Study Companion: S29 — Flexible Belts and Rolling Resistance

A rope round a post, a belt round a pulley and a band round a brake drum all press on a curved surface, and the pressure is not set by a weight or a clamp but by the rope's own tension. Friction then does something new: the more the rope is pulled, the harder it is pressed, and the more friction it gets. This lesson follows that feedback through a small piece of belt to an exponential law, uses it for holding, lowering and raising, joins it to the rigid bodies and blocks of earlier lessons, and closes with a resistance that is not friction at all: the rolling resistance of a wheel on a surface that gives.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 6, articles 6/8 ("Flexible Belts") and 6/9 ("Rolling Resistance"), with Sample Problems 6/9–6/10; Harper, Solving Statics Problems in MATLAB, "Flexible Belts" (its Sample Problem 6/9, the same in Meriam's 9th edition). Units SI throughout, g = 9.81\,\mathrm{m/s^2}; angles of contact in radians. As in S27–S28, \mu is the static coefficient when slip impends and the kinetic one while slipping.

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 boat's mooring line pulls 10\,\mathrm{kN} in a gust. A deckhand takes the line round a steel bollard (\mu = 0.30 between rope and bollard) and holds the free end.

With one full turn round the bollard the deckhand must hold the end with 1518\,\mathrm N, more than most people can. How much with two turns, and with three? The plausible guess is "each turn adds the same grip, so three turns hold three times what one does": one turn takes up 8.5\,\mathrm{kN}, so three would take up more than the whole 10\,\mathrm{kN} and the deckhand would need nothing. Sailors know the truth is stranger and better. The answer, 230.5\,\mathrm N and then 35.0\,\mathrm N, is in Nugget 02.

Nugget 01: Friction on a Wrapped Belt Grows With the Tension It Resists

Claim. On a belt wrapped round a drum, a short piece is pressed on the drum by the tension in its two ends, so its normal force, and with it its friction, is proportional to the tension. Each piece therefore adds tension in proportion to the tension already there: dT = \mu T\,d\theta.

The picture. Meriam's Fig. 6/11: a fixed drum with a belt in contact over the angle \beta, tensions T_1 and T_2 (T_2 > T_1), and the free-body diagram of an element r\,d\theta: tension T at angle \theta, T + dT at \theta + d\theta, the normal force dN and the friction \mu\,dN, opposing the impending slip towards T_2.

Check with numbers. Step round a half turn with \mu = 0.3, starting from T = 100\,\mathrm N, in 3142 steps of 0.001\,\mathrm{rad}, each multiplying T by (1 + 0.3 \times 0.001): the tension ends at 256.6\,\mathrm N, the value 100e^{0.3\pi} that Nugget 02 integrates to. No step adds much; every step adds a fixed fraction.

The idea. Meriam (6/8): along the tangent,

since the cosine of a differential angle is unity in the limit. Along the normal,

Step the source states in words: the sine of a differential angle equals the angle, and the product dT\,d\theta of two differentials is dropped against first-order terms. Together:

Unlike S26, the normal force is not given by a weight; it is the belt's own tension bent round the curve, T\,d\theta, which is why friction feeds on itself.

The equation (physical law: equilibrium of a belt element at impending slip): dT/T = \mu\,d\theta.

Assumptions. Flexible belt (no bending stiffness), weightless; slip impending (or occurring) over the whole contact; uniform \mu; drum fixed (or turning steadily, Nugget 02).

Worked number. Where the tension is 500\,\mathrm N, a 1^\circ (0.01745\,\mathrm{rad}) element presses with dN = 8.73\,\mathrm N and gains dT = 0.3 \times 8.73 = 2.62\,\mathrm N; where it is 1000\,\mathrm N, the same element gains twice as much.

Real example. A band brake on a winch drum: the band's own tension presses it on the drum, so a small lever force on the slack end becomes a large braking force.

Plausible wrong answer. "The normal force is the drum's reaction to the belt's weight." A weightless belt still presses with T\,d\theta per element; the pressure comes from the tension.

Check. Why does the result not contain the drum's radius r? (The element's length r\,d\theta never enters: both dN and dT depend on d\theta only.)

Narration spine. A belt over a drum, one short segment highlighted and pulled out. Its two end tensions tilt by half the small angle, and their inward parts add into dN; a friction arrow \mu\,dN appears. Zoom out: arrows all round the contact, growing towards the tight side.

