Study Companion: S28 — Journal Bearings, Thrust Bearings and Disk Friction

S26 and S27 put friction at a point or along a face of known direction. In a bearing, a clutch or a brake, two surfaces turn against each other, and what matters is not a friction force but a friction moment: how hard it is to turn a shaft in its bearing, or how much torque a clutch can pass before it slips. This lesson builds that moment two ways. For a shaft turning in a loose journal bearing the friction force acts at one line of contact, and a small circle, the friction circle, captures it. For flat surfaces pressed together over an area, the friction acts everywhere at once, and the moment is an integral over the contact.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 6, articles 6/6 ("Journal Bearings") and 6/7 ("Thrust Bearings; Disk Friction"), with Sample Problem 6/8. Neither MATLAB reference has a section on this topic; Marghitu's thrust-collar formula from S27 is reconciled here. Units SI throughout, g = 9.81\,\mathrm{m/s^2}. As in S27 (Meriam's Section B), \mu is the static or kinetic coefficient as the case needs, and \phi = \tan^{-1}\mu.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A car's disk brake (Meriam's Problem 6/78) squeezes a flat rotor between two pads. Each pad is pressed on by a piston force P = 5\,\mathrm{kN}, and covers the rotor between radii R_i = 100\,\mathrm{mm} and R_o = 150\,\mathrm{mm}; the pad material gives \mu = 0.4 on the rotor. A parts catalogue offers two pad shapes for the same caliper: a narrow pad spanning 40^\circ of the rotor and a wide one spanning 80^\circ.

With the same piston force, which pad stops the car harder? The plausible guess is "the wide one: twice the area rubbing, twice the friction". The pads of a heavy van and of a small car of the same rotor size really do differ in area, so the answer matters. It is in Nugget 04.

Nugget 01: A Turning Shaft Climbs Its Bearing Until the Reaction Touches the Friction Circle

Claim. A dry shaft in a loose bearing rolls up the bearing's inner surface as it starts to turn, and stops climbing where it slips. There the bearing's reaction leans from the radius by the friction angle, so its line misses the centre by r_f = r\sin\phi, and the moment to keep the shaft turning is L times that miss distance.

The picture. Meriam's Fig. 6/8: a shaft of radius r carrying a radial load L, turning in a bearing with exaggerated clearance. The contact point A has moved round the bearing; the reaction R at A is tilted \phi from the radius OA, so it is tangent to a small circle of radius r_f about O.

Check with numbers. A shaft of radius 25\,\mathrm{mm} carries L = 10\,\mathrm{kN} in a dry bearing with \mu = 0.15 (\phi = 8.53^\circ). The friction circle has r_f = 25\sin8.53^\circ = 3.709\,\mathrm{mm}, so turning the shaft steadily takes M = 10\,000 \times 0.003709 = 37.09\,\mathrm{N{\cdot}m}. The shortcut \mu Lr gives 37.5\,\mathrm{N{\cdot}m}, 1.1\,\% high.

The idea. Meriam (6/6): as the shaft begins to turn, it rolls up the inside of the bearing like a wheel on a slope, until it slips; it then stays at about that position. Three things act on the shaft: the load L, the driving moment M and the bearing's reaction R at A. For vertical equilibrium R = L in size and direction, but R is not collinear with L: the two form a couple that balances M. The normal part N of R points along the radius AO, and the friction part F, against the slip, tilts R from N by \phi. Moments about A, or about O:

For small \phi, \sin\phi \approx \tan\phi = \mu, so

This is for dry or partly lubricated journals; a fully lubricated one is a fluid-friction problem (S26, Nugget 01).

The equation (physical law with the friction limit; 6/4a an approximation): M = Lr\sin\phi \approx \mu Lr.

Assumptions. Line contact at one point; rigid shaft and bearing; steady, slow rotation; clearance small enough that L and R act at the same radius r.

Worked number. Meriam's Problem 6/66 (self-check 1) runs this backwards: a measured 3\,\mathrm{N{\cdot}m} on a shaft carrying 784.8\,\mathrm N gives r_f = 3.82\,\mathrm{mm}, and with r = 20\,\mathrm{mm}, \sin\phi = 0.191, \mu = 0.195.

Real example. The friction moment grows with the journal radius, which is why shafts are necked down to small journals at the bearings, and why a hand winch with a fat axle is hard work.

