Study Companion: S27 — Wedges and Screws

S26 treated friction as something that resists: it holds a ladder, it limits a push. This lesson turns it into a working part. A wedge and a screw are the two oldest machines for making a small effort move a large load through a small distance, and both work only because their surfaces slide against friction. The same friction that makes them hard to drive also makes them hold their load when the effort is removed, and one comparison of angles, the wedge or helix angle against the friction angle, decides which.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 6, the Section B introduction and articles 6/4 ("Wedges") and 6/5 ("Screws"), with Sample Problems 6/6–6/7; Marghitu, Dupac and Madsen, Statics with MATLAB, section 5.5 ("Technical Applications of Friction: Screws", with Example 5.3); Harper, Solving Statics Problems in MATLAB, "Friction on Wedges" (its Sample Problem 6/6, the same in Meriam's 9th edition) and "Friction on Screws" (its Problem 6/58, which is Meriam 9th edition Problem 6/52). Units SI throughout, g = 9.81\,\mathrm{m/s^2}. As in S26 and S30, \mu is the coefficient of friction and \phi = \tan^{-1}\mu the friction angle; from here on (Meriam's Section B) the subscript is dropped, and \mu is the static value when motion impends and the kinetic value while sliding. For screws, W is the axial load, L the lead, r the mean radius and \alpha the helix angle. Marghitu writes F, l, r_m, \lambda and \theta for these; they are translated.

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 C-clamp (Meriam's Problem 6/53) squeezes two boards with 600\,\mathrm N. Its screw has a mean diameter of 10\,\mathrm{mm} and advances 2.5\,\mathrm{mm} per turn; the coefficient of friction in the thread is 0.20, and the swivel pad at A turns freely. The handle is turned by a force F at 100\,\mathrm{mm} from the axis.

Tightening takes F = 8.52\,\mathrm N. Now let go. The boards push back on the screw with 600\,\mathrm N, along its axis. Do they drive the screw back out? The plausible guess is "yes, unless the screw is locked: 600\,\mathrm N pressing on a ramp has to slide down it", or, from the other side, "friction is the same both ways, so loosening will cost the same 8.52\,\mathrm N". Every clamp in a workshop says otherwise: it stays tight, and it loosens with less effort than it took to tighten, but not with none. The answer, 3.56\,\mathrm N to loosen, is in Nugget 05.

Nugget 01: A Wedge Lifts Through Two Sliding Faces, and Friction Sets the Force

Claim. A wedge driven under a load slides on two faces at once. On each face the reaction leans away from the normal by the friction angle \phi, against the slip, and the force to drive the wedge follows from two force triangles. For a thin wedge, friction, not the wedge angle, is most of that force.

The picture. Meriam's Fig. 6/3: a mass m, guided by a smooth vertical wall, sits on a wedge of angle \alpha resting on the floor; P drives the wedge in. The reactions R_2 (wedge on mass) and R_1 (floor on wedge) each lean by \phi; the wall gives only the horizontal R_3.

The mass rises and the wedge moves in, so on the upper face the mass slides up the slope relative to the wedge: friction on the mass points down the slope, and R_2 leans from the face normal, which is itself tilted \alpha from the vertical, by a further \phi: R_2 is at \alpha + \phi from the vertical. On the floor the wedge slides in the direction of P, so R_1 leans back from the vertical by \phi. Each body then has three forces, and each gives a triangle (Fig. 6/3c): the upper one, mg, R_3 and R_2, fixes R_2; the lower one, R_2, R_1 and P, fixes P.

Check with numbers. m = 2000\,\mathrm{kg} (mg = 19.62\,\mathrm{kN}), \alpha = 5^\circ, \mu = 0.25 on both faces (\phi = 14.04^\circ). Mass: the vertical component of R_2 carries mg, so its horizontal component is mg\tan(\alpha + \phi) = 19.62\tan19.04^\circ = 6.77\,\mathrm{kN}, which is R_3. Wedge: R_1's vertical part is again mg, and its horizontal part, mg\tan\phi = 4.91\,\mathrm{kN}, opposes P. So P = 6.77 + 4.91 = 11.67\,\mathrm{kN}. Without friction it would be mg\tan5^\circ = 1.72\,\mathrm{kN}: friction multiplies the effort by 6.8.

The idea. Meriam (6/4): when slip of a wedge impends, the resultant on each sliding face is inclined from the normal by the friction angle, in the direction that opposes the motion of the wedge relative to the mating surface. With the reactions' directions known, the unknowns are only magnitudes, and the polygons close. Neglecting the wedge's weight:

Step the source leaves to the figure: in the upper triangle R_2\cos(\alpha + \phi) = mg; in the lower one R_1\cos\phi = R_2\cos(\alpha + \phi) = mg, and P = R_2\sin(\alpha + \phi) + R_1\sin\phi.

The equation (physical law: equilibrium, with the friction limit on both faces): above, for the arrangement of Fig. 6/3.

Assumptions. Same \mu on both wedge faces; smooth guide; wedge weight neglected; slip impending on both faces at once.

Worked number. The same wedge driven by components instead: on the mass, the face's normal force N_2 = mg/(\cos\alpha - \mu\sin\alpha) = 20.14\,\mathrm{kN} and friction 0.25N_2; their horizontal parts on the wedge, N_2(\sin\alpha + \mu\cos\alpha) = 6.77\,\mathrm{kN}, plus the floor friction 0.25 \times 19.62 = 4.91\,\mathrm{kN}, give the same 11.67\,\mathrm{kN}.

Real example. Shims and folding wedges level machine bases and lift heavy columns a few millimetres; Meriam's Problem 6/54 (self-check 4) is two 5^\circ wedges raising a 5\,\mathrm{kN} column.

Plausible wrong answer. "A 5^\circ wedge multiplies force by \cot5^\circ = 11.4." Only without friction. With \mu = 0.25 it multiplies by 19.62/11.67 = 1.7.

Check. On which side of the normal does R_1 lean if the wedge is pulled out instead? (The other side: friction now opposes the outward slip.)

Narration spine. A crate on a wedge, a hammer arrow P. Two reaction arrows swing off their normals by the same small angle; the crate's three forces snap into a triangle, then the wedge's three; the P side of the second triangle is measured, then shrinks to its frictionless length for comparison.

Bridge. The polygon method works for one set of numbers. To see how the force depends on the angle and on each coefficient, keep them as symbols.

Nugget 02: The Wedge Force Grows Fastest With the Friction on the Wedge Faces

Claim. Written as a formula, the force to drive a wedge shows which friction matters: the coefficient on the wedge's own faces enters twice and can make the force grow without limit, while the wedge angle matters less.

The picture. Meriam's SP 6/6: a 500\,\mathrm{kg} concrete block is pushed sideways by a 5^\circ wedge driven down between it and a fixed abutment. \mu_1 = 0.30 on both wedge faces (\phi_1 = 16.70^\circ), \mu_2 = 0.60 between block and floor (\phi_2 = 30.96^\circ).

Check with numbers. Meriam solves along axes that skip unknowns. For the block, resolve along a–a, perpendicular to the floor reaction R_3: the weight's component 4905\sin30.96^\circ balances R_2's, whose angle to a–a is \phi_1 + \phi_2 = 47.66^\circ, so R_2 = 4905(0.5145)/0.6735 = 3747\,\mathrm N. For the wedge, resolve along b–b, perpendicular to R_1: R_2 makes 90^\circ - (2\phi_1 + 5^\circ) = 51.6^\circ with it and P makes \phi_1 + 5^\circ = 21.7^\circ, so P = 3747\cos51.6^\circ/\cos21.7^\circ = 2505\,\mathrm N (Meriam: 2500\,\mathrm N). A scaled drawing of the two polygons gives the same within drawing accuracy, which Meriam notes is enough given the uncertainty in \mu.

The idea. Harper writes the four component equations, block \Sigma F_x, \Sigma F_y and wedge \Sigma F_x, \Sigma F_y, with the wedge angle \theta and both coefficients as symbols, and eliminates N_1, N_2, N_3:

At \theta = 5^\circ, \mu_1 = 0.30, \mu_2 = 0.60 it gives 2505\,\mathrm N again. Read the formula. With frictionless wedge faces, \mu_1 = 0, it is \mu_2mg\tan\theta = 257.5\,\mathrm N: ten times less. The angle \theta appears only through \tan\theta, a small number for a thin wedge, while \mu_1 sits in the bracket and in both denominators. If \mu_1\mu_2 \to 1 or \mu_1\tan\theta \to 1 the force grows without bound: the wedge jams. Harper's plots for \mu_2 = 0.6 show P curving steeply up with \mu_1 for \theta = 5^\circ, 15^\circ, 25^\circ, to about 16\,\mathrm{kN} at \mu_1 = 0.8, \theta = 25^\circ.

The equation (physical law: equilibrium of block and wedge at impending slip): above.

Worked number. At \mu_1 = 0.30 the 15^\circ and 25^\circ wedges need 3.29 and 4.27\,\mathrm{kN}: going from 5^\circ to 25^\circ adds 1.8\,\mathrm{kN}. Greasing the wedge from \mu_1 = 0.30 to 0 at 5^\circ saves 2.25\,\mathrm{kN}.

Real example. Splitting wedges and the tapered cotters that join shafts (Meriam's Problem 6/62) are greased or polished for exactly this reason.

Plausible wrong answer. "A steeper wedge always needs much more force." It needs more, but much less extra than a rougher wedge face does.

Check. What does the formula give when \mu_2 = 0? (Zero: with nothing resisting the block's slide there is nothing to push against.)

Narration spine. The block and wedge, then three curves of P against \mu_1 rising and bending up, labelled 5^\circ, 15^\circ, 25^\circ; a dashed line marks the frictionless value at the bottom.

Bridge. Friction made the wedge hard to drive. It also decides whether the wedge stays in when the hammer stops.

Nugget 03: A Wedge Holds Itself if Its Angle Is Less Than Twice the Friction Angle

Claim. With the driving force removed, a wedge is held by only two reactions, so they must be equal, opposite and collinear. Each must lie inside the friction cone of its own face, and that is possible only if the wedge angle is less than 2\phi. Such a wedge is self-locking, and must be pulled out.

The picture. Meriam's Fig. 6/4b: the wedge with P removed. The normals of its two faces differ by the wedge angle \alpha; each reaction may lean up to \phi from its normal. A single line inside both ranges exists only if the two ranges overlap.

Check with numbers. For Nugget 01's wedge, \alpha = 5^\circ and \phi = 14.04^\circ: the two cones overlap over 2\phi - \alpha = 23.1^\circ, so any line in that band can carry R_1 = R_2 and the wedge stays. To pull it out (Fig. 6/5) both reactions swing to the other side of their normals, R_2 now at \phi - \alpha from the vertical and R_1 at \phi the other way, and P' = mg[\tan(\phi - \alpha) + \tan\phi] = 19.62(0.1591 + 0.2500) = 8.03\,\mathrm{kN}, a pull, not a push.

The idea. Meriam (6/4): if P is removed and the wedge stays, R_1 and R_2 are collinear. Slip impends at the upper face when the common line sits on that face's cone edge (Fig. 6/4a), at the lower face when it sits on the other (Fig. 6/4c). To slide out, the wedge must slip at both faces at once, which needs the line on both cone edges together. That is impossible if \alpha < 2\phi, and in between there is a finite range of positions for the line, every one of them holding: the wedge is self-locking. Meriam leaves the case \phi < \alpha < 2\phi as an exercise; the overlap argument covers it, since the cones of half-angle \phi about normals \alpha apart overlap whenever \alpha < 2\phi. The pull to remove it reverses every friction direction (Fig. 6/5).

The equation (geometric relationship: overlap of the friction cones): self-locking if \alpha < 2\phi.

Assumptions. Same \mu on both faces; weight of the wedge neglected; no other force on it.

Worked number. A plastic door wedge of 15^\circ on a tiled floor, \mu = 0.30: 2\phi = 33.4^\circ > 15^\circ, so it stays when kicked in. A 40^\circ wedge of the same plastic would spring out.

Real example. A splitting wedge stays in a log between hammer blows (Meriam's Problem 6/45 asks for the largest angle that does not pop out; self-check 1).

Plausible wrong answer. "A wedge stays put if its angle is less than the friction angle, \alpha < \phi." That is the condition with friction on one face only; with both faces rough the limit is 2\phi.

Check. A wedge with \mu = 0.10 on both faces: what is the largest self-locking angle? (2\tan^{-1}0.10 = 11.4^\circ.)

Narration spine. The wedge alone, two translucent cones on its faces. As the wedge angle grows the cones tilt apart; their overlap, shaded, narrows to a sliver and vanishes at 2\phi, when the wedge shoots out.

Bridge. Wrap a long, thin wedge around a cylinder and it becomes a screw thread, a wedge that can be driven indefinitely by turning.

Nugget 04: A Screw Is an Inclined Plane Wrapped Around a Cylinder

Claim. Unwrap one turn of a square thread and it is a ramp of slope \tan\alpha = L/(2\pi r). Turning the screw with a moment M pushes the load up that ramp with a force M/r, and the friction angle adds to the ramp angle: M = Wr\tan(\alpha + \phi).

The picture. Meriam's Figs. 6/6 and 6/7a: a square-threaded jack under an axial load W and a moment M; one turn of thread unrolled into a ramp of base 2\pi r and rise L, with the load pushed up it by P = M/r and the reaction R leaning \phi from the ramp's normal.

Check with numbers. The puzzle's clamp: r = 5\,\mathrm{mm}, L = 2.5\,\mathrm{mm}, so \tan\alpha = 2.5/(2\pi \times 5) = 0.0796, \alpha = 4.55^\circ, and \phi = \tan^{-1}0.20 = 11.31^\circ. To tighten, M = 600(0.005)\tan15.86^\circ = 0.852\,\mathrm{N{\cdot}m}, F = 0.852/0.1 = 8.52\,\mathrm N at the handle. Without friction, M = WL/2\pi = 0.239\,\mathrm{N{\cdot}m}: friction takes 72\,\% of the effort.

The idea. Meriam (6/5) takes a small piece of the thread. As the screw turns, its thread slides on the frame's thread, so the frame's reaction R on the piece leans from the thread's normal by \phi; the normal is itself tilted from the axis by the helix angle \alpha, so R is at \alpha + \phi from the axis. Every piece is alike, at the same radius r:

with \alpha = \tan^{-1}(L/2\pi r). The unwrapped thread gives the same from one force triangle: a block W on a ramp at \alpha, pushed horizontally by P = M/r, with R at \phi from the normal, needs P = W\tan(\alpha + \phi). The lead L is the advance per turn; the helix angle uses the lead, and the mean radius r, not the outer one.

Second view (Marghitu 5.5). Marghitu first sets out the vocabulary of a thread (Figs. 5.10–5.11): the pitch p is the axial distance between neighbouring threads; the major diameter d is the outside, the minor diameter d_r the root; the pitch (mean) diameter d_m, between them, is where thread and groove are equally wide (for a square thread d_r = d - p, d_m = d - p/2). With n threads cut side by side (a multiple-threaded screw) the lead is l = np. A thread is right-hand if it recedes when turned clockwise, and threads are right-hand unless stated. A metric thread is written M14 × 2: 14\,\mathrm{mm} major diameter, 2\,\mathrm{mm} pitch (the Unified system writes, for example, 5/8″-18 UNF: diameter in inches, threads per inch). Marghitu then puts the separate friction force F_f = \mu N on the unrolled thread (Fig. 5.14a) and writes \Sigma F_x = P_r - N\sin\alpha - \mu N\cos\alpha = 0, \Sigma F_y = W + \mu N\sin\alpha - N\cos\alpha = 0; eliminating N,

Step that reconciles the two: put \mu = \tan\phi and multiply top and bottom by \cos\phi; the fraction becomes \sin(\alpha + \phi)/\cos(\alpha + \phi), so P_r = W\tan(\alpha + \phi), and with d_m/2 = r Marghitu's Eq. 5.11 is Meriam's 6/3. His 5.5.2 says so directly, M_r = Wr_m\tan(\lambda + \theta).

The equation (physical law: equilibrium of the screw with the thread reaction at the friction angle): M = Wr\tan(\alpha + \phi), \alpha = \tan^{-1}(L/2\pi r).

Assumptions. Square thread (thread load parallel to the axis); load shared evenly round the mean radius; uniform \mu; slow, steady turning.

Worked number. An M14 × 2 thread has d_m = 14 - 1 = 13\,\mathrm{mm}; single-start, \tan\alpha = 2/(13\pi), \alpha = 2.80^\circ; double-start, L = 4\,\mathrm{mm}, \alpha = 5.59^\circ. Doubling the starts doubles the advance per turn, and steepens the ramp.

Real example. The scissor jack of S30 is driven by such a screw; S30 used Eq. 6/3 to find its efficiency.

Plausible wrong answer. "Use the pitch for the helix angle." For a multiple-start thread the advance per turn is the lead, L = np.

Check. Why the mean radius, not the outer radius? (The thread contact, and so R, sits on average at the mean radius; the outer edge carries little load.)

Narration spine. A screw turns; one turn of its thread peels off and unrolls into a ramp. A block rides up it pushed by a horizontal arrow M/r; the reaction arrow leans by \phi, and the force triangle closes on \tan(\alpha + \phi).

Bridge. Turning one way pushes the load up the ramp. What happens when the turning stops, or reverses?

Nugget 05: A Screw Holds Its Load by Itself When the Helix Angle Is Below the Friction Angle

Claim. Take away the moment and the load tries to slide back down the ramp. If \alpha < \phi, friction holds it: the screw is self-locking, and lowering needs a moment Wr\tan(\phi - \alpha) the other way. If \alpha > \phi, the load unwinds the screw, and a moment Wr\tan(\alpha - \phi) is needed to hold it.

The picture. Meriam's Fig. 6/7b and c: the unwrapped thread being lowered. The friction now acts up the ramp, so R leans from the normal on the other side, by \phi.

Check with numbers. The puzzle's clamp: \alpha = 4.55^\circ < \phi = 11.31^\circ, so it is self-locking. To loosen it, M = 600(0.005)\tan(11.31^\circ - 4.55^\circ) = 3\tan6.76^\circ = 0.356\,\mathrm{N{\cdot}m}, F = 3.56\,\mathrm N at the handle: less than the 8.52\,\mathrm N to tighten, because the load now helps, but not zero, because friction exceeds that help.

The idea. Meriam (6/5, conditions for unwinding): when M is removed, the friction reverses and R swings to the other side of the normal. The screw stays put as long as \alpha < \phi; it is on the verge of unwinding when \alpha = \phi. Lowering a self-locking screw needs a reversed moment

If \alpha > \phi the screw unwinds by itself, and the moment to stop it is

Marghitu reaches the same from his Eq. 5.12, M_l = \frac{Wd_m}{2}\,\frac{\pi\mu d_m - l}{\pi d_m + \mu l}: it is positive (self-locking) when \pi\mu d_m > l, that is \mu > \tan\alpha (5.13), which is \phi > \alpha.

Second view (Harper, the turnbuckle; Meriam 9e Problem 6/52). A turnbuckle carries a cable tension T = 10\,000\,\mathrm{lb} = 44.48\,\mathrm{kN} through two single-start square-threaded screws, one right-hand and one left-hand, of mean diameter 1.150\,\mathrm{in} (r = 14.605\,\mathrm{mm}) with five threads per inch (L = 5.08\,\mathrm{mm}). Turning the body works against both threads at once:

With \mu = 0.25, M_T = 402.3\,\mathrm{N{\cdot}m} and M_L = 249.5\,\mathrm{N{\cdot}m} (Meriam: 3560 and 2210\,\mathrm{lb{\cdot}in.}). Harper plots both against \mu: M_L is negative for \mu < \tan\alpha = 0.0554, the range in which the cable tension would unscrew the turnbuckle with no moment applied. (Harper prints this limit as 0.00553; (1/5)/(2\pi \times 0.575) = 0.0554, as its own plot shows.)

The equation (physical law with the friction limit): self-locking if \alpha < \phi; lower with Wr\tan(\phi - \alpha); hold an unlocking screw with Wr\tan(\alpha - \phi).

Assumptions. As Nugget 04; static friction coefficient for impending motion.

Worked number. A fast-lead screw, 20\,\mathrm{mm} per turn on a 10\,\mathrm{mm} mean diameter (\alpha = 32.48^\circ, \mu = 0.20): an axial load of 600\,\mathrm N spins it, and holding it still takes 3\tan(32.48^\circ - 11.31^\circ) = 1.16\,\mathrm{N{\cdot}m}. Pushing it axially turns it, which is how a push drill or a spinning top's plunger works.

Real example. Every jack, clamp and vice screw is cut with \alpha well below \phi, so the load stays up when the handle is let go. Rigging turnbuckles still carry lock nuts, because vibration lets threads creep even when static friction would hold.

Plausible wrong answer. "The load pushes along the screw, so it will drive it back unless the screw is locked." Only if \alpha > \phi. For the clamp, \alpha = 4.55^\circ and \phi = 11.31^\circ.

Check. A screw jack has \alpha = 3^\circ. Below what coefficient would it run down by itself? (\tan3^\circ = 0.052.)

Narration spine. The unwrapped ramp, the block resting on it, the hand off the handle. The reaction arrow swings across the normal and stops inside the cone: nothing moves. Steepen the ramp past \phi and the arrow cannot follow; the block slides and the screw spins.

Bridge. A real screw rarely works against its thread alone; the load usually turns on a collar too, and that changes both the effort and the efficiency.

Nugget 06: Collar Friction and Efficiency: Where the Turning Effort Goes

Claim. The moment on a power screw splits three ways: lifting the load (WL/2\pi), thread friction, and friction where the screw or nut bears on a collar. Efficiency is the first part over the total, and a self-locking thread alone can never exceed one half.

The picture. Meriam's SP 6/7: a vise whose screw bears on the movable jaw at B; the thread and the collar at B both resist turning. Marghitu's Fig. 5.16: a thrust collar carrying the axial load at a mean diameter d_c.

Check with numbers. Meriam's SP 6/7 in SI (the book's US units converted): a 266.9\,\mathrm N (60\,\mathrm{lb}) pull on a 0.2032\,\mathrm m (8\,\mathrm{in}) handle gives 4.448\,\mathrm{kN} (1000\,\mathrm{lb}) at the jaws, so moments about the jaw's pivot give a screw tension T = 4448 \times 16/10 = 7117\,\mathrm N. The screw: mean radius 12.7\,\mathrm{mm}, lead 5.08\,\mathrm{mm} (5 threads per inch), \mu = 0.20: \alpha = 3.64^\circ, \phi = 11.31^\circ. The applied 54.23\,\mathrm{N{\cdot}m} minus the thread's Tr\tan(\alpha + \phi) = 24.14\,\mathrm{N{\cdot}m} leaves M_B = 30.09\,\mathrm{N{\cdot}m} of collar friction at B (266\,\mathrm{lb{\cdot}in.}). To loosen, both frictions oppose the turning: M' = 30.09 + Tr\tan(\phi - \alpha) = 30.09 + 12.17 = 42.26\,\mathrm{N{\cdot}m}, a handle force of 42.26/0.2032 = 208.0\,\mathrm N (46.8\,\mathrm{lb}).

The idea. Marghitu (5.5.2) models a thrust collar with the axial load W concentrated at its mean diameter d_c; the friction \mu_cW there has an arm d_c/2:

The moment to raise is the thread's Wr\tan(\alpha + \phi) plus M_c; to lower, the thread's Wr\tan(\phi - \alpha) plus M_c (collar friction resists turning either way). With no friction at all, raising needs M_0 = WL/2\pi (5.14), and the efficiency is

For the thread alone this is \tan\alpha/\tan(\alpha + \phi), the form S30 found by virtual work. At the self-locking limit \alpha = \phi it is \tan\phi/\tan2\phi = (1 - \mu^2)/2, so a self-locking thread wastes at least half the effort (our consequence of the two equations).

Marghitu's Example 5.3. A square-thread screw, d = 38\,\mathrm{mm}, p = 6\,\mathrm{mm} (single start, l = 6\,\mathrm{mm}), d_r = 32\,\mathrm{mm}, d_m = 35\,\mathrm{mm}; \mu = 0.08, collar \mu_c = 0.1 at d_c = 45\,\mathrm{mm}; W = 9\,\mathrm{kN}. Raising: thread 21.29\,\mathrm{N{\cdot}m} (of which M_0 = 8.59 lifts) plus collar 20.25, M_r = 41.54\,\mathrm{N{\cdot}m}. Lowering: thread 3.99 plus collar 20.25, M_l = 24.24\,\mathrm{N{\cdot}m}. Self-locking: \pi\mu d_m - l = 2.796\,\mathrm{mm} > 0. Efficiency e = 9000(0.006)/(2\pi \times 41.54) = 0.207; the thread alone would give 0.404. (Marghitu prints these moments in "kN m"; with kN and mm they are N·m.)

The equation (definition: efficiency): e = WL/(2\pi M_r), with M_r including every friction moment.

Assumptions. Collar load at its mean diameter; same \mu_c throughout; slow turning.