Bridge. A rate of growth proportional to the amount is exponential growth.

Nugget 02: The Tension Ratio Is Exponential in the Wrap Angle, and the Radius Does Not Matter

Claim. Over a contact angle \beta the tension ratio is T_2/T_1 = e^{\mu\beta}. Each turn of a rope round a post multiplies the tension it can hold by the same factor, whatever the post's size.

The picture. The puzzle's bollard with n turns: the line tension falls by the same factor e^{2\pi\mu} for each turn, from 10\,\mathrm{kN} at the boat to the deckhand's hand.

Check with numbers. e^{0.30 \times 2\pi} = 6.586 per turn. One turn: 10\,000/6.586 = 1518\,\mathrm N. Two turns: 10\,000/6.586^2 = 230.5\,\mathrm N. Three turns: 35.0\,\mathrm N, the pull of a hand. The linear guess fails because the second turn works on a rope already relieved by the first: it takes the same fraction, not the same newtons.

The idea. Meriam (6/8) integrates the element law between the slack and tight ends:

\beta is the total angle of contact, in radians: 2\pi n for n turns. The radius never entered, so Eq. 6/7 holds for a noncircular section too, as long as \beta is the total turning of the belt over the contact. It also gives the ratio of belt tensions for slip in a belt drive running at constant speed; at high speed the belt tends to leave the rim and the equation is in error.

The equation (physical law, integrated): T_2 = T_1e^{\mu\beta}, T_2 > T_1.

Assumptions. As Nugget 01; at the point of slip everywhere on the contact.

Worked number. A belt drive (our numbers): the belt wraps 160^\circ (2.793\,\mathrm{rad}) of a 0.1\,\mathrm m motor pulley, \mu = 0.3, and the tight side may carry 1\,\mathrm{kN}. The ratio is e^{0.3 \times 2.793} = 2.311, so the slack side needs at least 432.7\,\mathrm N and the largest torque before slip is (1000 - 432.7)(0.1) = 56.7\,\mathrm{N{\cdot}m}.

Real example. Capstans, bollards, the turns of a climbing rope round a belay device, and Meriam's dockworker (Problem 6/85: 1\frac14 turns, 200\,\mathrm N holds 2.11\,\mathrm{kN}).

Plausible wrong answer. "A fatter bollard grips better." The radius cancels; only the angle and \mu count (a fat post is chosen for the rope's sake, to bend it gently).

Check. By what factor does one extra half turn change the holding force when \mu = 0.3? (e^{0.3\pi} = 2.57.)

Narration spine. The bollard, the line winding on turn by turn; a tension bar at the hand shrinks by the same proportion each turn, 10 to 1.52 to 0.23 to 0.035\,\mathrm{kN}, on a logarithmic staircase.

Bridge. Which end is T_2? The equation needs the larger tension on top, and that depends on which way the rope is about to slip.

Nugget 03: Holding, Lowering, Raising: the Larger Tension Is on the Side the Slip Heads For

Claim. Friction on the drum always helps the side that is losing. To raise a load the pull is T_2 and the load T_1; to hold or lower it slowly the load is T_2 and the pull T_1. The same rope therefore admits a whole range of pulls that leave the load at rest.

The picture. Meriam's SP 6/9: a 100\,\mathrm{kg} load (981\,\mathrm N) on a cable over a fixed drum, \mu = 0.30, the other end pulled by P at an angle \alpha below the horizontal; with \alpha = 0 the contact angle is \pi/2.

Check with numbers. (a) \alpha = 0, \beta = \pi/2: e^{0.3\pi/2} = 1.602. About to raise: P_{\max} = 981(1.602) = 1572\,\mathrm N. About to lower: 981 = P_{\min}(1.602), P_{\min} = 612\,\mathrm N. Any P between holds the load. (b) P = 500\,\mathrm N: the load is the tight side, 981/500 = e^{0.3\beta}, so \beta = \ln(1.962)/0.3 = 2.25\,\mathrm{rad} = 128.7^\circ, and \alpha = 128.7^\circ - 90^\circ = 38.7^\circ: with less wrap than that the load slips down.

