairs at the connections). Marghitu (6.1) names only two kinds, active forces and joint forces, the reactions being part of the constraints; it is the same principle.

The equation (physical law, from equilibrium: the principle of virtual work):

The virtual work done by external active forces on an ideal mechanical system in equilibrium is zero for any and all virtual displacements consistent with the constraints.

Assumptions. Rigid members; connections that store no energy; friction negligible; virtual displacements admissible.

Worked number. The hinged plate with P = 400\,\mathrm N at a = 0.30\,\mathrm m and the roller at b = 0.50\,\mathrm m: R = Pa/b = 240\,\mathrm N. Taking the jack apart instead (a second view): each link is a two-force member; at the top joint 2F\sin\theta = W, so F = 5/(2\sin15^\circ) = 9.66\,\mathrm{kN} in each link, and at a side joint T = 2F\cos\theta = 18.66\,\mathrm{kN}. The same answer, after a detour through four link forces.

Real example. Meriam's advice is to draw an active-force diagram: the system isolated with only the forces that do work, as in (a). For the jack that is two forces instead of a load, a reaction and eight pin-force components.

Plausible wrong answer. "The screw is horizontal, so it carries little of a vertical load." Near flat, a small rise of the top joint needs a large closing of the side joints, so the screw must pull hard: T = W\cot\theta.

Check. Why does the ground reaction on the jack's bottom joint never enter \delta U? (That joint cannot move, so its virtual displacement is zero.)

Narration spine. The jack under the sill. Three panels sort its forces: W and T glow as active; the ground reaction fades out under "\delta r = 0"; a pin's two forces cancel. The jack opens a hair, \delta h and \delta x appear with their signs, and T = W\cot\theta writes itself; a gauge reads 18.7\,\mathrm{kN} at 15^\circ and falls as the jack rises.

Bridge. The jack needed one angle to fix every point. How many coordinates does a mechanism need, and how are the virtual displacements of different points tied together?

Nugget 05: One Degree of Freedom: Write the Geometry, Then Differentiate

Claim. If one coordinate fixes the position of every part, one application of \delta U = 0 gives the equilibrium relation. Write each working point's position in that coordinate, differentiate, and let the signs fall out of the algebra.

The picture. Meriam's Sample Problem 7/1: two uniform bars of mass m and length l, hinged together at the bottom; the left top end is pinned, the right one runs in a horizontal guide and is pulled by P.

The angle \theta between the bars fixes everything: the ends are x = 2l\sin(\theta/2) apart and each centre of mass is h = (l/2)\cos(\theta/2) below the supports. Measure x positive to the right, with P, and h positive down, with mg: then each work is +force \times \deltacoordinate.

Check with numbers. \delta U = P\,\delta x + 2mg\,\delta h = 0, with \delta x = l\cos(\theta/2)\,\delta\theta and \delta h = -(l/4)\sin(\theta/2)\,\delta\theta. The minus sign says the centres rise as \theta opens, so the weights do negative work, as they must. Then Pl\cos(\theta/2) = (mgl/2)\sin(\theta/2) and

With m = 5\,\mathrm{kg} (mg = 49.05\,\mathrm N) and P = 20\,\mathrm N: \tan(\theta/2) = 0.8155, \theta = 78.40^\circ. Solving the same frame by dismembering, six equations for the two bars, gives the same 78.40^\circ, the guide reaction 49.05\,\mathrm N and a horizontal pin force of 20\,\mathrm N: three forces virtual work never needed.

The idea. Meriam (7/3): the number of degrees of freedom is the number of independent coordinates needed to fix the configuration. A slider-crank or a four-bar linkage has one; a double pendulum two. \delta U = 0 may be applied once per degree of freedom, letting one coordinate change at a time; the chapter treats one degree of freedom only. Relating the virtual displacements is usually the hardest part (Chapter Review): write the geometric relations, then take their differentials. Sample Problem 7/2: parallel links of length b carry a mass m, turned by a couple M; G is h = b\cos\theta + c below a fixed line, so M\,\delta\theta + mg\,\delta h = M\,\delta\theta - mgb\sin\theta\,\delta\theta = 0 and \theta = \sin^{-1}(M/mgb), possible only if M \le mgb. Sample Problem 7/3: a crank OA of length a, horizontal, is held by a couple M against a collar force P through a link AB of length b. With \delta y = a\,\delta\theta and b^2 = x^2 + y^2, 0 = 2x\,\delta x + 2y\,\delta y, so \delta x = -(y/x)a\,\delta\theta: the collar moves left when A drops. M\,\delta\theta + P\,\delta x = 0 gives P = Mx/(ya) = Mx/(ha).

The equation (mathematical identity: chain rule for virtual displacements):

Worked number. Sample Problem 7/2 with m = 20\,\mathrm{kg}, b = 0.4\,\mathrm m, M = 30\,\mathrm{N\cdot m}: mgb = 78.48\,\mathrm{N\cdot m}, \sin\theta = 0.3823, \theta = 22.47^\circ. Sample Problem 7/3 with a = 0.15, h = 0.12, b = 0.20\,\mathrm m (x = 0.16\,\mathrm m) and M = 30\,\mathrm{N\cdot m}: P = 30(0.16)/(0.12 \times 0.15) = 266.7\,\mathrm N; by statics the link carries 333.3\,\mathrm N, whose horizontal part is the same 266.7\,\mathrm N.

Real example. A lift platform or construction hoist of many crossed links has one degree of freedom; virtual work relates the cylinder force to the load in one equation, which is why Meriam opens the chapter with one.

Plausible wrong answer. "h goes down as \theta grows, so write -mg\,\delta h." With h measured positive down, mg acts along +h and its work is +mg\,\delta h; the differentiation supplies the minus sign. Choosing a sign by eye and again by algebra counts it twice.

Check. How many degrees of freedom has the scissor jack with its bottom joint fixed? (One: \theta fixes every joint.)

Narration spine. The two-bar frame of SP 7/1 opens slowly. Rulers mark x across the top and h down to each centre, each with a "+" arrow. \delta\theta ticks; \delta x grows to the right in green, \delta h shrinks upward in orange with its minus sign. The terms balance and \tan(\theta/2) = 2P/mg lands.

Bridge. Keep the geometry symbolic and the method gives more than one answer: it gives how the force changes with the design.

Nugget 06: Keep the Geometry as a Parameter: the Car Hoist

Claim. Written in symbols, the virtual-work relation shows how a force depends on the mechanism's proportions and position, and can expose a poor design.

The picture. Harper's car hoist (Meriam's Problem 7/27): two parallel links of length 2b hold the platform; a hydraulic cylinder runs from the midpoint B of the right link to a base A on the ground, a distance L from that link's pivot.

Right: the cylinder force divided by the car's weight against the link angle, for three proportions b/L. Every curve shoots up as the links lower towards the ground.

Check with numbers. For b/L = 0.5: C/mg = 5.841 at 10^\circ, 2.147 at 30^\circ, exactly 1 at 60^\circ, and 0.366 at 80^\circ. For b/L = 0.1 the 10^\circ value is 10.23; for 0.9, 2.192.

The idea. Harper (7.1): the cylinder's length a obeys a^2 = (b\sin\theta)^2 + (L - b\cos\theta)^2. Its variation is easier to take before the square root (a step Harper notes): 2a\,\delta a = 2Lb\sin\theta\,\delta\theta, so \delta a = (Lb/a)\sin\theta\,\delta\theta, positive: the cylinder lengthens as the hoist rises. The platform is at h = 2b\sin\theta + h_0, so \delta h = 2b\cos\theta\,\delta\theta. The compression C pushes B along the cylinder and does C\,\delta a; the weight does -mg\,\delta h. Then C\,\delta a - mg\,\delta h = 0 gives C = 2mga\cot\theta/L, or with \eta = b/L,

which is Meriam's answer to P 7/27. As \theta \to 0, \cot\theta \to \infty: the cylinder lies almost flat, its vertical component vanishes, and holding the car needs an unbounded push. Harper's second plot, C/mg against \eta at fixed angles, shows that retuning b/L does not cure it. (Ours: at a given \theta the best \eta is \cos\theta, giving C/mg = 2\cos\theta, but a hoist has one fixed \eta for every angle; \eta = 1 would need a cylinder that shrinks to zero length at the bottom.) Harper's remedy is to recess the cylinder's base below the floor so the cylinder never lies flat.

The equation (geometric relationship plus the principle; for this hoist):

Worked number. A 1500\,\mathrm{kg} car (mg = 14.72\,\mathrm{kN}) on a hoist with b/L = 0.5 at \theta = 20^\circ: 2\cot20^\circ = 5.495, \sqrt{1.25 - \cos20^\circ} = 0.5571, C/mg = 3.061, C = 45.0\,\mathrm{kN}. Recessing the base by d = 0.5L (ours, same method with a^2 = (b\sin\theta + d)^2 + (L - b\cos\theta)^2) gives C/mg = 2.83 even at \theta = 0, and 2.29 at 10^\circ instead of 5.84.

Real example. Hydraulic platforms, tipper bodies and excavator booms are sized this way: the cylinder force is plotted over the whole stroke, and the worst position, usually the lowest, sets the cylinder bore and pressure.

Plausible wrong answer. "The cylinder works hardest at the top, where the car is highest." Height costs nothing in force; geometry does. The force is largest at the bottom.

Check. At what angle does the b/L = 0.5 hoist's cylinder carry exactly the car's weight? (60^\circ: 2\cot60^\circ\sqrt{1.25 - 0.5} = 1.1547 \times 0.8660 = 1.)

Narration spine. The hoist rises and lowers with the car on it; beside it a point traces C/mg against \theta, climbing steeply as the platform comes down. Two more curves for other b/L climb too. The cylinder base sinks into a pit and the curve flattens.

Bridge. Every result so far assumed frictionless joints. Real machines lose some of their input to friction.

Nugget 07: Friction Makes a System Real: Mechanical Efficiency

Claim. Kinetic friction always does negative work, which is lost as heat. A machine's output work is therefore less than its input, and the ratio, the mechanical efficiency, can be found from one virtual displacement.

The picture. Meriam's Fig. 7/10: a block of mass m pulled up a rough slope at \theta by a force T.

During a virtual move \delta s up the slope, T puts in T\,\delta s. The useful output is the gain in potential of the block, mg\,\delta s\sin\theta. The rest goes to the kinetic friction force \mu_kmg\cos\theta, which points against the motion. (Kinetic friction, treated in full in Part 6: while surfaces slide, the friction force is \mu_kN, with N the normal force and \mu_k the coefficient of kinetic friction, and it opposes the sliding.)

Check with numbers. m = 50\,\mathrm{kg}, \theta = 20^\circ, \mu_k = 0.30. Per metre up the slope: output 490.5\sin20^\circ = 167.8\,\mathrm J; input T = 490.5(\sin20^\circ + 0.30\cos20^\circ) = 306.0\,\mathrm N, so 306.0\,\mathrm J. e = 167.8/306.0 = 0.548.

The idea. Meriam (7/3), systems with friction: a system with appreciable sliding friction is "real". Kinetic friction does -\mu_kN\,\delta x on a sliding block (Fig. 7/9a); the static friction on a wheel that rolls without slipping does no work, because its contact point does not slide (Fig. 7/9b); a friction moment M_f in a pin does -M_f(\delta\theta_1 + \delta\theta_2) = -M_f\,\delta\theta for a relative rotation \delta\theta (Fig. 7/9c). With internal friction the system must be taken apart to find the friction forces, so virtual work loses its main advantage. Mechanical efficiency:

For a square-threaded screw jack of mean radius r and helix angle \alpha, Meriam quotes the moment to raise a load W from his friction chapter (Art. 6/5), M = Wr\tan(\alpha + \phi), where \phi = \tan^{-1}\mu_k is the friction angle. In a small turn \delta\theta the input is M\,\delta\theta = Wr\,\delta\theta\tan(\alpha + \phi) and the output, lifting W by r\,\delta\theta\tan\alpha, is Wr\,\delta\theta\tan\alpha:

As friction falls, \phi \to 0 and e \to 1. The frictionless part is pure virtual work: one turn of the screw puts in 2\pi M and lifts W by the lead L = 2\pi r\tan\alpha, so M = Wr\tan\alpha.

The equation (definition: mechanical efficiency): e = U_{\text{out}}/U_{\text{in}} < 1.

Assumptions. Coulomb friction with constant \mu_k; uniform, slow operation of a one-degree-of-freedom machine.

Worked number. The puzzle's jack screw: mean radius 6\,\mathrm{mm}, lead 2\,\mathrm{mm}, \mu_k = 0.15 (our values). \tan\alpha = 2/(2\pi \times 6) = 0.05305, \alpha = 3.037^\circ; \phi = 8.531^\circ; e = 0.05305/\tan11.568^\circ = 0.259. At 15^\circ the screw pulls 18.66\,\mathrm{kN}: the ideal torque is 18\,660 \times 0.002/2\pi = 5.94\,\mathrm{N\cdot m}, the real one 5.94/0.259 = 22.9\,\mathrm{N\cdot m}, a push of 91.7\,\mathrm N on a 0.25\,\mathrm m handle. At 60^\circ it is 3.55\,\mathrm{N\cdot m}: the first turns are \cot15^\circ/\cot60^\circ = 6.46 times harder, which is what the hand feels.

Real example. Meriam's Problem 7/12: a 40 : 1 speed reducer takes 30\,\mathrm{N\cdot m} in and gives 1180\,\mathrm{N\cdot m} out. In one input radian the output turns 1/40\,\mathrm{rad}: e = 1180/(40 \times 30) = 0.983.

Plausible wrong answer. "Friction is a force, so put it in \delta U like any other and the method works unchanged." Kinetic friction inside the system must first be found, which means dismembering; and static friction at a rolling contact does no work at all.

Check. Why is a low-efficiency screw jack still useful? (Its friction also holds the load: with \alpha = 3.0^\circ less than \phi = 8.5^\circ, the car cannot run the jack back down when the handle is released.)

Narration spine. A block climbs a ramp; an input bar T\,\delta s splits into a green output part mg\,\delta s\sin\theta and an orange friction part. The slope steepens and the green share grows. Cut to the jack's screw: an ideal torque arrow, then the real one four times longer.

Synthesis

The jack's screw carries 18.7\,\mathrm{kN} under a 5\,\mathrm{kN} load when the jack is nearly flat, and the answer came from one line: imagine the jack opening a hair, and add the work of the only two forces that move. The ground reaction, four link forces and four pin forces never appeared, because a fixed point does no work and equal and opposite pin forces cancel. For an ideal system in equilibrium the active forces do zero virtual work in every admissible virtual displacement; with one degree of freedom that is one equation relating the active forces, found by writing the geometry and differentiating it.

• Central idea: \delta U = 0 replaces dismembering when a system can move; reactions and pin forces drop out.
• Key relationship: \delta U = \Sigma\mathbf F\cdot\delta\mathbf r + \Sigma M\,\delta\theta = 0; for the jack T = W\cot\theta, for the hoist C/mg = 2\cot\theta\sqrt{1 + \eta^2 - 2\eta\cos\theta}.
• Most important assumption: an ideal system: rigid members and negligible friction; otherwise friction work must be included and efficiency e < 1 follows.
• Open question: springs and weights do work that can be stored and given back. Writing it as a change in potential energy leads to a test of whether an equilibrium is stable (S31).

Computed views

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

Python (numpy, scipy, sympy). S30_virtual_work.py lets sympy take the virtual displacements. Write the position of each working force in terms of \theta, differentiate, and solve \delta U = 0. That gives the jack's T = W\cot\theta with no pin forces in sight. The view then tabulates Harper's hoist and puts friction back through the efficiency.

Part 1: work
  ring: U1 = 60 J, U2 = 40 J, resultant 100 J
  spring 2 kN/m, 20 to 50 mm: U = -2.10 J
  40 N m through 60 deg: U = 41.89 J
Part 2: virtual work with sympy
  scissor jack: T = W/tan(theta), at 15 deg 18.66 kN
  jack 15 to 60 deg lifts 242.9 mm: 1214 J
  SP 7/2: sin theta = M / (mgb) = 0.3823, theta = 22.47 deg
  SP 7/3: P = M x / (h a) = 266.7 N
Part 3: Harper's hoist, C / mg = 2 cot(theta) sqrt(1 + eta^2 - 2 eta cos)
  theta   eta 0.5   eta 1.0
     10     5.841     1.977
     20     3.061     1.908
     30     2.147     1.793
     60     1.000     1.155
  1500 kg car at 20 deg: C = 45.0 kN
Part 4: efficiency
  jack screw: e = 0.259; at 15 deg ideal 5.94 N m, real 22.9 N m, push 91.7 N
  at 60 deg it needs 3.55 N m, 6.46 times less
  gearbox 40:1, 30 N m in, 1180 N m out: e = 0.983

MATLAB (Octave-compatible). S30_virtual_work.m takes the jack's virtual displacement numerically, as a small finite step in \theta. It needs no symbolic toolbox and gets the same T.

spring 2 kN/m, 20 to 50 mm: U = -2.10 J
jack at 15 deg: T = 18.66 kN (W cot = 18.66 kN)
eta 0.5: C/mg =    5.841   3.061   2.147   1.000
eta 1.0: C/mg =    1.977   1.908   1.793   1.155
screw: e = 0.259, real torque 22.9 N m, push 91.7 N

Traps

• Multiplying the whole force by the distance. Only the component along the displacement works: U = Fs\cos\alpha, and a perpendicular force does none. (From Nugget 01.)
• Writing work in N·m. Work is in joules; N·m is kept for moment. (From Nugget 02.)
• Choosing a displacement the supports forbid. A virtual displacement must be admissible; moving a pinned point brings its reaction into \delta U. (From Nugget 03.)
• Putting reactions or pin forces into \delta U. They do no work in an admissible move; only active forces belong on the active-force diagram. (From Nugget 04.)
• Expecting virtual work to give a reaction directly. It eliminates reactions; to find one, free that support and treat its reaction as active, or use the equilibrium equations. (From Nugget 04.)
• Fixing a sign twice. Measure each coordinate in a stated positive sense, write the work as force times its \delta, and let differentiation give the sign. (From Nugget 05.)
• Differentiating only the obvious term. In b^2 = x^2 + y^2 with b constant, \delta b = 0 links \delta x to \delta y with a minus sign. (From Nugget 05.)
• Judging a mechanism by one position. The hoist's cylinder force is moderate at 60^\circ and unbounded near 0^\circ. (From Nugget 06.)
• Applying \delta U = 0 with appreciable friction left out. Kinetic friction does negative work that must be included, or the result is the ideal one. (From Nugget 07.)

Self-check

1. A rope pulls a sledge 8\,\mathrm m with 250\,\mathrm N at 25^\circ above the direction of motion, against a 60\,\mathrm N drag. Find the work of each force.
2. A constant force \mathbf F = 3\mathbf i - 2\mathbf j\,\mathrm{kN} acts while its point moves \mathbf s = 0.4\mathbf i + 0.5\mathbf j\,\mathrm m. Find the work.
3. A valve handwheel is turned two and a half turns against a steady 12\,\mathrm{N\cdot m}. What work is done, and in what unit?
4. A spring of k = 4\,\mathrm{kN/m} is stretched 30\,\mathrm{mm}. Compare its virtual work F\,\delta x for \delta x = 2\,\mathrm{mm} with the exact work, and name the difference.
5. Meriam's hinged plate (Fig. 7/6) carries P = 600\,\mathrm N at a = 0.25\,\mathrm m from the hinge; the roller is at b = 0.6\,\mathrm m. Find R by virtual work.
6. How many degrees of freedom have (a) the scissor jack, (b) the car hoist, (c) a double pendulum? How many times can \delta U = 0 be applied to each?
7. Meriam's Problem 7/1: a block on a vertical guide is pushed down by P; two links of length r join it to a shaft directly below, each link at \theta to the vertical, and a moment M on the lower link supports the load. Find M, and evaluate it for P = 2\,\mathrm{kN}, r = 0.1\,\mathrm m, \theta = 30^\circ.
8. Meriam's Problem 7/2: an A-frame of two uniform links (mass m, length b) stands with its right foot pinned and its left foot on rollers, both links at \theta to the horizontal. A horizontal force P acts to the right at the middle of the left link. Find \theta, and evaluate for m = 10\,\mathrm{kg}, P = 50\,\mathrm N.
9. Meriam's Sample Problem 7/3 with M = 40\,\mathrm{N\cdot m}, a = 0.2\,\mathrm m, h = 0.15\,\mathrm m and b = 0.3\,\mathrm m: find P.
10. Harper's hoist with b/L = 0.5 carries a 1500\,\mathrm{kg} car at \theta = 20^\circ. Find C.
11. Meriam's Problem 7/12: a 40 : 1 reducer takes in 30\,\mathrm{N\cdot m} and delivers 1180\,\mathrm{N\cdot m}. Find e.
12. Meriam's Problem 7/18 (US units in the book): 12 turns of a jack's handle raise its pad 1\,\mathrm{in}; a force of 10\,\mathrm{lb} on a 6\,\mathrm{in} crank raises 2700\,\mathrm{lb}. Convert to SI and find the efficiency.