Worked number. The puzzle's clamp, if its pad did not swivel: a collar of d_c = 12\,\mathrm{mm} with \mu_c = 0.20 would add 600(0.20)(0.006) = 0.72\,\mathrm{N{\cdot}m}, nearly doubling the 0.852\,\mathrm{N{\cdot}m}. That is why clamp pads swivel and jacks use ball thrust bearings.

Real example. A tensile-testing machine or a lathe lead screw uses a thrust ball bearing at the collar, removing most of M_c; a car jack keeps a plain collar and relies on the low efficiency to hold the car.

Plausible wrong answer. "Collar friction helps when loosening." It resists turning in either direction; it adds to the loosening moment.

Check. Why is the lowering moment in Example 5.3 positive though the thread alone needs only 3.99\,\mathrm{N{\cdot}m}? (Both the thread friction and the collar resist; the thread is self-locking anyway.)

Narration spine. A stacked bar for the raising moment: a short green block for the useful M_0, an orange block for thread friction, a grey one for the collar. A ball bearing slides under the collar and the grey block vanishes.

Synthesis

The clamp holds its 600\,\mathrm N when let go and loosens with 3.56\,\mathrm N rather than 8.52\,\mathrm N, and both facts come from one picture: a reaction leaning off a sloping face by the friction angle. Driving, it leans against the motion and adds to the slope, \tan(\alpha + \phi); releasing, it swings across the normal, and the load can move it only if the slope is steeper than friction allows, \alpha > \phi for a screw, \alpha > 2\phi for a wedge rough on both faces. Wedges and screws are inclined planes whose reactions lean by \phi: the effort is set by \alpha + \phi, the holding by whether \alpha is below \phi (or 2\phi), and the same friction that holds the load costs at least half the effort.

• Central idea: draw each sliding reaction at \phi from its normal, on the side that opposes the slip.
• Key relationship: M = Wr\tan(\alpha \pm \phi) for a screw, \alpha = \tan^{-1}(L/2\pi r); wedge self-locking if \alpha < 2\phi.
• Most important assumption: Coulomb friction with one \mu per surface; real threads and wedges vary with lubrication and vibration, so self-locking margins are kept generous.
• Open question: the screw's collar is a flat ring rubbing on a flat ring. How does friction spread over an area of contact, as in a thrust bearing, a clutch or a disk brake (S28)?

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). S27_wedges_and_screws.py drives and withdraws the 5^\circ wedge and tabulates Harper's wedge formula against \mu_1. It then treats every screw as an unrolled wedge with M = Wr\tan(\phi \pm \alpha): the puzzle's clamp, the M14 thread, a fast-lead screw, Harper's turnbuckle against \mu, and Meriam's vise with its collar.

Part 1: wedges
  2 t on a 5 deg wedge, mu 0.25: P = 11.67 kN; to pull out 8.03 kN; frictionless 1.72 kN
  SP 6/6: P = 2505 N
  Harper, mu_2 = 0.6 (kN):  theta  mu1 0  mu1 0.3  mu1 0.6  mu1 0.8
                               5    0.26    2.50    6.10    9.93
                              15    0.79    3.29    7.51   12.22
                              25    1.37    4.27    9.57   15.96
  door wedge 15 deg, mu 0.3: 2 phi = 33.4 deg
Part 2: screws
  clamp: alpha = 4.55 deg, tighten 0.852 N m, loosen 0.356 N m, no friction 0.239
  M14 x 2, 1 start(s): alpha = 2.80 deg
  M14 x 2, 2 start(s): alpha = 5.59 deg
  fast lead, 20 mm on 10 mm: holding takes 1.16 N m
  Harper's turnbuckle (two threads), T = 44.48 kN:
    mu 0.0000: M_T =   71.9 N m, M_L =  -71.9 N m
    mu 0.0554: M_T =  144.3 N m, M_L =    0.1 N m
    mu 0.1500: M_T =  269.0 N m, M_L =  122.0 N m
    mu 0.2500: M_T =  402.3 N m, M_L =  249.4 N m
  SP 6/7 vise: T = 7117 N, thread 24.14, collar 30.10, loosen 42.26 N m, 208.0 N at the handle

At \mu = \tan\alpha = 0.0554 the loosening moment is zero, which is the self-locking boundary. Below it, M_L is negative and the cable unscrews the turnbuckle by itself. The turnbuckle's 249.4 and the vise's collar 30.10\,\mathrm{N\cdot m} carry one more unrounded figure than the lesson's 249.5 and 30.09.

MATLAB (Octave-compatible). S27_wedges_and_screws.m does the same with two anonymous functions.

SP 6/6: P = 2505 N
 theta   mu1 0 mu1 0.3 mu1 0.6 mu1 0.8  (kN)
     5    0.26    2.50    6.10    9.93
    15    0.79    3.29    7.51   12.22
    25    1.37    4.27    9.57   15.96
clamp: tighten 0.852 N m, loosen 0.356 N m
turnbuckle mu 0.0000: M_T =   71.9, M_L =  -71.9 N m
turnbuckle mu 0.0554: M_T =  144.3, M_L =    0.1 N m
turnbuckle mu 0.1500: M_T =  269.0, M_L =  122.0 N m
turnbuckle mu 0.2500: M_T =  402.3, M_L =  249.4 N m

Traps

• Drawing the wedge reaction on the wrong side of the normal. It leans against the relative slip of that face; withdrawing reverses every friction. (From Nugget 01.)
• Leaving friction out of a wedge. Frictionless analysis can be ten times too low; SP 6/6 needs 2505\,\mathrm N, not 257.5\,\mathrm N. (From Nugget 02.)
• Using \alpha < \phi for a wedge rough on both faces. The self-locking limit is 2\phi. (From Nugget 03.)
• Computing the helix angle from the pitch or the outer diameter. Use the lead L = np and the mean radius: \tan\alpha = L/(2\pi r). (From Nugget 04.)
• Using \tan(\alpha - \phi) for lowering a self-locking screw. If \alpha < \phi it is \tan(\phi - \alpha), a moment in the lowering direction; \tan(\alpha - \phi) is the holding moment of a screw that unwinds. (From Nugget 05.)
• Forgetting the second thread of a turnbuckle. Both screws resist, so M = 2Tr\tan(\phi \pm \alpha). (From Nugget 05.)
• Letting collar friction change sign. It opposes the turning, raising or lowering. (From Nugget 06.)
• Mixing kN and mm. kN × mm is N·m, not kN·m (Marghitu's Example 5.3). (From Nugget 06.)

Self-check

1. Meriam's Problem 6/45: a steel wedge is driven into a stump; \mu = 0.20 on both faces. What is the largest wedge angle that will not pop out?
2. Meriam's Problem 6/46: a 10^\circ wedge is driven under a wheel on a spring-loaded strut; the wheel turns freely. What is the least \mu_s between wedge and floor for the wedge to stay in place?
3. Meriam's Problem 6/48: a 100\,\mathrm{kg} door, its mass centre 1.2\,\mathrm m from corner A (held by a ledge), is lifted at corner B, 2.2\,\mathrm m from A, by a 5^\circ wedge pushed in horizontally; \mu_s = 0.60 at both wedge faces. Find P.
4. Meriam's Problem 6/54: two 5^\circ wedges, pushed together by equal forces P, raise a guided column carrying 5\,\mathrm{kN}; \mu = 0.40 on every sliding surface. Find P.
5. Meriam's Problem 6/55: the same column is lowered. Find the forces P' to withdraw the wedges.
6. Meriam's Problems 6/60–6/61 (US units in the book): a 100\,\mathrm{lb} body in smooth vertical guides rests on a 15^\circ wedge, \mu_s = 0.20 between them. Convert to SI and find the force to start raising it, and to start lowering it, (a) with frictionless rollers under the wedge, (b) with \mu_s = 0.20 under the wedge too.
7. Harper's wedge formula with \theta = 15^\circ, \mu_1 = 0.30, \mu_2 = 0.60, m = 500\,\mathrm{kg}: find P, and P with frictionless wedge faces.
8. Meriam's Problem 6/52 in SI (Harper's turnbuckle): T = 44.48\,\mathrm{kN}, r = 14.605\,\mathrm{mm}, L = 5.08\,\mathrm{mm}, \mu = 0.25. Find the moments to tighten and to loosen, and the coefficient below which it loosens by itself.
9. Marghitu's Problem 5.7: a double square thread, mean diameter 50\,\mathrm{mm}, pitch 8\,\mathrm{mm}, \mu = 0.15. Is it self-locking?
10. Marghitu's Problem 5.6: a double square thread, d_m = 30\,\mathrm{mm}, pitch 6\,\mathrm{mm}, \mu = 0.1; collar \mu_c = 0.2 at d_c = 40\,\mathrm{mm}; load 8\,\mathrm{kN}. Find the moment to lower the load and the overall efficiency.
11. Meriam's Problem 6/63: a 100\,\mathrm{kg} block in vertical guides rests on a 10^\circ wedge pushed by a single-start screw (d_m = 30\,\mathrm{mm}, lead 10\,\mathrm{mm}, \mu = 0.25); \mu = 0.40 at the block's and wedge's mating surfaces, guides included. Find the moment to raise the block.
12. An M14 × 2 thread with \mu = 0.15: find the helix angle single-start and double-start, and say whether each self-locks.

Fully worked solutions

1. The wedge has two rough faces and no driving force once struck: self-locking if \alpha < 2\phi = 2\tan^{-1}0.20 = 22.6^\circ.

2. The wheel turns freely, so it pushes on the wedge only along the normal to the 10^\circ face. The wedge then has two forces, and the floor's reaction must be collinear with that normal, 10^\circ from the vertical. It stays if 10^\circ \le \phi_s: \mu_s \ge \tan10^\circ = 0.1763.

3. Moments about A for the door (the wedge force at B is level with A, so only its vertical part has a moment): R_{By}(2.2) = 981(1.2), R_{By} = 535.1\,\mathrm N. The door slides up the wedge's top face relative to it, so friction on the door points down the slope: N(\cos5^\circ - 0.60\sin5^\circ) = 535.1, N = 566.9\,\mathrm N; its horizontal push on the wedge is N(\sin5^\circ + 0.60\cos5^\circ) = 388.2\,\mathrm N. The floor adds 0.60 \times 535.1 = 321.1\,\mathrm N of friction. P = 388.2 + 321.1 = 709\,\mathrm N.