The idea. Meriam derives Eq. 6/7 with T_2 > T_1, the tight side being the one towards which the belt moves, or would move, relative to the drum. Harper keeps \mu as a variable: P_{\max} = 981e^{\mu\pi/2} and P_{\min} = 981e^{-\mu\pi/2} fan out from 981\,\mathrm N at \mu = 0 (a frictionless drum only redirects) to 4719 and 204\,\mathrm N at \mu = 1. For P = 500\,\mathrm N, \alpha_{\min} = \ln(981/500)/\mu - \pi/2: it is 0 at \mu = 0.429, negative above that (the cable may leave the drum above the horizontal line through its top, with less than a quarter turn of contact), and above 360^\circ for \mu below 0.086, more than a whole extra turn.

The equation (physical law): raising P = We^{\mu\beta}, holding or lowering P = We^{-\mu\beta}.

Assumptions. Fixed drum; slow, steady motion; cable weight neglected.

Worked number. Harper's version at \mu = 0.5: P_{\max} = 2152\,\mathrm N, P_{\min} = 447.3\,\mathrm N, and for P = 500\,\mathrm N, \alpha_{\min} = -12.8^\circ.

Real example. A rock climber lowered over an edge, a tree surgeon letting himself down on a rope over a limb (Problem 6/93, self-check 6): the rope's contact gives the lowering party a large advantage, and works against them when raising.

Plausible wrong answer. "Put the pull on top: P = We^{\mu\beta} always." That is only raising; for holding, the load is the tight side.

Check. If \mu doubles, how does the range P_{\max}/P_{\min} change? (It squares: e^{\mu\pi} becomes e^{2\mu\pi}.)

Narration spine. The drum, the load, the pull. The pull slider moves: green zone between 612 and 1572\,\mathrm N, the load creeping up above, down below. Then the pull fixed at 500\,\mathrm N while its angle swings, and the wrap angle grows until the load holds at 38.7^\circ.

Bridge. So far the belt met only a drum and a weight. In a machine it also meets bars, pins and sliding blocks.

Nugget 04: Belts Join the Rest of Statics: a Cord, a Bar and a Block

Claim. A problem with a belt is solved like any other: free-body diagrams for each body, the belt equation across each curved contact, the block friction law at each sliding face, and a decision about which way motion impends, once for each limit.

The picture. Meriam's SP 6/10: a uniform 9\,\mathrm{kg} bar pinned at O, inclined 25^\circ, held by a cord at A (2L/3 from O) that rises at 30^\circ above the horizontal, 35^\circ from the normal to the bar. The cord passes over the two rounded corners of a fixed post (\mu = 0.20, turning through 30^\circ and then 40^\circ) and runs down parallel to a 40^\circ incline to a block of mass m (\mu = 0.40).

Check with numbers. Bar, moments about O: T_A(\frac23L\cos35^\circ) = 9(9.81)(\frac L2\cos25^\circ), T_A = 73.3\,\mathrm N. Over the curve \beta = 70^\circ = 1.222\,\mathrm{rad}, e^{0.2 \times 1.222} = 1.277. Block, normal force N = mg\cos40^\circ = 0.766mg. Block about to move up: the block's side is slack, T_1 = 73.3/1.277 = 57.4\,\mathrm N, and 57.4 = mg(\sin40^\circ + 0.40\cos40^\circ) gives m = 6.16\,\mathrm{kg}. Block about to move down: the block's side is tight, T_2 = 73.3(1.277) = 93.5\,\mathrm N, and 93.5 = mg(\sin40^\circ - 0.40\cos40^\circ) gives m = 28.3\,\mathrm{kg}. Range 6.16 \le m \le 28.3\,\mathrm{kg}.

The idea. Meriam's hints: only the total angle of contact enters Eq. 6/7, so the surface's radius and position do not matter; and the cord's tension is the bar's T_A whichever case, but it is the larger tension in one case and the smaller in the other. Meriam then asks for the same problem with the incline at 20^\circ, promising a surprise. Case I: 57.4 = mg(\sin20^\circ + 0.40\cos20^\circ), m = 8.15\,\mathrm{kg}. Case II needs mg(\sin20^\circ - 0.40\cos20^\circ) = 93.5, but \sin20^\circ - 0.40\cos20^\circ = -0.034 < 0: since \tan20^\circ = 0.364 < 0.40 the block holds itself on the incline (S26's angle of repose), so no mass, however large, slides it down. The range is m \ge 8.15\,\mathrm{kg}, with no upper limit.

The equation (physical law: equilibrium of each body, with Eq. 6/7 on the curve and F = \mu N on the incline at impending motion).

Assumptions. Frictionless pin O; light cord; slip impending on the curve and on the incline together.

Worked number. If the rounded surface were frictionless, T would be 73.3\,\mathrm N on both sides and the 40^\circ range would shrink to 7.87–22.2\,\mathrm{kg} (our numbers).

Real example. Cables over rounded edges of decks and walls (Problem 6/106, self-check 7) and band brakes are always part of a larger free-body problem like this one.

Plausible wrong answer. "Use the same tension on the block in both cases." The curved contact makes the block's side the slack side in one case and the tight side in the other.

Check. Why does Case II at 20^\circ fail rather than give a negative mass? (It needs friction to resist motion down the slope that would never start.)

Narration spine. The bar, the cord over the rounded corner, the block on the incline. The block creeps up: the cord tension drops across the corner. It creeps down: the tension rises across it. The incline flattens to 20^\circ and the "down" case grey-out appears.

Bridge. Every resistance so far came from surfaces sliding, or about to slide. A rolling wheel does not slide at all, yet it still needs a push.

Nugget 05: Rolling Resistance: the Ground Pushes Back Ahead of the Wheel

Claim. A wheel rolling at constant speed on a surface that deforms needs a horizontal push, not because of sliding friction but because the ground's resultant force acts slightly ahead of the wheel's lowest point. Moments about that point give P = (a/r)L.

The picture. Meriam's Fig. 6/12: a wheel of radius r with load L on its axle and a push P at its centre. The contact flattens; the pressure is larger on the leading side, and its resultant R acts at a point A a distance a ahead of the centre line, passing through the wheel's centre.

Check with numbers. A 40\,\mathrm t railway wagon on steel wheels of r = 0.45\,\mathrm m, with a = 0.5\,\mathrm{mm} (an illustrative value): \mu_r = 0.0005/0.45 = 0.00111, and P = 0.00111 \times 40\,000 \times 9.81 = 436\,\mathrm N keeps it rolling. Dragging it on locked wheels with \mu = 0.7 would take about 275\,\mathrm{kN}.

The idea. Meriam (6/9): three forces act on the wheel, L, P and the ground's resultant R, so R passes through the centre. Moments about A, taking the arm of P as r:

\mu_r = a/r is the coefficient of rolling resistance: like a friction coefficient it is the ratio of the resisting force to the normal force, but nothing slips, and the mechanism is the elastic and plastic deformation of wheel and surface. The distance a has the dimension of length (some books call a the coefficient of rolling friction, which is misleading); it depends on the materials, the wheel radius, the speed and the roughness, varies little with r, and has no complete theory. Strictly the arm of P is \sqrt{r^2 - a^2}, about r since a \ll r.

The equation (physical law of moments with a material model for a): P = (a/r)L.

Assumptions. Steady rolling; a known from tests; a \ll r.

Worked number. A 1500\,\mathrm{kg} car on tyres of r = 0.3\,\mathrm m with a = 4.5\,\mathrm{mm} (illustrative): \mu_r = 0.015, P = 0.015 \times 14\,715 = 220.7\,\mathrm N. Using the exact arm changes this by 0.01\,\%.

Real example. Rail transport is efficient because steel on steel deforms so little; a wheelbarrow in soft soil is hard work because a becomes large.

Plausible wrong answer. "Rolling resistance is static friction at the contact." Static friction at a non-slipping contact does no work (S30, Nugget 07) and is whatever equilibrium needs; the resistance comes from where the normal pressure sits.

Check. For the same a, does a bigger wheel roll more easily? (Yes: P = aL/r falls as r grows.)

Narration spine. A wheel on a soft surface, exaggerated; the pressure diagram under it leans forward, its resultant arrow tilts back through the centre, and a small horizontal push appears to balance it.

Synthesis

The deckhand holds a 10\,\mathrm{kN} pull with 35\,\mathrm N after three turns, because a rope round a post presses on it with its own tension, so each part of the contact sheds the same fraction of the tension: T_2 = T_1e^{\mu\beta}. The rest of the lesson is bookkeeping with that law: put T_2 on the side the slip heads for, and solve the other bodies as usual. Rolling resistance closes the chapter as a contrast: the resisting force there comes not from slip but from where the ground's push sits. Belt friction is friction whose normal force is the belt's own tension, so the tension ratio grows exponentially with the angle of wrap, independently of the radius.

