Classical-Mechanics · Unit 9 · Video 5 · Interactive Practice

One Hand Holds the Ship: The Capstan Equation

IKey Formulas

Formula	Name	What it says
$T_B = T_A\, e^{-\mu_s \theta_{BA}}$	Capstan (belt-friction) equation	Held tension from load, $\mu_s$ , wrap angle $\theta_{BA}$ (radians)
$\dfrac{dT}{d\theta} = -\mu_s T$	Governing equation	Rate of tension loss is proportional to the tension
$N = T\,\Delta\theta$	Normal force on a slice	From the radial force balance
$f_s = \mu_s N$	Maximum static friction	The on-the-verge-of-slipping condition

Key Insight: Each bit of contact removes a fixed fraction of the tension, not a fixed amount — so the losses compound and tension collapses geometrically, wrap after wrap.

IIVisualization 1 — Tension Falls Around the Drum

How far does tension fall as the rope wraps farther around the drum?

Friction μ_s

💡 In the capstan equation $\theta_{BA}$ is measured in radians, and each full wrap adds $2\pi$ .

IIIVisualization 2 — The Power of Extra Wraps

How small is the grip after one, two, or three full wraps of the rope?

Full wraps (n) 2 Grip guess (fraction) 0.20

IVVisualization 3 — Why the Loss Compounds

Each equal slice of contact removes the same fraction of tension — not the same amount.

Slice (k) 0

VQuiz Questions

Problem 1 · Apply the Capstan Equation

Given: a rope over a fixed post with $\mu_s = 0.3$ , wrapped through a total angle $\theta = \pi$ . A load pulls with $T_A = 600\text{ N}$ . Find the minimum holding tension $T_B$ .

✅ Correct!

T_B = 600\,e^{-0.3\pi} = 600(0.390) \approx 234\text{ N}

— friction lets one hand hold what the load pulls with.

❌ Check the exponent's sign. You computed

600\,e^{+0.3\pi}

. Friction reduces the tension you must supply, so the exponent is negative.

❌ That treats the loss as linear.

\mu_s\theta

is not the fraction removed — the decay is exponential,

T_B = T_A e^{-\mu_s\theta}

❌ Not quite. Substitute directly:

T_B = 600\,e^{-0.3\pi}

Show solution

The load side is $T_A$ ; the held side is $T_B$ . With $\mu_s = 0.3$ and $\theta = \pi$ :

T_B = T_A\, e^{-\mu_s \theta} = 600\, e^{-0.3\pi} = 600\, e^{-0.9425}

= 600 \times 0.390 \approx 234\text{ N}

Any grip below $234\text{ N}$ lets the rope slip.

Problem 2 · Radians, Not Degrees

Given: $\mu_s = 0.2$ and the rope makes 2 full turns around the drum. Find the wrap angle $\theta$ and the holding ratio $T_B/T_A$ .

✅ Correct! Two turns give

\theta = 2(2\pi) = 4\pi

, so

T_B/T_A = e^{-0.2\cdot4\pi} = e^{-2.513} \approx 0.081

❌ Those are degrees. The exponent needs

\theta

in radians:

720\degree = 4\pi \approx 12.57

, giving a ratio of

0.081

, not

0

❌ That's only one turn. Two full turns give

\theta = 4\pi

and a ratio of

e^{-0.2\cdot4\pi} \approx 0.081

❌ Not quite. One full turn is

2\pi

radians, so two turns are

4\pi

Show solution

Each full turn contributes $2\pi$ radians of contact, so two turns give

\theta = 2 \times 2\pi = 4\pi \approx 12.57.

Then

\frac{T_B}{T_A} = e^{-\mu_s\theta} = e^{-0.2\,(4\pi)} = e^{-2.513} \approx 0.081.

The grip needed is about one-twelfth of the load.

Problem 3 · Solve for the Wrap Angle

Given: you must hold $T_A = 5000\text{ N}$ using at most $T_B = 500\text{ N}$ of grip, with $\mu_s = 0.25$ . Find the minimum wrap angle $\theta$ (radians) from $T_B = T_A e^{-\mu_s\theta}$ .

First, what is $\ln(T_A/T_B)$ ?

Then, what is the required $\theta$ ?

✅ Correct!

\theta = \tfrac{1}{\mu_s}\ln(T_A/T_B) = 4\ln 10 \approx 9.21\text{ rad}

— about

1.5

turns.

❌ Check the ratio.

T_A/T_B = 5000/500 = 10

, and you need its natural log,

\ln 10 \approx 2.30

❌ Check the algebra. Solving

T_B = T_A e^{-\mu_s\theta}

gives

\theta = \tfrac{1}{\mu_s}\ln(T_A/T_B)

❌ You multiplied by $\mu_s$ . Solving for

\theta

divides by

\mu_s

\theta = \ln 10 / 0.25

❌ Don't forget the $1/\mu_s$ factor.

\theta = \ln 10 / 0.25

, not

\ln 10

alone.

Show solution

Start from the capstan equation and isolate $\theta$ :

T_B = T_A e^{-\mu_s\theta} \;\Rightarrow\; \frac{T_A}{T_B} = e^{\mu_s\theta} \;\Rightarrow\; \theta = \frac{1}{\mu_s}\ln\!\frac{T_A}{T_B}.

With $T_A/T_B = 10$ and $\mu_s = 0.25$ :

\theta = \frac{\ln 10}{0.25} = \frac{2.303}{0.25} \approx 9.21\text{ rad} = \frac{9.21}{2\pi} \approx 1.47\text{ turns}.

So one and a half wraps already does the job.

Problem 4 · The Multiplier per Wrap (Video Values)

Given: $\mu_s = 0.2$ . Adding one extra full wrap multiplies the holding ratio $T_B/T_A$ by what factor?

✅ Excellent! One more wrap adds

2\pi

\theta

, multiplying the ratio by

e^{-\mu_s\,2\pi} = e^{-0.4\pi} \approx 0.285

. Each turn squares your advantage.

❌ You dropped the $2\pi$ . A full wrap adds

2\pi

radians, so the factor is

e^{-\mu_s\cdot2\pi} = e^{-0.4\pi}

, not

e^{-\mu_s}

❌ The advantage is multiplicative, not linear. Wraps compound geometrically by a constant factor

e^{-0.4\pi}

❌ It's a factor, not a subtraction. An extra wrap multiplies the ratio by

e^{-0.4\pi}

; it does not subtract from it.

Show solution

Compare the ratios for $n$ and $n+1$ wraps ( $\theta = 2\pi n$ ):

\frac{e^{-\mu_s\,2\pi(n+1)}}{e^{-\mu_s\,2\pi n}} = e^{-\mu_s\,2\pi} = e^{-0.2\,(2\pi)} = e^{-0.4\pi} \approx 0.285.

So every extra turn scales the required grip by the same factor $0.285$ : this is exactly why $1 \to 2 \to 3$ wraps drop the ratio through $0.285 \to 0.081 \to 0.023$ .

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