Belt friction and rolling resistance

A rope wrapped around a post can hold a load hundreds of times heavier than the force you apply to the free end — not by magic, but because friction, distributed along the whole wrap, compounds exponentially rather than adding up. This is belt friction: the physics behind capstans, bollards, and belt drives, captured in one clean relation between the tension on the taut side of a wrapped belt and the tension on the slack side. The same chapter closes with a very different kind of resistance. A wheel rolling on a surface is not frictionless — it meets a small rolling resistance, caused not by sliding but by tiny deformation at the contact patch, and it behaves nothing like ordinary Coulomb friction. This article builds both ideas from the equilibrium tools you already have, keeps the belt-friction derivation tight enough to reproduce on an exam, and works two numbers all the way through: how little force holds a two-tonne load on a post, and how much smaller rolling resistance is than sliding friction for the same wheel.

The core insight

Wrapping a belt or rope around a drum turns friction into an exponential amplifier of holding force, $T_1/T_2=e^{\mu\beta}$ , while a rolling wheel meets a much smaller, physically distinct resistance $F_r\approx aN/r$ caused by contact-patch deformation rather than sliding.

The core idea

Belt friction — the setup. Picture a flat belt, rope, or cable wrapped around a fixed cylindrical drum or post through a total wrap angle $\beta$ , measured in radians. One end carries tension $T_1$ , the other $T_2$ , and the belt is on the verge of slipping — impending slippage, the same idea used throughout Coulomb-friction problems, just applied along a curved contact instead of a flat one. Because friction acts all along the wrap rather than at a single point, the two tensions cannot simply differ by a fixed amount; the relation between them turns out to be exponential in the wrap angle.

The governing relation. At impending slippage, with $T_1$ the larger ("tight side") tension and $T_2$ the smaller ("slack side") tension:

\frac{T_1}{T_2} = e^{\mu\beta}

where: $\mu$ is the coefficient of static friction between belt and drum, $\beta$ is the total angle of wrap in radians (not degrees), and $T_1 > T_2$ always, with $T_1$ on the side toward which slipping is impending. Two features are worth sitting with. First, the relation contains no drum radius at all — a bigger post does not "hold better"; only $\mu$ and $\beta$ matter, because both the friction force and the normal force scale together with radius and cancel out of the ratio. Second, because $\beta$ sits in an exponent, each extra wrap multiplies the achievable tension ratio rather than adding to it — this is the capstan effect, and it is why sailors take just one or two turns of rope around a bollard to hold a ship that could otherwise tear the rope from their hands.

Figure

A belt wrapped around a drum showing tight side tension T one at the top and slack side tension T two at the bottom, with a magnified wedge of the wrap showing friction acting along the contact arc

Rolling resistance — a different animal. When a wheel rolls without slipping, the wheel and the surface both deform very slightly at the contact patch, and the ground's normal reaction shifts a small distance ahead of the wheel's centre, in the direction of rolling. That shifted reaction creates a small resisting moment about the axle, modelled through a coefficient of rolling resistance $a$ , which has units of length (not a dimensionless ratio like $\mu$ ):

F_r \approx \frac{aN}{r}

where: $N$ is the normal load on the wheel and $r$ is the wheel radius. Rolling resistance is not sliding friction relabelled — it arises from deformation, not from surfaces sliding past each other, and for a given load it is typically far smaller than the equivalent Coulomb friction force would be. In this course you use $F_r\approx aN/r$ in simple equilibrium and treat $a$ as a given quantity; you are not expected to derive it from contact mechanics.

Derivation AThe belt-friction tension ratio from equilibrium of a belt element

The goal is to show, from nothing more than force equilibrium of one small piece of belt and an integration over the wrap angle, why the tight-side and slack-side tensions on a belt at impending slippage are related by the exponential $T_1/T_2=e^{\mu\beta}$ rather than by any linear or additive rule, and why the drum's radius drops out entirely along the way.

Step 1. Isolate a small element of belt subtending an angle $d\theta$ at the drum's centre, with tension $T$ entering one side and $T+dT$ leaving the other. The element presses against the drum with a small normal force $dN$ from the drum, and because slipping is impending, friction along the contact acts at its limiting value $\mu\,dN$ , directed to oppose the tendency to slip — that is, opposing the pull of the larger tension.

Step 2. Resolve forces along the direction of the belt (tangential) and perpendicular to it (radial). Tangentially, equilibrium requires $(T+dT) = T + \mu\,dN$ , so $dT = \mu\,dN$ — the tension increases across the element by exactly the friction force it experiences. Radially, for a small element the two tension components each contribute approximately $T\,d\theta/2$ toward the drum, so $dN \approx T\,d\theta$ (second-order terms in $dT\,d\theta$ are dropped as negligible).

Step 3. Substitute the radial result into the tangential one: $dT = \mu\,T\,d\theta$ , which rearranges to $dT/T = \mu\,d\theta$ — a separable differential equation relating the fractional change in tension to the angle swept. Notice that the drum radius $r$ never appears in this equation at all: it cancelled when relating arc length to angle, which is why the final answer cannot depend on radius.