• Central idea: dN = T\,d\theta, so friction grows with the tension and dT/T = \mu\,d\theta.
• Key relationship: T_2 = T_1e^{\mu\beta}, \beta in radians; rolling P = (a/r)L.
• Most important assumption: a flexible, weightless belt slipping (or about to) over the whole contact, at modest speed.
• Open question: a V-belt wedges into its groove and so presses harder than a flat belt for the same tension; how much harder (self-check 8)?

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). S29_belts_and_rolling.py steps dT = \mu T\,d\beta round a half turn and compares the result with Te^{\mu\beta}. It also does the capstan, the belt drive, Harper's hoist at two coefficients, the bar and block over the rough curve, and rolling resistance.

Part 1: the belt equation, stepped and integrated
  3142 steps of 0.001 rad: T = 256.6 N; 100 e^(0.3 pi) = 256.6 N
Part 2: capstans, drives and Harper's hoist
  10 kN held with 1, 2, 3 turns at mu 0.3: 1518.4, 230.5, 35.0 N
  drive, 160 deg wrap: ratio 2.311, slack 432.7 N, torque 56.7 N m
  SP 6/9, mu 0.3: P from 612.4 to 1572 N; P = 500 N needs alpha >= 38.7 deg
  SP 6/9, mu 0.5: P from 447.3 to 2152 N; P = 500 N needs alpha >= -12.8 deg
Part 3: bar and block over a rough curve
  T_A = 73.3 N, e^(mu beta) = 1.277: m from 6.16 to 28.3 kg
Part 4: rolling resistance, P = (a / r) W
  car: mu_r = 0.01500, P = 220.7 N
  wagon: mu_r = 0.00111, P = 436.0 N

MATLAB (Octave-compatible). S29_belts_and_rolling.m steps the belt in a loop and evaluates Harper's hoist.

stepped 256.6 N, exact 256.6 N
mu 0.3: P from 612.4 to 1572 N; P = 500 N needs alpha >= 38.7 deg
mu 0.5: P from 447.3 to 2152 N; P = 500 N needs alpha >= -12.8 deg

Traps

• Taking the normal force on a belt from a weight. It is the tension bent round the curve, T\,d\theta per element. (From Nugget 01.)
• Putting \beta in degrees. e^{\mu\beta} needs radians: 2\pi n for n turns. (From Nugget 02.)
• Adding a fixed grip per turn. Each turn multiplies; three turns of \mu = 0.3 give a factor 286, not 3 \times 6.6. (From Nugget 02.)
• Expecting the drum radius to matter. It cancels. (From Nugget 02.)
• Putting the pull on the tight side when holding or lowering. The load is T_2 then. (From Nugget 03.)
• Measuring the contact angle from a figure's labelled angle. It is the angle turned by the belt while in contact: SP 6/9's \alpha adds to \pi/2; SP 6/10's two angles add. (From Nugget 03.)
• Forgetting that a limit may not exist. A block that holds itself on a gentle slope has no upper limit (SP 6/10 at 20^\circ). (From Nugget 04.)
• Treating a as a dimensionless coefficient. It is a length; \mu_r = a/r. (From Nugget 05.)