Plausible wrong answer. "Friction is \mu L acting at the shaft surface, so M = \mu Lr exactly." The friction force is L\sin\phi, not \mu L, because the normal force is L\cos\phi: M = Lr\sin\phi.

Check. At what \mu does the shortcut \mu Lr overestimate by 5\,\%? (1/\cos\phi = 1.05: \phi = 17.8^\circ, \mu = 0.32.)

Narration spine. A shaft in an exaggerated bearing. It starts to turn and rolls up the bore; the contact point stops climbing. The reaction arrow at the contact leans by \phi, and a small dashed circle appears, just touched by the arrow's line.

Bridge. The friction circle reduces a bearing to one geometric fact, the line of the reaction. That is all a three-force problem needs.

Nugget 02: The Friction Circle Turns a Bearing Problem Into Three Concurrent Forces

Claim. A part that pivots on a rough pin carries the pin's reaction along a line tangent to the friction circle. With two other known lines of action, the three forces must meet at a point, and the force triangle gives the answer, once for each direction of impending rotation.

The picture. Meriam's SP 6/8: a bell crank on a fixed 100\,\mathrm{mm} shaft, \mu = 0.20. A vertical P = 100\,\mathrm N acts 180\,\mathrm{mm} to the right of the shaft centre O, a horizontal T acts 120\,\mathrm{mm} above it; their lines meet at C.

Check with numbers. \phi = \tan^{-1}0.20 = 11.31^\circ and r_f = 50\sin11.31^\circ = 9.81\,\mathrm{mm}. The line OC is \sqrt{120^2 + 180^2} = 216.3\,\mathrm{mm} long at \theta = \tan^{-1}(120/180) = 33.7^\circ to the horizontal. The reaction R passes through C and touches the friction circle, so it is turned from OC by \beta = \sin^{-1}(9.81/216.3) = 2.60^\circ. Counter-clockwise rotation impending (large T): R is at \theta - \beta to the horizontal and the triangle gives T_1 = P\cot(\theta - \beta) = 100\cot31.1^\circ = 165.8\,\mathrm N. Clockwise impending (small T): R swings to the other side, T_2 = 100\cot36.3^\circ = 136.2\,\mathrm N. Without friction R would pass through O and T = 150\,\mathrm N.

The idea. Meriam (SP 6/8): impending rotation of the crank on the fixed shaft makes the reaction lean \phi from the normal, so it is tangent to the friction circle, on the side whose moment about O opposes the impending rotation. Three forces in equilibrium are concurrent (S11), so R goes through C. Check by moments about O: the reaction's moment is Rr_f; with R = \sqrt{T^2 + P^2}, 100(180) - 165.83(120) = -1899\,\mathrm{N{\cdot}mm} = -Rr_f and 100(180) - 136.20(120) = +1656\,\mathrm{N{\cdot}mm} = +Rr_f (with R = 193.6 and 169.0\,\mathrm N).

The equation (geometric relationship: three-force concurrency with R tangent to the friction circle): T = P\cot(\theta \mp \beta), \sin\beta = r_f/OC.

Assumptions. Crank's weight neglected; dry pin, contact on one line.

Worked number. On a well-greased pin, \mu = 0.05: r_f = 2.50\,\mathrm{mm}, \beta = 0.66^\circ, and T ranges only from 146.3 to 153.8\,\mathrm N (our numbers, same method).

Real example. Pulleys on fixed axles: the tension needed to raise a load exceeds the load's tension by the bearing friction, and lowering needs less (Problems 6/68–6/69, self-check 3).

Plausible wrong answer. "Pin friction acts at the pin, so it cannot change the crank's force balance." It moves the reaction's line off the centre, which changes the moments.

Check. If the shaft were frictionless, where would R's line pass? (Through O and C.)

Narration spine. The crank with its two arrows meeting at C. A line from C to O, then the friction circle; the line pivots about C until it just grazes the circle on one side, the force triangle redraws, T reads 165.8\,\mathrm N; it pivots to the other side, 136.2\,\mathrm N.

Bridge. A journal touches along a line. A thrust bearing, a clutch or a brake presses flat surfaces together over an area, and the friction is spread over every part of it.

Nugget 03: Friction Over a Flat Contact: Sum the Moments of Thin Rings

Claim. When two flat circular surfaces are pressed together by an axial force P, each small patch carries friction \mu p\,dA at its own radius. Adding their moments over the contact gives the moment to slip one surface on the other; for uniform pressure it is \frac23\mu PR.