Step 4. Integrate from the slack side, where $T=T_2$ at $\theta=0$ , to the tight side, where $T=T_1$ at $\theta=\beta$ : $\int_{T_2}^{T_1}\frac{dT}{T} = \int_0^\beta \mu\,d\theta$ , giving $\ln(T_1/T_2) = \mu\beta$ , and exponentiating both sides yields $T_1/T_2 = e^{\mu\beta}$ . The exponential form is the direct consequence of friction compounding continuously along the arc rather than acting once: each infinitesimal slice of belt multiplies, not adds to, the tension carried by the previous slice.

Worked example

A rope is wrapped $1.5$ turns around a fixed horizontal post to hold a load of weight $W=2000$ N hanging from one end, with coefficient of static friction $\mu_s=0.3$ between rope and post. Find the minimum holding force $P$ at the free end, then check how much larger a force would be needed with only one full turn instead of $1.5$ .

Step 1. Convert the wrap to radians. $1.5$ turns is $1.5\times360°=540°$ , and $\beta = 540°\times\pi/180° = 9.425$ rad.

Step 2. Identify the tensions. The hanging load tends to pull the rope and cause slipping, so $W=2000$ N is the tight-side tension $T_1$ , and the applied holding force $P$ is the slack-side tension $T_2$ .

Step 3. Apply the belt-friction relation. $\dfrac{T_1}{T_2}=e^{\mu\beta}$ gives $\dfrac{2000}{P}=e^{0.3(9.425)}=e^{2.827}=16.9$ , so $P=\dfrac{2000}{16.9}=118.3$ N. A holding force of only about $118$ N supports the $2000$ N load — roughly a seventeen-fold amplification from friction alone.

Step 4. Compare with one full turn. $\beta$ becomes $360°=2\pi=6.283$ rad, so $e^{\mu\beta}=e^{0.3(6.283)}=e^{1.885}=6.58$ , and $P=2000/6.58=304.0$ N. Half a turn fewer nearly triples the required holding force — the exponential sensitivity to $\beta$ is exactly why an extra wrap matters so much more than an extra few degrees would in a linear relation.

Answer: with $1.5$ turns, $P\approx 118$ N; with one turn, $P\approx 304$ N — both far below the $2000$ N load, but the difference between them shows how strongly the holding force depends on wrap angle.

Competitive stretch

If the free end is limited to a maximum pull of $P=150$ N, the same $1.5$ -turn wrap could hold $W_{max}=P\times16.9\approx2535$ N — the ratio works in either direction, from load to holding force or from a capped holding force to a maximum sustainable load.

Common traps

Using degrees instead of radians for $\beta$ . The exponent $\mu\beta$ requires radians; forgetting the conversion (as in Step 1 above, $540°\to9.425$ rad) silently produces a wildly wrong ratio.
Swapping $T_1$ and $T_2$ . $T_1$ is always the larger tension, on the side toward which slipping is impending; reversing them inverts the exponential and gives an answer greater than one where it should be less than one, or vice versa.
Assuming a fatter drum "grips better." The derivation shows radius cancels entirely — only $\mu$ and $\beta$ set the tension ratio, contrary to intuition built from everyday friction problems where normal force and area seem to matter.
Treating rolling resistance as $F_r=\mu N$ . Rolling resistance uses a length coefficient $a$ and the formula $F_r\approx aN/r$ ; plugging in a dimensionless friction coefficient in its place mixes two different physical mechanisms and gives answers wrong by orders of magnitude.

Ask the tutor: ask the AI tutor to walk through a capstan problem with a different number of wraps, or to set up a full equilibrium problem for a wheel meeting rolling resistance when the pulling force is applied above axle height.

Why does the belt-friction ratio not depend on the drum's radius at all?

Because both the friction force and the normal force pressing the belt against the drum grow together with radius in the same way, they cancel when you take their ratio. In the derivation, the radial equilibrium step gives $dN\approx T\,d\theta$ with no radius in sight — the geometry only cares about the angle subtended, not how far out that angle reaches. A skinny post and a fat drum with the same wrap angle and the same $\mu$ hold exactly the same tension ratio.

Why does wrapping a rope one extra time around a post make such a huge difference?

Because $\beta$ sits in an exponent, not a multiplier. Going from one turn to $1.5$ turns only adds $180°$ , but that addition happens inside $e^{\mu\beta}$ , so the tension ratio does not just increase by a fixed amount — it multiplies by another factor of $e^{\mu\pi}$ . Small additions to the wrap angle produce large multiplicative gains in holding power, which is precisely the mechanical advantage sailors and riggers exploit with a capstan or bollard.

Is rolling resistance just friction that happens to be small?

No — it is a genuinely different mechanism, even though both oppose motion. Sliding friction comes from two dry surfaces resisting relative sliding, scaling with normal force through a dimensionless $\mu$ . Rolling resistance comes from the wheel and surface deforming slightly at the contact patch, shifting the normal reaction forward and creating a resisting moment; it scales with a length coefficient $a$ and depends on wheel radius through $F_r\approx aN/r$ . Both get called resistive forces, but the mechanisms and formulas are unrelated.

Belt friction and rolling resistance

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads