7.6 Rolling Friction, Belt-Pulley Drives, Brakes, Clutches & Vehicle Dynamics

Rolling Friction

A wheel rolling without slipping still meets resistance, but its origin differs from sliding friction. Both wheel and surface deform slightly, so the normal reaction acts a small distance $a$ ahead of the geometric contact point. Taking moments about the contact patch, the force required to maintain rolling is

$F = \frac{N a}{r}$

where $a$ is the coefficient of rolling resistance, carrying units of length, and $r$ is the wheel radius. The dimensionless rolling resistance coefficient is then $\mu_r = a/r$.

The practical consequence for a coal mine is direct: rolling resistance on a poorly maintained haul road can exceed that on a well-graded surface several times over, which is why haul-road maintenance is a first-order cost lever for dumper fleets. Note also that larger wheels reduce rolling resistance for a given $a$, which is one reason mine haulers use very large diameter tyres.

Rolling resistance is typically one to two orders of magnitude smaller than sliding friction — the reason the wheel is worth having at all.

Belt and Rope Friction

For a flexible belt wrapped around a pulley over an angle of lap $\theta$ (in radians), on the point of slipping:

$\frac{T_1}{T_2} = e^{\mu\theta}$

where $T_1$ is the tight-side tension and $T_2$ the slack-side tension. This exponential relation explains why a few extra turns of rope around a bollard hold an enormous load.

Power transmitted

$P = (T_1 - T_2)\,v$

where $v$ is the belt speed.

Centrifugal tension

At speed, the belt's own mass generates an additional tension in every part of the belt:

$T_c = m v^2$

with $m$ the mass per unit length. Centrifugal tension adds to both sides equally, so it does not increase the power transmitted, but it does consume part of the allowable maximum tension. Including it,

$\frac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$

Condition for maximum power

Differentiating the power with respect to velocity and setting the result to zero gives the classical result

$T_{\max} = 3T_c \quad\Longrightarrow\quad v_{\text{opt}} = \sqrt{\frac{T_{\max}}{3m}}$

That is, maximum power is transmitted when the centrifugal tension equals one third of the maximum permissible tension. This is one of the most frequently examined single results in the whole friction bullet.

V-belts

For a V-belt seated in a groove of total included angle $2\beta$, the wedging action increases the normal reaction, and the relation becomes

$\frac{T_1}{T_2} = e^{\mu\theta/\sin\beta}$

Since $\sin\beta < 1$, the effective friction coefficient $\mu' = \mu/\sin\beta$ is substantially larger. For a typical half-angle of 17 degrees, $\sin\beta \approx 0.292$, so the effective coefficient is over three times the flat-belt value.

Angle of lap

For an open belt drive between pulleys of radii $r_1$ and $r_2$ at centre distance $C$, the lap angle on the smaller pulley is

$\theta = \pi - 2\sin^{-1}\!\left(\frac{r_1 - r_2}{C}\right)$

For a crossed belt drive, both pulleys share the same lap angle

$\theta = \pi + 2\sin^{-1}\!\left(\frac{r_1 + r_2}{C}\right)$

A crossed belt therefore always has a larger lap and can transmit more power, at the cost of belt wear from rubbing at the crossover.

Brakes

A brake absorbs kinetic or potential energy through friction and dissipates it as heat.

Single block or shoe brake

For a block pressed against a drum of radius $r$ with normal force $N$, the friction force is $F = \mu N$ and the braking torque is

$T_b = \mu N r$

The key design consideration is whether the brake is self-energising. If the moment of the friction force about the lever pivot acts in the same sense as the applied effort, the friction assists the application, reducing the required effort. If the friction moment equals or exceeds the restoring moment, the brake becomes self-locking — it grips without any applied effort, which is dangerous in a service brake but desirable in a holding brake.

Band brake

A flexible band wrapped over the drum obeys the same belt-friction relation:

$\frac{T_1}{T_2} = e^{\mu\theta}, \qquad T_b = (T_1 - T_2)\,r$

A simple band brake anchors one end to the frame; a differential band brake attaches both ends to the lever at different distances from the fulcrum and can be made strongly self-energising.

Heat considerations

Braking energy appears as heat, and brake capacity is often limited by temperature rather than torque. For a vehicle of mass $M$ descending a gradient at constant speed, the power to be dissipated is

$P = M g v \sin\alpha$

On long descents in an opencast pit this is precisely why haul trucks use retarders — engine, hydraulic or electric — rather than relying on friction brakes, which would overheat.

Clutches

A clutch transmits torque between coaxial shafts through friction.

Single plate clutch

For an annular friction surface of outer radius $R_o$ and inner radius $R_i$ with axial force $W$:

Uniform pressure theory (new, rigid, well-supported linings):

$T = \frac{2}{3}\mu W\,\frac{R_o^3 - R_i^3}{R_o^2 - R_i^2}$

Uniform wear theory (worn-in linings, where $pr$ is constant):

$T = \frac{1}{2}\mu W\left(R_o + R_i\right) = \mu W R_m$

Uniform wear always gives the lower torque, so it is the safer basis for design. The mean radius $R_m$ is simply the arithmetic mean of the two radii.

For a clutch with $n$ pairs of contacting surfaces, multiply by $n$. A single plate clutch with friction material on both faces has $n = 2$.

Cone clutch

A cone clutch of semi-cone angle $\alpha$ gains the same wedging benefit as a V-belt:

$T = \frac{\mu W R_m}{\sin\alpha}$

Vehicle Dynamics on a Gradient

The syllabus names vehicles explicitly. For a vehicle of mass $M$ on a gradient of angle $\alpha$, moving at constant speed, the tractive effort required is

$F = Mg\sin\alpha + \mu_r Mg\cos\alpha + \tfrac{1}{2}\rho C_d A v^2$

the three terms being grade resistance, rolling resistance and aerodynamic drag. For a slow, heavy mine hauler the drag term is negligible and the first two dominate.

Limit of adhesion

Whatever the engine can deliver, tractive effort cannot exceed what friction at the tyres permits:

$F_{\max} = \mu W_{\text{driven}}$

where $W_{\text{driven}}$ is the load carried by the driven wheels. This is why a fully loaded rear-dump truck climbs better than an empty one on a slippery haul road — more weight sits on the driving axle.

Maximum gradient without slipping

Setting tractive effort equal to the adhesion limit and neglecting drag,

$\tan\alpha_{\max} \approx \mu - \mu_r \;\approx\; \mu \text{ for small } \mu_r$

so the steepest climbable gradient is governed essentially by the coefficient of adhesion between tyre and road.