Self-check

1. Meriam's Problem 6/83: a rope over a fixed horizontal shaft carries a weight on each end, one twice the other (both ends hang vertically). What is the least \mu that keeps them still?
2. Meriam's Problem 6/84: a cord over a fixed cylinder (both ends vertical) holds a 40\,\mathrm{kg} cylinder, \mu = 0.30. Find the force P on the free end to (a) raise and (b) lower it at slow steady speed.
3. Meriam's Problem 6/85: a rope with 1\frac14 turns round a mooring bit (\mu = 0.30) is pulled with 200\,\mathrm N. What tension can it hold?
4. Meriam's Problem 6/86: a rope over a boulder (both ends vertical) holds a 50\,\mathrm{kg} package; lowering it at a steady rate needs P = 70\,\mathrm N on the other end. Find \mu, and the force to raise it.
5. Meriam's Problem 6/87: a force of 4\,\mathrm{kN} raises a mass m over a fixed cylinder and 1.6\,\mathrm{kN} lowers it slowly. Find m.
6. Meriam's Problem 6/93 (US units in the book): a 180\,\mathrm{lb} tree surgeon hangs from one end of a rope that passes over a limb (\mu = 0.60, both ends vertical) and holds the other end. Convert to SI and find the pull needed to let himself down slowly.
7. Meriam's Problem 6/106: a 100\,\mathrm{kg} crate hangs on a cable that turns through 90^\circ over a ship's railing, \mu = 0.20. Find the tension to (a) lower it and (b) raise it at slow steady speed.
8. Meriam's Problem 6/103: a V-belt of groove angle \alpha replaces the flat belt of Fig. 6/11. Derive the tension relation, and the factor by which a 35^\circ groove multiplies the effective coefficient.
9. Harper's drum with \mu = 0.5: find P_{\max} and P_{\min} for \alpha = 0, and \alpha_{\min} for P = 500\,\mathrm N.
10. Meriam's SP 6/10 with the incline at 20^\circ: find the range of m.
11. A 100\,\mathrm{kg} cart rides on wheels of radius 0.2\,\mathrm m. Find the push to keep it rolling if a = 10\,\mathrm{mm} (soft ground) and if a = 1\,\mathrm{mm} (concrete) (our numbers).

Fully worked solutions

1. About to slip, the heavy side is T_2: 2 = e^{\mu\pi}, \mu = \ln2/\pi = 0.221.

2. \beta = \pi, e^{0.3\pi} = 2.566. (a) P = 392.4(2.566) = 1007\,\mathrm N. (b) P = 392.4/2.566 = 152.9\,\mathrm N.

3. \beta = 2.5\pi; T = 200e^{0.3 \times 2.5\pi} = 200(10.55) = 2.11\,\mathrm{kN}.

4. Lowering, the package is the tight side: 490.5 = 70e^{\mu\pi}, \mu = \ln(7.007)/\pi = 0.620. Raising: P' = 490.5(7.007) = 3.44\,\mathrm{kN}.

5. 4000 = mge^{\mu\beta} and 1600 = mge^{-\mu\beta}; multiplying, (mg)^2 = 6.4 \times 10^6, mg = 2530\,\mathrm N, m = 258\,\mathrm{kg}.

6. W = 800.7\,\mathrm N. Both rope ends hold him up: harness end T_2 and hand end T_1, with T_1 + T_2 = W. Letting himself down, he moves with the harness end, which is therefore the tight side: T_2 = T_1e^{0.6\pi} = 6.586T_1. So T_1 = 800.7/7.586 = 105.5\,\mathrm N (23.7\,\mathrm{lb}).

7. \beta = \pi/2, e^{0.2\pi/2} = 1.369. (a) Lowering, the crate is the tight side: T = 981/1.369 = 717\,\mathrm N. (b) T = 981(1.369) = 1343\,\mathrm N.

8. In a groove of angle \alpha the belt touches the two sides; on an element each side presses with dN' normal to it, and the radial balance is 2\,dN'\sin(\alpha/2) = T\,d\theta. Friction acts on both sides: dT = 2\mu\,dN' = \mu T\,d\theta/\sin(\alpha/2). So T_2 = T_1e^{\mu\beta/\sin(\alpha/2)}, and the factor is n = 1/\sin17.5^\circ = 3.33.

9. P_{\max} = 981e^{0.25\pi} = 2152\,\mathrm N; P_{\min} = 981e^{-0.25\pi} = 447.3\,\mathrm N; \alpha_{\min} = \ln(1.962)/0.5 - \pi/2 = -0.223\,\mathrm{rad} = -12.8^\circ (the cable may leave a little above the drum's top tangent).

10. T_A = 73.3\,\mathrm N and the curve factor 1.277 are unchanged. Up: 57.4 = mg(\sin20^\circ + 0.4\cos20^\circ), m = 8.15\,\mathrm{kg}. Down: \sin20^\circ - 0.4\cos20^\circ < 0, the block cannot slide down. m \ge 8.15\,\mathrm{kg}, no upper limit.

11. L = 981\,\mathrm N. Soft: P = (0.010/0.2)(981) = 49.1\,\mathrm N. Concrete: 4.9\,\mathrm N.