Study Companion: S26 — Dry Friction

Every earlier lesson treated contact as smooth: a surface could push on a body only at right angles to itself. Real surfaces also push along the contact, and that tangential force, friction, is what lets a ladder stand, a car brake and a wedge hold. It is an odd force. It is zero when nothing tries to slide, it grows to match whatever tries, and it stops growing at a limit set by how hard the surfaces are pressed together. This lesson builds that model from one experiment, puts it into a single geometric object, the friction angle, and then uses it on the questions friction actually asks: will it slip, which way, where, and at what load.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 6, articles 6/1 ("Introduction"), 6/2 ("Types of Friction") and 6/3 ("Dry Friction"), with Sample Problems 6/1–6/5; Marghitu, Dupac and Madsen, Statics with MATLAB, sections 5.1 ("Introduction"), 5.2 ("Static Coefficient of Friction"), 5.3 ("Kinetic Coefficient of Friction") and 5.4 ("Angle of Friction") with Examples 5.1–5.2; Harper, Solving Statics Problems in MATLAB, "Friction" (its Sample Problem 6/5 and Problem 6/31, the same numbers in Meriam's 9th edition). Units SI throughout; g = 9.81\,\mathrm{m/s^2}. Marghitu writes the friction force F_f and the friction angle \theta; here they are F and \phi, as in Meriam.

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 painter of mass 90\,\mathrm{kg} climbs a 4\,\mathrm m, 15\,\mathrm{kg} ladder. Its foot A stands 1.5\,\mathrm m out from the wall, so the ladder leans at \theta = 67.98^\circ to the floor. The top has a small roller, so the wall can push only horizontally; all the grip is at the foot, where the coefficient of static friction is 0.25. The empty ladder stands. The painter starts up.

How far up the ladder can the painter climb before the foot slides? The plausible guess is "all the way: the ladder stood empty, and the painter's weight presses the foot harder into the floor, which only adds grip". Ladder labels disagree: they tell you to set the foot one unit out for every four up. The answer, 2.55\,\mathrm m of the 4\,\mathrm m, is in Nugget 05.

Nugget 01: Static Friction Matches the Push Up to a Limit, Then Kinetic Friction Takes Over

Claim. A surface resists a sideways push on a resting body with a friction force exactly equal to the push, up to a maximum \mu_sN proportional to the normal force. Once the body slides, the friction drops to a smaller, roughly constant \mu_kN.

The picture. Meriam's Fig. 6/1: a block of mass m on a rough floor, a horizontal force P raised slowly from zero, and the friction F recorded against P.

On the free-body diagram the floor gives two forces: the normal force N, here equal to mg, and the friction force F along the surface, pointing against the motion the push tries to cause. Their resultant is the total reaction R. Under a magnifier the surfaces are humps meshing with humps (Fig. 6/1c): each contact reaction R_1, R_2, \dots is tilted by the local slope, N is the sum of their normal parts and F the sum of their tangential parts. While nothing moves, F is whatever equilibrium needs, F = P: the static range, the rising line. At a certain P the block breaks free and F falls a little, then stays nearly level: once sliding, the contacts ride nearer the tops of the humps, the reactions tilt less, and their tangential parts are smaller. That is why it takes more force to start a block than to keep it going.

Check with numbers. A 50\,\mathrm{kg} crate on a floor with \mu_s = 0.50, \mu_k = 0.40: N = 50 \times 9.81 = 490.5\,\mathrm N. Push with 150\,\mathrm N: F = 150\,\mathrm N, not 0.50 \times 490.5. The largest friction the floor can supply is F_{\max} = 0.50 \times 490.5 = 245.25\,\mathrm N, so any push up to that holds. Once it slides, F_k = 0.40 \times 490.5 = 196.2\,\mathrm N, and a push of 196.2\,\mathrm N keeps it moving at constant speed.

The idea. Meriam (6/3) and Marghitu (5.2–5.3) state the experimental laws:

\mu_s is the coefficient of static friction, \mu_k the coefficient of kinetic friction, and \mu_k is usually somewhat less than \mu_s. Equation 6/1 gives only the limit: below impending motion F < \mu_sN, and F comes from the equilibrium equations. \mu_k falls a little as the speed rises, and both coefficients depend on the exact state of the surfaces, so they are uncertain. Friction always acts along the surface against the motion or the tendency to move; on the other surface it is the equal and opposite force (Marghitu's Fig. 5.2: the floor feels the sliding block's friction in the direction of the slide). Books often write just F = \mu N; you must decide from the problem whether it is the static limit or the kinetic value.

Historically (Marghitu 5.1) the laws are Amontons' (1699) rediscovery of Leonardo's notes, checked by Coulomb (1781) and extended by Morin (1830–1834): friction is proportional to the normal force, independent of the apparent contact area, and independent of the sliding speed. Meriam (6/2) calls this dry or Coulomb friction, to tell it from fluid friction between layers of a fluid moving at different speeds (it depends on the velocity gradient and the viscosity, and belongs to fluid mechanics) and from internal friction in a solid under cyclic loading (the energy lost in deformation). A machine in which friction can be neglected is ideal; one where it must be counted is real, and its sliding parts turn energy into heat and wear (6/1).

The equation (material model: Coulomb friction): F \le \mu_sN at rest, with equality only when slip impends; F = \mu_kN while sliding.

Assumptions. Dry, unlubricated surfaces; coefficients constant over the contact; N compressive (a contact that would have to pull has separated).

Worked number. The crate is pushed with P = 300\,\mathrm N. Assume it stays put: F = 300\,\mathrm N > 245.25\,\mathrm N, so it cannot; it slides, and the friction is F_k = 196.2\,\mathrm N, leaving 300 - 196.2 = 103.8\,\mathrm N to accelerate it.

Real example. The tree surgeon in Meriam's photograph and every rope brake rely on static friction holding until a chosen load, then kinetic friction letting the rope slip under control.

Plausible wrong answer. "Friction equals \mu_sN." Only at the moment of slipping; a box nobody pushes has zero friction on it.

Check. A 10\,\mathrm{kg} box rests on a floor with \mu_s = 0.6. What friction acts if you push it with 20\,\mathrm N? (20\,\mathrm N: the limit 58.9\,\mathrm N is not reached.)

Narration spine. A block sits on a floor and an arrow P grows. A matching friction arrow grows beneath it while a graph draws the rising line. At the peak the block jerks forward, the friction arrow shrinks a little and the graph drops to its level kinetic line.

Bridge. Normal force and friction always come together. Their resultant has a direction, and that direction carries all the information about how close the contact is to slipping.

Nugget 02: The Friction Angle Folds Normal Force and Friction Into One Reaction

Claim. The total reaction of a rough surface leans away from the normal by an angle \alpha with \tan\alpha = F/N. At impending slip it leans by the friction angle \phi_s = \tan^{-1}\mu_s, and it can never lean further: every possible reaction lies inside a cone of half-angle \phi_s.

The picture. Meriam's Figs. 6/1b and 6/2 (Marghitu's Fig. 5.3): N and F combine into R; the limiting directions of R sweep out the cone of static friction.

Check with numbers. For the crate of Nugget 01, \phi_s = \tan^{-1}0.50 = 26.57^\circ and \phi_k = \tan^{-1}0.40 = 21.80^\circ. Pushed with 150\,\mathrm N, the reaction leans by \tan^{-1}(150/490.5) = 17.0^\circ: inside the cone, so it holds.

The idea. Meriam (6/3) measures the direction of R from the normal by \tan\alpha = F/N. When F reaches \mu_sN, \alpha reaches its largest value \phi_s:

\phi_s is the angle of static friction, \phi_k the angle of kinetic friction; Marghitu (5.4) defines the same angles as the static and kinetic angles of friction, F = R\sin\phi, N = R\cos\phi. If the push can come from any direction along the surface, the limiting reactions form a right circular cone about the normal of vertex angle 2\phi_s: the cone of static friction. At impending motion R lies on the cone; if motion is not impending, R is inside it; while sliding, R lies on the slightly narrower cone of kinetic friction, 2\phi_k. The friction angle is a geometric picture of the roughness: rough surfaces have steeper hump slopes and larger angles.

The coefficient belongs to a pair of surfaces, never to one surface alone, and it does not depend on the apparent area (Meriam, factors affecting friction): only the peaks touch, the true contact area is much smaller and grows with the load, which is why F follows N and not the area. Molecular attraction, local heating and adhesion, and films of oxide, oil or dirt also matter, so tabulated values (Meriam's Table D/1; Marghitu's Table 5.1, for example metal on metal 0.15–0.20, rubber on concrete 0.50–0.90) are guides only; a design value should come from a test that copies the real surfaces.

Angle of repose (Meriam SP 6/1). Tilt a plane under a block until it slips. The reaction must balance the weight alone, so it is vertical; it makes the plane's tilt angle with the plane's normal, so at impending slip that tilt is \phi_s:

The equation (definition: friction angle): \tan\phi = \mu, with \phi_s for the static limit, \phi_k for sliding.

Worked number. A phone on a book slides when the book is tilted to 17^\circ: \mu_s = \tan17^\circ = 0.306. A heavier phone of the same case slides at the same angle, because \theta_{\max} does not contain the mass.

Real example. Pulling a crate with a rope (Marghitu's Problem 5.1): at impending slip the floor's reaction lies on the cone edge at \phi_s, so the pull P needed to close the force triangle with the weight is shortest when it is perpendicular to R, which is at the angle \phi_s above the floor. For the 50\,\mathrm{kg} crate with \mu_s = 0.40: pulling at 21.8^\circ needs P = mg\sin\phi_s = 490.5 \times 0.3714 = 182.2\,\mathrm N; pulling horizontally needs 196.2\,\mathrm N.

Plausible wrong answer. "A bigger contact area gives more friction, so wide tyres grip more." For dry Coulomb friction the area cancels; tyres grip by other effects (rubber's adhesion and deformation) that this model leaves out.

Check. A ramp is tilted until a box slides at 31^\circ. What is \mu_s? (\tan31^\circ = 0.601.)

Narration spine. A tilting board, a block, and the reaction arrow under it. As the board tilts the arrow stays vertical while the normal tilts, the angle between them opening until it meets the edge of a translucent cone, and the block goes.

Bridge. Knowing the limit is half the job. In a given problem you usually do not know whether the limit has been reached, and the method must find out.

Nugget 03: Three Kinds of Friction Problem: Assume, Solve, Then Check

Claim. Every dry-friction problem is one of three kinds: impending motion is known (use F = \mu_sN), sliding is known (use F = \mu_kN), or neither is known, and then you assume equilibrium, solve for the friction needed, and compare it with \mu_sN.

The picture. Meriam's SP 6/3: a 100\,\mathrm{kg} block on a 20^\circ incline, a horizontal push P, \mu_s = 0.20, \mu_k = 0.17. The friction direction is guessed (up the plane); the sign of the answer corrects it.

Check with numbers. Axes along and normal to the plane, friction F assumed up the plane:

Case I, P = 500\,\mathrm N. F = 335.5 - 469.8 = -134.3\,\mathrm N: the minus sign says that if the block is in equilibrium, friction points down the plane. N = 171.0 + 921.8 = 1093\,\mathrm N, so the surfaces can supply F_{\max} = 0.20 \times 1093 = 219\,\mathrm N > 134.3\,\mathrm N. The assumption holds: F = 134.3\,\mathrm N down the plane, and the block stays.

Case II, P = 100\,\mathrm N. F = 335.5 - 94.0 = 242\,\mathrm N and N = 956\,\mathrm N, but F_{\max} = 0.20 \times 956 = 191.2\,\mathrm N < 242\,\mathrm N. Equilibrium cannot exist: the block slides down, and the friction on it is kinetic, opposing the motion: F = 0.17 \times 956 = 162.5\,\mathrm N up the plane. \Sigma F_y = 0 still holds, because the block does not move normal to the plane, so N = 956\,\mathrm N is right even while it slides.

The idea. Meriam's Key Concepts list the three types:

1. Impending motion known: the body is on the verge of slipping, F = F_{\max} = \mu_sN, and the equilibrium equations hold.
2. Neither impending motion nor motion known: assume equilibrium, solve for the F it needs, then (a) F < \mu_sN: equilibrium, and F is the actual friction; (b) F = \mu_sN: motion impends; (c) F > \mu_sN: impossible, so the body slides and F = \mu_kN.
3. Relative motion known: kinetic friction, F = \mu_kN directly.

The direction of an unknown friction force may be assumed; a negative answer reverses it. Only in type 1 and type 3 is the direction forced on you, by the motion.

The equation (material model, applied as a test): equilibrium valid only if \lvert F\rvert \le \mu_sN.

Worked number. Case II solved for its slip: the net force down the plane is 241.6 - 162.5 = 79.0\,\mathrm N on 100\,\mathrm{kg}, so the block accelerates at 0.790\,\mathrm{m/s^2}; dynamics, not statics, takes over.

Real example. A loaded pallet on a tilted truck bed: whether it creeps is exactly a type 2 question, and the answer depends on both the tilt and any strap force.

Plausible wrong answer. "Set F = \mu_sN and solve." In case I that gives 219\,\mathrm N of friction the block does not need, and a wrong answer for everything else.

Check. In case I, why is the friction down the slope although the block's weight pulls it down? (The 500\,\mathrm N push has a component up the slope, 469.8\,\mathrm N, larger than the weight's 335.5\,\mathrm N.)

Narration spine. A block on a slope with a dashed friction arrow guessed up the slope. Case I: the computed value comes out negative and the arrow flips; a bar for \mu_sN shows room to spare. Case II: the needed friction bar overshoots the \mu_sN bar, a red cross, the block slides and the arrow shrinks to \mu_kN.

Bridge. When motion impends, the friction is at its limit, but in which direction? Often the body could slip either way, and each way gives a different answer.

Nugget 04: The Direction of Impending Motion Fixes the Direction of Friction

Claim. At impending slip the friction is \mu_sN against the motion that would happen. A body that could slip either way therefore has two limits, and the safe values of the load lie between them.

The picture. Meriam's SP 6/2: a 100\,\mathrm{kg} block on a 20^\circ incline with \mu_s = 0.30, held by a cord parallel to the plane that runs over a pulley to a hanging mass m_0.

Check with numbers. N = 981\cos20^\circ = 921.8\,\mathrm N in both cases, so F_{\max} = 0.30 \times 921.8 = 276.6\,\mathrm N. The weight's pull down the plane is 981\sin20^\circ = 335.5\,\mathrm N. Case I, m_0 large, the block about to slide up: friction acts down, so 9.81m_0 = 335.5 + 276.6 and m_0 = 62.4\,\mathrm{kg}. Case II, m_0 small, about to slide down: friction acts up, 9.81m_0 = 335.5 - 276.6 and m_0 = 6.01\,\mathrm{kg}. Any m_0 from 6.01 to 62.4\,\mathrm{kg} leaves the block at rest.

The idea. In both cases the equilibrium equations are the same, \Sigma F_y = 0 for N and \Sigma F_x = 0 along the plane; only the sign of F_{\max} changes, because friction must oppose the slip that is about to happen. Write the two cases out: there is no single formula with a \pm that sorts itself. Meriam's hint: without the cord the block would slide anyway, since \tan20^\circ = 0.364 > 0.30 (SP 6/1), which is why even the lower limit is positive. In both cases the friction, normal force, weight and cord tension are concurrent, so moments give nothing new for a block.

The equation (physical law with the friction model): case I m_0g = mg(\sin\theta + \mu_s\cos\theta); case II m_0g = mg(\sin\theta - \mu_s\cos\theta).

Assumptions. Frictionless pulley (the cord tension is m_0g on both sides); the block slides without tipping.

Worked number. The same block on a gentler 10^\circ plane: mg\sin10^\circ = 170.3\,\mathrm N, F_{\max} = 0.30 \times 966.1 = 289.8\,\mathrm N. Case II would need m_0 = (170.3 - 289.8)/9.81 < 0: the block holds itself, and m_0 may be anything up to (170.3 + 289.8)/9.81 = 46.9\,\mathrm{kg}.

Real example. A car parked on a hill with a trailer on a rope over a pulley is the same range: too little load and it slides down, too much and it is dragged up.

Plausible wrong answer. "The limiting friction is \mu_sN, so there is one limiting load." There are two, one for each direction of impending slip.

Check. What is the range if \mu_s were 0.364, so that \tan20^\circ = \mu_s? (From 0 to 2 \times 335.5/9.81 = 68.4\,\mathrm{kg}.)

Narration spine. The block, the cord, the hanging mass growing. The friction arrow points down the slope as the mass is heavy, then the mass shrinks and the arrow swings to point up the slope. A number line from 6.01 to 62.4\,\mathrm{kg} fills in green.

Bridge. In those problems the friction acted at one contact of a block whose forces all met at a point. A ladder has two contacts, and the load can stand anywhere along it.

Nugget 05: The Ladder: Where the Load Sits Decides Whether It Slips

Claim. The grip at a ladder's foot grows with the load, but so does the moment the load makes about the foot, and that moment grows with how high the load stands. Slip starts at a height fixed by the angle and the coefficient, not by the weight alone.

The picture. The puzzle's ladder, free of the floor and the wall: weight 15g at the middle, the painter's 90g at distance s along the ladder from A, the roller's horizontal push N_B at the top, and at the foot the normal force N_A and the friction \mu N_A toward the wall, opposing the slide outward.

Check with numbers. \Sigma F_y = 0: N_A = (90 + 15)(9.81) = 1030.05\,\mathrm N, whatever s is. \Sigma F_x = 0: N_B = 0.25N_A = 257.5\,\mathrm N at impending slip. Moments about A (\cos\theta = 0.375, 4\sin\theta = 3.708\,\mathrm m):

so s = (954.9 - 110.4)/331.1 = 2.55\,\mathrm m. Above that the ladder needs more than 257.5\,\mathrm N of friction.

The idea. Harper keeps \mu and \theta as symbols. The force sums give N_A = 105g and N_B = \mu N_A = 105\mu g; the moment equation is then solved for s:

The weight does make the foot grip harder, N_A = 105g, but the friction needed is N_B, set by moments about A, and those grow with s. With the painter at the top the ladder needs \mu = 390\cot\theta/420 = 0.376; empty it needs only \cot\theta/2 = 0.202. Harper's plot of s(\theta) for \mu = 0.2, 0.4, 0.6 reads: s = 0 where \tan\theta = 1/(14\mu), a ladder so flat that it slips under its own weight before anyone steps on it (15.95^\circ for \mu = 0.25); s \ge 4\,\mathrm m where \tan\theta \ge 13/(14\mu), a ladder steep enough to climb to the top (74.93^\circ for \mu = 0.25). The ladder rule, one out for every four up, is \theta = \tan^{-1}4 = 75.96^\circ, which gives s = 4.33\,\mathrm m here: the whole ladder is safe even on a floor as slick as \mu = 0.25.

Second view (Marghitu Ex 5.2). A 1\,\mathrm m, 10\,\mathrm{kg} rod rests at \beta = 30^\circ with its upper end A on a face inclined at \alpha = 45^\circ and its lower end B on the floor, \mu = 0.2 at both, with a force F along the rod pulling A up the face. Marghitu writes each force as a vector, \mathbf N_A = N_A(\sin\alpha, \cos\alpha), \mathbf F_{fA} = \mu N_A(-\cos\alpha, \sin\alpha), and so on, and solves \Sigma\mathbf F = \mathbf 0 with moments about A symbolically. Both ends are rough, and a rigid rod touching two surfaces can move in only one way, so both ends slip together and each friction is \mu N. Taking the friction at B into the moment equation (it acts l\sin\beta below A), the least pull that stops the rod sliding down is F = 14.46\,\mathrm N, and the greatest before it slides up is F = 55.68\,\mathrm N. (Marghitu's moment equation leaves out that friction term and prints F = 19.085\,\mathrm N, which is the least pull, not the greatest.)

The equation (physical law: equilibrium with the friction limit at the foot): N_B = \mu N_A and \Sigma M_A = 0 give s(\theta, \mu) above.

Assumptions. Ideal roller at the top (no friction there); painter's centre of mass above the feet; ladder uniform and rigid.

Worked number. On a better floor, \mu = 0.40, at the same 67.98^\circ: s = (14 \times 0.40 \times 0.9270 - 0.375)/(3 \times 0.375) = 4.28\,\mathrm m. The painter reaches the top.

Real example. Ladder safety rules (the 4-to-1 rule, and the foot held by a second person or a cleat) are this calculation, done once for a cautious \mu.

Plausible wrong answer. "If the ladder stands empty, it will stand with a painter on it, because friction grows with the weight." The friction that is needed grows faster, with the painter's height times the weight.

Check. At what angle does the empty puzzle ladder (\mu = 0.25) just slip? (\tan\theta = 1/3.5, \theta = 15.95^\circ.)

Narration spine. The ladder against the wall. The painter climbs; a bar of needed friction at the foot rises with each step towards the fixed bar of \mu N_A. They meet at 2.55\,\mathrm m and the foot skids. Rewind, set the foot in to the 4-to-1 angle, and the needed bar stays below to the top.

Bridge. The ladder slid; a heavy box pushed too high does not slide at all, it tips. Which happens first is again decided by where a force acts.

Nugget 06: Slip or Tip: Follow the Line of the Total Reaction

Claim. A block pushed sideways slides if the total reaction can stay under its base, and tips if it cannot. Since the reaction leans at the friction angle, pushing above a certain height tips the block before it can slide.

The picture. Meriam's SP 6/4: a block of mass m, width b, height H, pushed at height h by a horizontal force P that moves it at constant speed (coefficient \mu_k). Three forces act, P, mg and the floor's reaction R, so they must meet at one point.

Check with numbers. Our crate: b = 0.6\,\mathrm m, H = 1.2\,\mathrm m, m = 60\,\mathrm{kg}, \mu_k = 0.40. Sliding needs P = \mu_kmg = 0.40 \times 588.6 = 235.4\,\mathrm N wherever it is pushed. On the verge of tipping all the floor's reaction sits at the front corner A, leaning back at \phi_k = 21.8^\circ; it must pass through the point where P crosses the line of mg, so that point is at height h = (b/2)/\tan\phi_k = 0.3/0.4 = 0.75\,\mathrm m. Push lower and it slides; push higher and it tips.

The idea. With the block sliding, F_k = \mu_kN and N = mg, so P = \mu_kmg. Three-force equilibrium: R (at angle \phi_k) and P cross on the line of mg. At the tipping limit, R acts at A, b/2 from that line:

The same by moments about A: Ph - mg\,b/2 = 0 with P = \mu_kmg. If h is less, the reaction's line meets the base a distance x behind G's line: for h = H/2, x/(H/2) = \tan\phi_k, so x = \mu_kH/2 (Meriam's hint: moments about G, F(H/2) - Nx = 0, give the same). With \mu_s in place of \mu_k the same formulas give the verge of tipping and the verge of slipping from rest.

The equation (geometric relationship: three-force concurrency): tipping before slipping when h > b/(2\mu).

Assumptions. Rigid block and floor; reaction distributed over the base, its resultant anywhere between the corners; uniform block, G at the centre.

Worked number. Pushed at mid-height, h = 0.60\,\mathrm m: it slides, with the reaction crossing the base x = 0.40 \times 1.2/2 = 0.24\,\mathrm m ahead of the centre line, inside the 0.30\,\mathrm m half-width.

Real example. Furniture movers push a tall wardrobe low down for exactly this reason, and a car's tipping-or-skidding limit on a slope is the same comparison (self-check 9).

Plausible wrong answer. "Friction acts at the centre of the base." Its line moves forward as the push rises, reaching the corner at the tipping limit.

Check. If \mu_k doubles to 0.80, at what height does the crate tip? (0.3/0.8 = 0.375\,\mathrm m.)

Narration spine. A tall block, a pushing hand sliding up its side. The reaction arrow under the block leans back at a fixed angle, its foot running forward along the base as the hand rises, until it reaches the corner; the block rocks onto its edge.

Bridge. One contact, one question. A stack of blocks has several contacts, and any one of them could be the one that gives.

Nugget 07: Several Contact Surfaces: Find Which One Slips First

Claim. With several rough contacts, guess which surfaces slip, put \mu_sN there, solve, and check every other contact. If a check fails, the guess was wrong; the true mode is the one that needs the least load.

The picture. Meriam's SP 6/5: three flat blocks on a 30^\circ plane, 30\,\mathrm{kg} on top held by a wire to the support, 50\,\mathrm{kg} in the middle pushed down the plane by P, 40\,\mathrm{kg} at the bottom. \mu_s = 0.30 between the top two, 0.40 between the lower two, 0.45 under the bottom one.

Check with numbers. Normal forces from \Sigma F_y = 0 on each block, top down: N_1 = 30(9.81)\cos30^\circ = 254.9\,\mathrm N, N_2 = N_1 + 50(9.81)\cos30^\circ = 679.7\,\mathrm N, N_3 = N_2 + 40(9.81)\cos30^\circ = 1019.5\,\mathrm N. Guess 1: the 50 kg slides alone. F_1 = 0.30N_1 = 76.5\,\mathrm N, F_2 = 0.40N_2 = 271.9\,\mathrm N, and \Sigma F_x = 0 on it: P = 76.5 + 271.9 - 245.3 = 103.1\,\mathrm N. Check the 40 kg: it needs F_3 = 271.9 + 196.2 = 468.1\,\mathrm N from the plane, but the plane can give only 0.45 \times 1019.5 = 458.8\,\mathrm N. The guess fails. Guess 2: the 50 and 40 kg slide together. F_3 = 458.8\,\mathrm N, so the 40 kg block gives F_2 = 458.8 - 196.2 = 262.6\,\mathrm N < 271.9\,\mathrm N (no slip between them, as assumed), and the 50 kg block gives P = 262.6 + 76.5 - 245.3 = 93.8\,\mathrm N.

The idea. Harper does not guess. It writes all the equations with the incline angle \theta as a parameter, N_1 = 30g\cos\theta, N_2 = 80g\cos\theta, N_3 = 120g\cos\theta, and for the two possible modes:

The largest P without slip is the smaller of the two at each \theta. They cross at 0.503\,\mathrm{rad} (28.8^\circ, P = 116.0\,\mathrm N): flatter than that, the middle block slips alone; steeper, the lower two go together, and from 35.0^\circ on they slide with no push at all. (Solving the same equations as a linear programme, maximising P with every \lvert F\rvert \le \mu N and no mode assumed, gives exactly \min(P_1, P_2).)

Second view (Marghitu Ex 5.1). A prism of 10\,\mathrm{kg} with a 10^\circ face carries a 5\,\mathrm{kg} slider held between vertical guides; a horizontal F pushes the prism. \mu = 0.2 on all three contacts. Marghitu writes every force as a vector on each body, \mathbf F_{12} = F_{12}(-\sin\alpha, \cos\alpha), \mathbf F_{f12} = \mu F_{12}(\cos\alpha, \sin\alpha), and solves the four scalar equations: F_{12} = 47.89\,\mathrm N, F_{02} = 1.117\,\mathrm N, F_{01} = 146.93\,\mathrm N and the greatest push F = 30.50\,\mathrm N. Here there is no choice of mode: the prism moving right lets the slider drop, so all three contacts slip at once, and each friction opposes its own relative motion.

The equation (physical law: equilibrium of each body, with F = \mu_sN at the assumed slipping surfaces and \lvert F\rvert \le \mu_sN checked at the others).

Worked number. Harper's version at 20^\circ: P_1 = 402.2(0.9397) - 490.5(0.3420) = 210.2\,\mathrm N, P_2 = 278.8\,\mathrm N: the middle block slips alone at 210.2\,\mathrm N.

Real example. Stacked loads on a sloping conveyor or a tilted pallet: whether the top layer slides off or the whole stack goes is this comparison of modes.

Plausible wrong answer. "The surface with the smallest coefficient slips first." It depends on the normal forces and the other loads; here the 0.45 surface gives way, because it carries both lower blocks' weight components.

Check. In guess 2, why must F_2 come out below 0.40N_2? (The 50 and 40 kg move together only if the surface between them does not slip.)

Narration spine. Three stacked blocks on a slope; P pushes the middle one. Guess 1 draws the middle block sliding out; a gauge on the bottom surface overflows red. Guess 2 moves the lower two as one, every gauge stays green, and P = 93.8\,\mathrm N. Then the slope tilts and the plot of P_1 and P_2 draws, the lower curve highlighted.

Synthesis

The painter can climb 2.55\,\mathrm m, not the whole ladder, and the reason runs through every nugget. Friction is not a fixed force but a response, equal to what equilibrium needs, up to \mu_sN; the needed friction at the foot grows with the moment of the load about it, while the available friction grows only with the load. Each problem is then the same three steps: decide the direction of the impending motion, put friction against it, and check the assumption, whether that is F \le \mu_sN at the other contacts, the reaction staying under the base, or the other modes needing more load. Dry friction is a limit, not a value: the reaction may lean anywhere inside the friction cone, \tan\phi_s = \mu_s, and a body slips, tips or holds according to whether equilibrium can keep it there.

• Central idea: equilibrium decides the friction; \mu_sN only bounds it.
• Key relationship: F \le \mu_sN at rest, F = \mu_kN sliding, \tan\phi = \mu; for the ladder s = (14\mu\sin\theta - \cos\theta)/(3\cos\theta).
• Most important assumption: Coulomb's model with constant coefficients, which are uncertain to tens of percent; a design keeps clear of the limit.
• Open question: a wedge or a screw is a machine built on friction: the friction angle tells you the force to drive it and whether it will hold its load by itself (S27).

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). S26_dry_friction.py checks the crate and the incline. It tabulates Harper's ladder s(\theta, \mu) and solves Harper's three blocks two ways. The first is his: the two slip modes and their crossing, found with brentq. The second needs no guess at all. It is a linear programme (linprog) that maximises P, with every friction force allowed anywhere between -\mu N and +\mu N.

Part 1: one block
  crate: N = 490.5 N, F_max = 245.25 N, F_k = 196.2 N
  pushed with 150 N the reaction leans 17.0 deg, inside phi_s = 26.57 deg
  least pull, at phi_s = 21.8 deg: 182.2 N
  100 kg on 20 deg: m0 from 6.01 to 62.4 kg
  crate pushed: tips if pushed above 0.75 m
Part 2: Harper's ladder, s in metres up the 4 m ladder
  theta    mu 0.2   mu 0.25  mu 0.4   mu 0.6
   60.00    1.28    1.69    2.90    4.52
   67.98    1.97    2.55    4.28    6.59
   75.96    3.40    4.33    7.13   10.86
  mu 0.25: slips empty below 15.95 deg, safe to the top above 74.93
Part 3: Harper's three blocks on an incline
  modes cross at 0.503 rad (28.8 deg), P = 116.1 N; P2 = 0 at 35.0 deg
  theta   P1 (50 alone)   P2 (50 + 40)   min   linprog
   20.0          210.2          278.8   210.2    210.2
   28.8          116.2          116.2   116.2    116.2
   30.0          103.1           93.8    93.8     93.8
   32.0           81.2           56.2    56.2     56.3

Values of s above 4\,\mathrm m mean the painter can reach the top. The linear programme agrees with the smaller mode at every angle. Its last figure differs at 32^\circ only because it uses the exact coefficients, while P_1 and P_2 use their four-figure values. The same rounding makes the crossing 116.1\,\mathrm N, against Nugget 07's 116.0.

MATLAB (Octave-compatible). S26_dry_friction.m does the same with fzero and min. linprog needs the Optimization Toolbox, so the MATLAB view keeps Harper's two-mode route.

m0 from 6.01 to 62.4 kg
  theta  mu 0.2 mu 0.25  mu 0.4  mu 0.6
  60.00    1.28    1.69    2.90    4.52
  67.98    1.97    2.55    4.28    6.59
  75.96    3.40    4.33    7.13   10.86
modes cross at 0.503 rad (28.8 deg), P = 116.1 N
theta 20.0: P1  210.2, P2  278.8, largest P  210.2 N
theta 28.8: P1  116.2, P2  116.2, largest P  116.2 N
theta 30.0: P1  103.1, P2   93.8, largest P   93.8 N
theta 32.0: P1   81.2, P2   56.2, largest P   56.2 N

Traps

• Writing F = \mu_sN by default. Only at impending slip; otherwise solve for F and check F \le \mu_sN. Counterexample: an unpushed box has F = 0. (From Nugget 01.)
• Using \mu_s for a body that is sliding. Sliding friction is \mu_kN; SP 6/3 case II has 162.5\,\mathrm N, not 191.2\,\mathrm N. (From Nugget 03.)
• Taking N = mg on a slope or under a slanted pull. N comes from the normal equation: 1093\,\mathrm N, not 981\,\mathrm N, in SP 6/3 case I. (From Nugget 03.)
• Giving one limit when slip can go two ways. SP 6/2 has 6.01 \le m_0 \le 62.4\,\mathrm{kg}. (From Nugget 04.)
• Friction without a normal force. A rough wall against a smooth floor cannot hold a ladder: with no horizontal force at the floor, the wall's normal force and so its friction are zero. (From Nugget 05.)
• Dropping a friction force from the moment equation. A friction force that does not pass through the moment centre has a moment; leaving it out is Marghitu's Ex 5.2 slip. (From Nugget 05.)
• Assuming a block slides when it may tip. Check where the reaction must act; it must lie under the base. (From Nugget 06.)
• Assuming the slip surface without checking the others. SP 6/5's first guess fails at the bottom surface. (From Nugget 07.)
• Quoting a coefficient to three figures as exact. Coefficients vary with the surfaces; three figures are for checking arithmetic only (Meriam 6/10). (From Nugget 02.)

Self-check

1. Meriam's Problem 6/2: a 700\,\mathrm N force pulls a 100\,\mathrm{kg} block, at rest, at 30^\circ above the horizontal; \mu_s = 0.80, \mu_k = 0.60. Find the magnitude and direction of the friction force.
2. Meriam's Problem 6/3: a 50\,\mathrm{kg} block rests on a 15^\circ incline, \mu_s = 0.25, \mu_k = 0.20; a force P acts at 20^\circ above the incline, pointing up it. Find the friction for (a) P = 0, (b) P = 200\,\mathrm N, (c) P = 250\,\mathrm N, and (d) the P to start the block up the incline.
3. Meriam's Problem 6/4: skis on snow have \mu_s = 0.10, \mu_k = 0.08. What slope keeps a skier's speed constant?
4. Meriam's Problem 6/11 (US units in the book): a 100\,\mathrm{lb} block is pressed against a vertical wall (\mu_s = 0.50, \mu_k = 0.40) by a 120\,\mathrm{lb} force pushing up towards the wall at \theta above the horizontal. Convert to SI and find the friction force for (a) \theta = 15^\circ, (b) \theta = 30^\circ.
5. Marghitu's Problem 5.1: at what angle \theta above the floor should a rope pull a block of mass m (\mu_s = 0.4) to start it with the smallest force, and how large is that force?
6. Meriam's SP 6/2 with new numbers: the 100\,\mathrm{kg} block on a 25^\circ incline, \mu_s = 0.35. Find the range of m_0.
7. Meriam's Problem 6/10: a uniform 7\,\mathrm m, 100\,\mathrm{kg} pole rests on the corner A of a 3\,\mathrm m high ledge and on the floor at B, 4\,\mathrm m horizontally from the ledge, \mu_s = 0.40 at both. Find the horizontal force P at B, pushing towards the ledge, that starts the pole moving.
8. Harper's ladder with \mu = 0.4 at \theta = 60^\circ: how far can the painter climb?
9. Marghitu's Problem 5.2: a car (m = 2000\,\mathrm{kg}, wheelbase l = 1.75\,\mathrm m, centre of mass midway and h = 0.5\,\mathrm m above the road, \mu = 0.3) drives slowly down a slope. Does it slip or tip first, and at what slope?
10. Meriam's Problem 6/30: a 5\,\mathrm{kg} block B sits on the 15^\circ face of a 10\,\mathrm{kg} block A (the face rises in the direction of the push) and is pushed by a horizontal P = 50\,\mathrm N; A rests on the floor. Where does slip occur if (a) \mu_1 = 0.40 (floor), \mu_2 = 0.50 (A–B); (b) \mu_1 = 0.30, \mu_2 = 0.60? Take \mu_k = 0.75\mu_s.
11. Harper's three blocks at \theta = 20^\circ and at 32^\circ: find the largest P and the slip mode.
12. Marghitu's Problem 5.3: a uniform bar leans with its top A on a rough wall (\mu = 0.4) and its foot B on a smooth floor, 0.3\,\mathrm m from the wall. Will it stay when released?

Fully worked solutions

1. Assume rest: N = 981 - 700\sin30^\circ = 631.0\,\mathrm N; needed F = 700\cos30^\circ = 606.2\,\mathrm N; available 0.80 \times 631.0 = 504.8\,\mathrm N. Not enough: the block slides right, and F = 0.60 \times 631.0 = 378.6\,\mathrm N to the left (Meriam: 379\,\mathrm N left).

2. mg\sin15^\circ = 126.95\,\mathrm N, mg\cos15^\circ = 473.8\,\mathrm N. (a) N = 473.8\,\mathrm N, F_{\max} = 118.45 < 126.95\,\mathrm N: it slides down, F = 0.20 \times 473.8 = 94.8\,\mathrm N up the incline. (b) N = 473.8 - 200\sin20^\circ = 405.4\,\mathrm N; needed F = 200\cos20^\circ - 126.95 = 61.0\,\mathrm N down; F_{\max} = 101.3\,\mathrm N: it holds, F = 61.0\,\mathrm N down the incline. (c) N = 388.3\,\mathrm N; needed 108.0\,\mathrm N down, F_{\max} = 97.1\,\mathrm N: it slides up, F = 0.20 \times 388.3 = 77.7\,\mathrm N down. (d) P\cos20^\circ = 126.95 + 0.25(473.8 - P\sin20^\circ), P = 245.4/1.0252 = 239.4\,\mathrm N (Meriam: 239\,\mathrm N).

3. At constant speed the skier slides, so kinetic friction balances the slope: \tan\theta = \mu_k = 0.08, \theta = 4.57^\circ. (\mu_s only tells you the skier, once stopped, stays stopped below 5.71^\circ.)

4. W = 444.8\,\mathrm N, P = 533.8\,\mathrm N. (a) N = P\cos15^\circ = 515.6\,\mathrm N; needed F = 444.8 - 533.8\sin15^\circ = 306.7\,\mathrm N up; F_{\max} = 257.8\,\mathrm N: it slides down, F = 0.40 \times 515.6 = 206.2\,\mathrm N up (46.4\,\mathrm{lb}). (b) N = 462.3\,\mathrm N; needed 444.8 - 266.9 = 177.9\,\mathrm N up; F_{\max} = 231.1\,\mathrm N: holds, F = 177.9\,\mathrm N up (40\,\mathrm{lb}).

5. P\cos\theta = \mu(mg - P\sin\theta) gives P = \mu mg/(\cos\theta + \mu\sin\theta), smallest when the denominator is largest, at \tan\theta = \mu: \theta = \phi_s = 21.8^\circ, and P = mg\sin\phi_s = 0.371mg. (In the friction-angle picture, P is perpendicular to R.)

6. N = 981\cos25^\circ = 889.1\,\mathrm N, F_{\max} = 311.2\,\mathrm N, mg\sin25^\circ = 414.6\,\mathrm N. Up: m_0 = (414.6 + 311.2)/9.81 = 73.98\,\mathrm{kg}. Down: m_0 = (414.6 - 311.2)/9.81 = 10.54\,\mathrm{kg}. Range 10.5–74.0\,\mathrm{kg}.

7. AB = 5\,\mathrm m (a 3-4-5 triangle), unit vector B \to A (-0.8, 0.6), G 3.5\,\mathrm m from B at (-2.8, 2.1)\,\mathrm m. The pole slides up towards A: friction at B points away from the ledge, at A down along the pole. The normal at A is perpendicular to the pole, (0.6, 0.8)N_A. Moments about B (friction at A passes through B): 5N_A = 2.8 \times 981, N_A = 549.4\,\mathrm N. \Sigma F_y: N_B = 981 - 0.8(549.4) + 0.6(0.40)(549.4) = 673.4\,\mathrm N. \Sigma F_x: P = 0.40(673.4) + 549.4(0.6 + 0.40 \times 0.8) = 774.8\,\mathrm N (Meriam: 775\,\mathrm N).

8. s = (14 \times 0.4 \times 0.8660 - 0.5)/(3 \times 0.5) = (4.850 - 0.5)/1.5 = 2.90\,\mathrm m.

9. Slip when \tan\theta = \mu = 0.3: 16.7^\circ. Tip when the weight's line passes the lower wheels, \tan\theta = (l/2)/h = 1.75: 60.3^\circ. It slips first, at 16.7^\circ.

10. On B, normal to the face: N = 49.05\cos15^\circ + 50\sin15^\circ = 60.32\,\mathrm N; along the face, B needs 35.60\,\mathrm N of friction down the face to stay. (a) 0.50 \times 60.32 = 30.16 < 35.60\,\mathrm N: B slides up A. Then B pushes A horizontally with 60.32\sin15^\circ + 0.375(60.32)\cos15^\circ = 37.5\,\mathrm N, and the floor can hold 0.40(98.1 + 52.4) = 60.2\,\mathrm N: A stays. Slip between A and B. (b) 0.60 \times 60.32 = 36.19 > 35.60\,\mathrm N: B holds on A. Together: 50\,\mathrm N > 0.30 \times 147.15 = 44.1\,\mathrm N: both slide on the floor.

11. 20^\circ: P_1 = 210.2\,\mathrm N, P_2 = 278.8\,\mathrm N; P = 210.2\,\mathrm N, the middle block alone. 32^\circ: P_1 = 81.2\,\mathrm N, P_2 = 56.3\,\mathrm N; P = 56.3\,\mathrm N, the middle and bottom blocks together.

12. No. The smooth floor gives only a vertical force, so \Sigma F_x = 0 makes the wall's normal force zero, and with it the wall friction. The bar then has only two vertical forces, its weight at the middle and the floor's push at B, 0.15\,\mathrm m apart: they form a couple and the bar slides down, whatever \mu is.