The picture. Meriam's Fig. 6/9: two disks on shafts, pressed together by P. On one face a patch of area dA = r\,dr\,d\theta at radius r carries the friction \mu p\,dA, tangent to its circle, with moment \mu pr\,dA about the axis.

Check with numbers. A clutch disk of radius R = 100\,\mathrm{mm}, pressed with P = 2\,\mathrm{kN}, \mu = 0.3. If the friction acted at the rim it would give 0.3 \times 2000 \times 0.1 = 60\,\mathrm{N{\cdot}m}; at the centre, nothing. Spread uniformly, the total friction \mu P = 600\,\mathrm N has moment \frac23 \times 600 \times 0.1 = 40\,\mathrm{N{\cdot}m}: it acts, in effect, at \frac23R = 66.7\,\mathrm{mm}.

The idea. Meriam (6/7): the maximum torque the clutch can pass is the moment M needed to slip one disk on the other:

To evaluate it, \mu and p must be known as functions of r. For new, flat, well-supported surfaces take \mu constant and p uniform, p = P/\pi R^2:

Step the source compresses: the inner integral is R^3/3, the outer 2\pi, so M = (\mu P/\pi R^2)(2\pi R^3/3). The rings near the rim, having both more area and a longer arm, dominate. For a collar bearing, a ring from R_i to R_o (Fig. 6/10), the same integral between those limits, with p = P/\pi(R_o^2 - R_i^2), gives

The equation (physical law summed over the contact, with the material model of uniform pressure): (6/5), (6/5a).

Assumptions. Constant \mu; uniform pressure (new, flat, well supported surfaces); slip over the whole face at once.

Worked number. A collar from R_i = 30 to R_o = 60\,\mathrm{mm} carrying 5\,\mathrm{kN}, \mu = 0.2: M = \frac23(0.2)(5000)(0.000189/0.0027) = 46.67\,\mathrm{N{\cdot}m}. S27's simpler model, with the load concentrated at the mean diameter d_c = 90\,\mathrm{mm} (Marghitu's Eq. 5.18, \mu Pd_c/2), gives 45.0\,\mathrm{N{\cdot}m}, 3.7\,\% less; Nugget 05 shows why the two differ.

Real example. A drill-press quill or a vertical turbine shaft rests on a flat pivot or collar bearing; the moment above is what the motor loses to it.

Plausible wrong answer. "Friction \mu P acts at the mean radius R/2, so M = \frac12\mu PR." For uniform pressure the outer rings carry more of the force and the effective radius is \frac23R.

Check. Where does the moment of a full disk come from? (Rings beyond 0.794R give half of it: (r/R)^3 = \frac12.)

Narration spine. A disk face fills with tiny friction arrows, all tangent to their circles, longer arms towards the rim. They gather into one arrow \mu P sitting on the circle of radius \frac23R.

Bridge. A brake pad covers only part of the ring. Does the part it misses matter?

Nugget 04: Brakes and Clutches: the Clamping Force and the Radii Set the Torque

Claim. For a pad or plate under uniform pressure, the friction moment depends on the clamping force and the inner and outer radii only. The pad's angular span cancels, so a larger pad gives no more torque; it only lowers the pressure.

The picture. The puzzle's rotor with a pad of angular span \beta between R_i and R_o. The pressure is the force over the pad area, p = P/[\frac12\beta(R_o^2 - R_i^2)]; a wider pad spreads the same P thinner.

Check with numbers. For one pad the integral of Nugget 03 over \theta from 0 to \beta instead of 2\pi:

The \beta in p cancels the \beta from the integral: it is the collar formula (6/5a) again. With (R_o^3 - R_i^3)/(R_o^2 - R_i^2) = 0.19\,\mathrm m, one pad gives \frac23(0.4)(5000)(0.19) = 253.3\,\mathrm{N{\cdot}m}, two pads 506.7\,\mathrm{N{\cdot}m}, for the 40^\circ pad and for the 80^\circ pad alike. The pressure is 1.146\,\mathrm{MPa} under the narrow pad and 0.573\,\mathrm{MPa} under the wide one.

The idea. This is the puzzle's answer, and S26's Coulomb law at a new scale: friction depends on the normal force, not the apparent area. Doubling the area halves the pressure, and the moment, summed over twice as many patches each pressed half as hard, is unchanged. What does change the moment is where the force sits: the effective radius here is 506.7/(2 \times 0.4 \times 5000) = 126.7\,\mathrm{mm}, a little outside the mid-radius 125\,\mathrm{mm}. (Meriam's Problem 6/78 asks for exactly this.) In a multiple-plate clutch each pair of rubbing faces adds the same moment, so n pairs pass n times the torque of one.

The equation (physical law summed over the pad, uniform pressure): two pads, M = \frac43\mu P(R_o^3 - R_i^3)/(R_o^2 - R_i^2), independent of \beta.

Assumptions. Uniform pressure over each pad (Meriam's problem asks what changes if it varies with \theta: the moment depends on p only through its total over each radius, so a \theta-variation alone changes nothing); constant \mu.

Worked number. To stop the rotor with 800\,\mathrm{N{\cdot}m} the caliper needs P = 800/(\frac43 \times 0.4 \times 0.19) = 7.89\,\mathrm{kN} per piston, whatever the pad size.

Real example. Larger pads exist for heat and wear: the same work is spread over more material and the pressure is lower. Torque comes from a larger rotor (larger radii) or a harder clamp.

Plausible wrong answer. "Twice the pad area, twice the braking." The friction follows the force, not the area (S26, Nugget 02).

Check. If the rotor's outer radius grew to 170\,\mathrm{mm} with the same pads' inner radius, would the moment rise or fall? (Rise: (0.17^3 - 0.1^3)/(0.17^2 - 0.1^2) = 0.2070\,\mathrm m.)

Narration spine. A rotor with a narrow pad; the friction arrows over it, and the torque gauge. The pad widens to twice the angle; the arrows halve in length, double in number, and the gauge does not move.

Bridge. All of this assumed uniform pressure, true of new surfaces. Surfaces that have run for a while have worn, and wear changes the pressure.

Nugget 05: Worn Surfaces Settle to a Pressure That Falls With Radius, and Lose a Quarter of Their Moment

Claim. Wear is fastest where the rubbing distance and the pressure are largest, so the outer parts of a face wear away until the product of pressure and radius is the same everywhere. That shifts the load inwards, and the friction moment falls to \frac12\mu PR, three quarters of the new value.

The picture. Pressure across a disk face: flat for new surfaces; after wearing-in, p = K/r, high at the centre and low at the rim. The total load P is the same in both.

Check with numbers. The clutch of Nugget 03, worn in: K = P/2\pi R = 2000/(2\pi \times 0.1) = 3183\,\mathrm{N/m}, so p = 0.318\,\mathrm{MPa} at r = 10\,\mathrm{mm} and less further out. The moment is \frac12(0.3)(2000)(0.1) = 30\,\mathrm{N{\cdot}m}, against 40\,\mathrm{N{\cdot}m} new: \frac34 as much.

The idea. Meriam (6/7): after the initial wearing-in the surfaces keep their shape, so further wear is the same everywhere. Wear depends on the circumferential distance travelled, which is proportional to r, and on the pressure: so rp = K, a constant. The axial balance fixes K:

With pr = K = P/2\pi R:

that is \frac12 \big/ \frac23 = \frac34 of the new value. For a worn collar the same limits R_i to R_o give

Reconciliation with S27. With the mean diameter d_c = R_o + R_i, Eq. 6/6a is \mu Pd_c/2: Marghitu's collar formula (Eq. 5.18) is exactly Meriam's worn-in collar. For Nugget 03's collar that is 45.0\,\mathrm{N{\cdot}m}, against 46.67\,\mathrm{N{\cdot}m} new; for the puzzle's brake, 500.0\,\mathrm{N{\cdot}m} worn against 506.7 new. A narrow ring hardly cares which model is used; a full disk loses a quarter.

Meriam adds that other pressure laws are handled the same way: write p(r), fix its constant from P = \int p\,dA, and integrate \mu pr\,dA (self-checks 9 and 10).

The equation (material model: uniform wear, pr = K; then physical law summed): (6/6), (6/6a).

Assumptions. Wear proportional to pressure times sliding distance; surfaces kept in full contact; constant \mu.

Worked number. A worn clutch must still pass 40\,\mathrm{N{\cdot}m}: the spring must now press with P = 40/(\frac12 \times 0.3 \times 0.1) = 2.67\,\mathrm{kN} instead of 2\,\mathrm{kN}.

Real example. Clutch designers rate torque capacity on the worn-in formula, the conservative one, since a clutch spends nearly all its life worn in.

Plausible wrong answer. "Wear smooths the surfaces, so the coefficient drops and that is why the moment falls." The coefficient was held constant; the fall comes from the load moving towards the centre, where its arm is short.

Check. Why is the worn pressure infinite at r = 0, and does it matter? (The model pr = K; the load in a small central circle, 2\pi Kr_0, still tends to zero, so the integrals are finite. Real pivots are often relieved at the centre.)

Narration spine. A pressure profile across the disk, flat. As the disk spins, the rim glows with wear and the profile sinks at the edges and rises at the centre into a 1/r curve; the effective-radius marker slides from \frac23R to \frac12R.

Synthesis

The wide brake pad stops the car no harder than the narrow one, because friction follows the clamping force, not the area, and the moment depends only on how far out that force acts. Every result here is that one idea, the moment of the friction about the axis, summed over wherever the contact is: on a journal it sits on one line, tangent to the friction circle; on a flat face it is spread over the area and integrates to an effective radius, \frac23R new and \frac12R worn. A friction moment is the friction force times an effective arm: r\sin\phi for a journal, \frac23R or \frac12R for a flat disk, set by where the normal pressure sits, never by the contact area.

• Central idea: sum \mu pr\,dA over the contact; for a journal the sum is Lr_f.
• Key relationship: M = Lr\sin\phi \approx \mu Lr; M = \frac23\mu PR new, \frac12\mu PR worn; collars (6/5a), (6/6a).
• Most important assumption: the pressure law, uniform or uniform wear; the truth for a given bearing lies between, which is why the two answers bracket the design.
• Open question: friction can also act along a curved contact that wraps round a drum, where the normal force grows with the tension itself: belts and ropes (S29).

Computed views

Neither MATLAB reference covers bearings or disk friction. These views are our own, built on the lesson's Meriam examples, in both languages so you can work in whichever you prefer. Both print the same numbers.

Python (numpy, scipy). S28_bearings_and_disks.py uses the friction circle for the journal bearing and the pulley on a pin, at two coefficients. It integrates \mu p r\,dA for the clutch, the collar and the brake, with uniform pressure for a new surface and pr constant for a worn one.

Part 1: journal bearings
  25 mm shaft, 10 kN, mu 0.15: r_f = 3.709 mm, M = 37.09 N m (mu L r = 37.5)
  P 6/66 backwards: r_f = 3.82 mm, mu = 0.195
  pulley on a pin, mu 0.2: T from 136.2 to 165.8 N (no friction 150)
  pulley on a pin, mu 0.05: T from 146.3 to 153.8 N (no friction 150)
Part 2: disks and collars (N m)
  clutch R = 100 mm, 2 kN, mu 0.3: new 40.00, worn 30.00
  collar 30 to 60 mm, 5 kN, mu 0.2: 46.67 (mean-radius model 45.00)
  brake 800 N m at mu 0.4, r 0.19 m: P = 7.89 kN
  worn clutch passing 40 N m: P = 2.67 kN

MATLAB (Octave-compatible). S28_bearings_and_disks.m does the same with integral.

journal: r_f = 3.709 mm, M = 37.09 N m
pulley mu 0.20: T from 136.2 to 165.8 N
pulley mu 0.05: T from 146.3 to 153.8 N
clutch: new 40.00 N m, worn 30.00 N m
collar: 46.67 N m

Traps

• Writing the journal friction as \mu L. The friction is L\sin\phi; M = Lr\sin\phi, and \mu Lr only approximates it. (From Nugget 01.)
• Putting the friction-circle tangent on the wrong side. The reaction's moment about the centre must oppose the impending rotation; each direction gives its own answer. (From Nugget 02.)
• Placing disk friction at the mean radius for a new full disk. Uniform pressure gives \frac23R. (From Nugget 03.)
• Expecting a larger pad or plate area to raise the torque. Only force and radii count. (From Nugget 04.)
• Forgetting each rubbing pair in a multi-plate clutch or a two-pad brake. Each pair of faces adds its moment. (From Nugget 04.)
• Using the new-surface formula for a worn bearing. Worn surfaces give \frac34 as much for a full disk. (From Nugget 05.)
• Choosing the pressure constant before checking the axial balance. Fix it from P = \int p\,dA first. (From Nugget 05.)

Self-check

1. Meriam's Problem 6/66: two 40\,\mathrm{kg} flywheels on a shaft of 40\,\mathrm{mm} diameter run in one journal bearing between them; 3\,\mathrm{N{\cdot}m} keeps them turning slowly. Find \mu and the radius of the friction circle.
2. Meriam's Problem 6/65: a hoisting drum (drum and shaft 100\,\mathrm{kg}) raises 500\,\mathrm{kg} at constant speed on a cable at 300\,\mathrm{mm} radius, with 1510\,\mathrm{N{\cdot}m} on its 50\,\mathrm{mm} shaft. Find \mu for the bearing.
3. Meriam's Problem 6/68: an 800\,\mathrm{kg} load hangs from a pulley of 300\,\mathrm{mm} diameter on a 30\,\mathrm{mm} bearing, \mu = 0.25; one end of the cable is fixed above (T_0), the other is pulled up (T). Find T and T_0 to raise the load.
4. Meriam's Problem 6/69: the same, lowering the load.
5. Meriam's Problems 6/74–6/75: two pulleys fixed together (radii r and kr) turn on a bearing of radius r_0; a cylinder of mass m hangs from the small one, the tension T pulls down on the large one; the unit has mass m_0. Derive T to raise and to lower m steadily (use \mu for \sin\phi) and evaluate for m = 50\,\mathrm{kg}, m_0 = 30\,\mathrm{kg}, r = 0.3\,\mathrm m, k = \frac12, r_0 = 25\,\mathrm{mm}, \mu = 0.15.
6. Meriam's Problem 6/70: a 20\,\mathrm{kg} ring (inner radius 50, outer 60\,\mathrm{mm}) hangs on a fixed horizontal shaft of 40\,\mathrm{mm} radius. A downward P = 150\,\mathrm N at the ring's outer edge, level with its centre, just makes it slip. Find \mu.
7. Meriam's Problem 6/67 (US units in the book): disk A (9\,\mathrm{in} diameter) is pressed on disk B (12\,\mathrm{in}) by 80\,\mathrm{lb}, uniform pressure; \mu = 0.40 between them. Convert to SI; find the moment to slip A on B, and the least \mu between B and its support C to stop B turning.
8. Meriam's Problem 6/76: a disk sander of radius R presses on the end of a thin board, with force P, uniform pressure, the board end spanning radii R/4 to 3R/4. Find the moment to drive the disk.
9. Meriam's Problem 6/77: two mating disks, contact from r_i to r_o, pressure p = k/r^2. Find M in terms of the axial load L.
10. Meriam's Problem 6/79: a sanding disk of radius a under axial force L, with pressure falling linearly from p_0 at the centre to p_0/2 at the rim. Find M.
11. Meriam's Problem 6/73: a 10\,\mathrm{Mg} crate on a 3\,\mathrm{Mg} platform is lowered by two screws (0.9\,\mathrm{Mg} each, mean diameter 120\,\mathrm{mm}, single square thread, lead 11\,\mathrm{mm}), each hanging from a flat collar bearing (250\,\mathrm{mm} outside, 125\,\mathrm{mm} inside, uniform pressure); \mu = 0.15 throughout. Find the moment on each screw to raise and to lower.
12. Meriam's Problem 6/82: a conical pivot of vertex angle \alpha carries a thrust L on its surface between diameters d_1 and d_2, with uniform pressure. Find M.

Fully worked solutions

1. L = 80(9.81) = 784.8\,\mathrm N; r_f = M/L = 3/784.8 = 3.82\,\mathrm{mm}. \sin\phi = 3.82/20 = 0.1911, \phi = 11.02^\circ, \mu = \tan\phi = 0.1947.

2. Lifting needs 500(9.81)(0.3) = 1471.5\,\mathrm{N{\cdot}m}, so friction takes 38.5\,\mathrm{N{\cdot}m}. The bearing carries the cable's 4905\,\mathrm N and the drum's 981\,\mathrm N, both vertical: L = 5886\,\mathrm N. \sin\phi = 38.5/(5886 \times 0.025) = 0.2616, \mu = \tan\phi = 0.271 (the shortcut would give 0.262).

3. Both cable ends are vertical, so the bearing load is T + T_0 = 7848\,\mathrm N, and r_f = 15\sin14.04^\circ = 3.638\,\mathrm{mm}. Raising, the reaction's moment opposes rotation: (T - T_0)(0.150) = 7848(0.003638), T - T_0 = 190.3\,\mathrm N. So T = 4020\,\mathrm N, T_0 = 3829\,\mathrm N (Meriam: 3830).