4. Upper wedge (moving towards the column's side, up the interface): column friction 0.40 \times 5000 = 2000\,\mathrm N opposes its slide; at the 5^\circ interface N(\cos5^\circ - 0.40\sin5^\circ) = 5000, N = 5201\,\mathrm N, with horizontal part N(\sin5^\circ + 0.40\cos5^\circ) = 2526\,\mathrm N. P = 2000 + 2526 = 4.53\,\mathrm{kN}; the lower wedge gives the same.

5. Every friction reverses: N(\cos5^\circ + 0.40\sin5^\circ) = 5000, N = 4849\,\mathrm N; P' = 2000 - N(\sin5^\circ - 0.40\cos5^\circ) = 2000 + 1510 = 3.51\,\mathrm{kN}.

6. W = 444.8\,\mathrm N. Raising, the body slides up the face: N = 444.8/(\cos15^\circ - 0.2\sin15^\circ) = 486.6\,\mathrm N, horizontal push on the wedge N(\sin15^\circ + 0.2\cos15^\circ) = 219.9\,\mathrm N. (a) P = 219.9\,\mathrm N (49.4\,\mathrm{lb}); (b) +0.2 \times 444.8: P = 308.9\,\mathrm N (69.4\,\mathrm{lb}). Lowering, the friction reverses: N = 444.8/(\cos15^\circ + 0.2\sin15^\circ) = 437.1\,\mathrm N, and the body pushes the wedge out with N(\sin15^\circ - 0.2\cos15^\circ) = 28.7\,\mathrm N. (a) Rollers: since 15^\circ > \phi = 11.3^\circ the wedge is driven out; hold it back with 28.7\,\mathrm N in the original pushing direction (6.45\,\mathrm{lb}). (b) Floor friction (89.0\,\mathrm N) holds: 15^\circ < 2\phi, self-locking; pull it out with 89.0 - 28.7 = 60.3\,\mathrm N the other way (13.55\,\mathrm{lb}).

7. P = 0.6(4905)[0.6 + 0.91\tan15^\circ]/[(1 - 0.3\tan15^\circ)(0.82)] = 3.29\,\mathrm{kN}. With \mu_1 = 0: 0.6(4905)\tan15^\circ = 788.6\,\mathrm N.

8. \alpha = \tan^{-1}(5.08/(2\pi \times 14.605)) = 3.17^\circ, \phi = 14.04^\circ. M_T = 2(44\,480)(0.014605)\tan17.20^\circ = 402.3\,\mathrm{N{\cdot}m}; M_L = 1299.3\tan10.87^\circ = 249.5\,\mathrm{N{\cdot}m}. It loosens by itself if \mu < \tan\alpha = 0.0554.

9. Lead l = 2 \times 8 = 16\,\mathrm{mm}; \tan\alpha = 16/(50\pi) = 0.1019 (\alpha = 5.82^\circ) < \mu = 0.15 (\phi = 8.53^\circ): self-locking.

10. l = 12\,\mathrm{mm}, \tan\alpha = 12/(30\pi) = 0.1273 > \mu = 0.1: the thread alone is not self-locking. Lowering: thread \frac{8(30)}{2}\,\frac{3\pi - 12}{30\pi + 1.2} = -3.24\,\mathrm{N{\cdot}m} (it would unwind) plus collar 8(0.2)(40)/2 = 32.0\,\mathrm{N{\cdot}m}: M_l = 28.8\,\mathrm{N{\cdot}m}, the collar holds it. Raising: thread 120(12 + 3\pi)/(30\pi - 1.2) = 27.63 plus 32.0, M_r = 59.63\,\mathrm{N{\cdot}m}; e = 8000(0.012)/(2\pi \times 59.63) = 0.256.

11. Block (rising, pushed sideways against the left guide): wedge-face normal N, friction 0.40N down the face; guide normal N_w = N(\sin10^\circ + 0.40\cos10^\circ) with friction 0.40N_w downward. Vertically, N(\cos10^\circ - 0.40\sin10^\circ) - 0.40N_w = 981: N = 1425\,\mathrm N, N_w = 808.9\,\mathrm N. Wedge: base normal 1304.6\,\mathrm N, base friction 521.8\,\mathrm N; screw thrust 808.9 + 521.8 = 1331\,\mathrm N. Screw: \alpha = \tan^{-1}(10/30\pi) = 6.06^\circ, \phi = 14.04^\circ; M = 1331(0.015)\tan20.09^\circ = 7.30\,\mathrm{N{\cdot}m}. (With smooth guides it would be 5.49\,\mathrm{N{\cdot}m}; Meriam's answer counts the guides.)

12. d_m = 13\,\mathrm{mm}. Single: \alpha = \tan^{-1}(2/13\pi) = 2.80^\circ; double: L = 4\,\mathrm{mm}, \alpha = 5.59^\circ. \phi = \tan^{-1}0.15 = 8.53^\circ: both self-lock, the double-start with less margin.