Fully worked solutions

1. Rope: U = 250 \times 8 \times \cos25^\circ = 1812.6\,\mathrm J. Drag: U = -60 \times 8 = -480\,\mathrm J.

2. U = \mathbf F\cdot\mathbf s = 3(0.4) + (-2)(0.5) = 1.2 - 1.0 = 0.2\,\mathrm{kJ} = 200\,\mathrm J.

3. \theta = 2.5 \times 2\pi = 15.71\,\mathrm{rad}; U = 12 \times 15.71 = 188.5\,\mathrm J, in joules, not N·m.

4. F = 4000 \times 0.030 = 120\,\mathrm N; F\,\delta x = 120 \times 0.002 = 0.240\,\mathrm J. Exact: \frac12(4000)(0.032^2 - 0.030^2) = 2000 \times 0.000124 = 0.248\,\mathrm J. The difference 0.008\,\mathrm J = \frac12k\,\delta x^2 is second order and vanishes as \delta x \to 0.

5. Turn the plate \delta\theta about the hinge: P does -Pa\,\delta\theta, R does +Rb\,\delta\theta; the hinge force does none. -600(0.25) + R(0.6) = 0, R = 250\,\mathrm N.

6. (a) One (\theta). (b) One (\theta). (c) Two (one angle per link). \delta U = 0 applies once per degree of freedom, letting one coordinate change at a time: once, once, twice.

7. The block is y = 2r\cos\theta above the shaft, \delta y = -2r\sin\theta\,\delta\theta. P (down) does -P\,\delta y = 2Pr\sin\theta\,\delta\theta; the moment, opposing the opening, does -M\,\delta\theta. \delta U = 0: M = 2Pr\sin\theta = 2(2000)(0.1)(0.5) = 200\,\mathrm{N\cdot m}.

8. Put the pinned foot at the origin. The left link's middle is at x = -1.5b\cos\theta, y = 0.5b\sin\theta; both centres are at y = 0.5b\sin\theta. \delta U = P\,\delta x - 2mg\,\delta y = P(1.5b\sin\theta)\,\delta\theta - mgb\cos\theta\,\delta\theta = 0, so \tan\theta = 2mg/(3P) = 2(98.1)/150 = 1.308 and \theta = 52.6^\circ (Meriam: \theta = \tan^{-1}(2mg/3P)).

9. x = \sqrt{0.3^2 - 0.15^2} = 0.2598\,\mathrm m; P = Mx/(ha) = 40(0.2598)/(0.15 \times 0.2) = 346.4\,\mathrm N.

10. C/mg = 2\cot20^\circ\sqrt{1 + 0.25 - \cos20^\circ} = 5.495 \times 0.5571 = 3.061; C = 3.061 \times 1500 \times 9.81 = 45.0\,\mathrm{kN}.

11. Output per input radian 1180/40 = 29.5\,\mathrm J; input 30\,\mathrm J; e = 29.5/30 = 0.983.

12. 10\,\mathrm{lb} = 44.48\,\mathrm N, 6\,\mathrm{in} = 0.1524\,\mathrm m, 2700\,\mathrm{lb} = 12.01\,\mathrm{kN}, 1\,\mathrm{in} = 0.0254\,\mathrm m. Input over 12 turns: 12 \times 2\pi \times 0.1524 \times 44.48 = 511.1\,\mathrm J. Output: 12\,010 \times 0.0254 = 305.1\,\mathrm J. e = 305.1/511.1 = 0.597, as Meriam.