4. The rotation reverses and so does the friction moment: T_0 - T = 190.3\,\mathrm N, so T = 3830\,\mathrm N, T_0 = 4020\,\mathrm N.

5. The bearing carries mg + m_0g + T (all vertical). Raising: Tr = mg\,kr + \mu r_0(mg + m_0g + T), so T = [kmg + \mu(r_0/r)(m + m_0)g]/(1 - \mu r_0/r) = (245.25 + 9.81)/0.9875 = 258\,\mathrm N. Lowering, the friction term changes sign: T = [kmg - \mu(r_0/r)(m + m_0)g]/(1 + \mu r_0/r) = 235.44/1.0125 = 233\,\mathrm N.

6. The ring has three vertical forces: its weight 196.2\,\mathrm N at its centre G, P = 150\,\mathrm N at 60\,\mathrm{mm} to one side, and the shaft's reaction R = 346.2\,\mathrm N upward. Moments about G: R acts 150(60)/346.2 = 26.0\,\mathrm{mm} to the same side as P. R acts at the contact point on the ring's inner circle (50\,\mathrm{mm} from G), where the normal runs through G; a vertical line 26.0\,\mathrm{mm} from G meets that circle where the normal is \sin^{-1}(26.0/50) = 31.3^\circ from vertical. So \phi = 31.3^\circ and \mu = 0.609.

7. P = 355.9\,\mathrm N, R_A = 0.1143\,\mathrm m: M = \frac23(0.40)(355.9)(0.1143) = 10.85\,\mathrm{N{\cdot}m} (96\,\mathrm{lb{\cdot}in.}). B is held if C can supply that moment over B's face (R_B = 0.1524\,\mathrm m): \frac23\mu(355.9)(0.1524) = 10.85, \mu = 0.30.

8. The board end is a thin strip along a radius with uniform pressure, so the friction is uniform along r from R/4 to 3R/4: its resultant \mu P acts at the middle, R/2. M = \mu PR/2.

9. L = \int_{r_i}^{r_o}(k/r^2)2\pi r\,dr = 2\pi k\ln(r_o/r_i); M = \int\mu(k/r^2)r\,2\pi r\,dr = 2\pi\mu k(r_o - r_i). So M = \mu L(r_o - r_i)/\ln(r_o/r_i).

10. p = p_0(1 - r/2a). L = \int_0^ap\,2\pi r\,dr = 2\pi p_0(a^2/2 - a^2/6) = \frac23\pi p_0a^2; M = \int_0^a\mu p\,2\pi r^2\,dr = 2\pi\mu p_0(a^3/3 - a^3/8) = \frac5{12}\pi\mu p_0a^3. M = 5\mu La/8.

11. Each thread carries half the crate and platform, 6500(9.81) = 63\,765\,\mathrm N; each collar carries that plus its screw, 7400(9.81) = 72\,594\,\mathrm N. Thread: \alpha = \tan^{-1}(11/120\pi) = 1.67^\circ, \phi = 8.53^\circ; raise 63\,765(0.06)\tan10.20^\circ = 688.5\,\mathrm{N{\cdot}m}, lower 63\,765(0.06)\tan6.86^\circ = 460.2\,\mathrm{N{\cdot}m}. Collar (6/5a): \frac23(0.15)(72\,594)(0.14583) = 1058.7\,\mathrm{N{\cdot}m}. Raise 1747\,\mathrm{N{\cdot}m}; lower 1519\,\mathrm{N{\cdot}m}.

12. The surface makes \alpha/2 with the axis. A ring of the cone at radius r has area 2\pi r\,dr/\sin(\alpha/2); its normal force p\,dA has axial part p\,dA\sin(\alpha/2) = 2\pi pr\,dr, so L = \pi p(r_2^2 - r_1^2). Its friction moment is \mu p\,r\,dA = 2\pi\mu pr^2\,dr/\sin(\alpha/2), so M = 2\pi\mu p(r_2^3 - r_1^3)/[3\sin(\alpha/2)]. Eliminating p and writing r = d/2: M = \frac{\mu L}{3\sin(\alpha/2)}\,\frac{d_2^3 - d_1^3}{d_2^2 - d_1^2}, the flat collar's moment divided by \sin(\alpha/2).