Mining Methods & MachineryFREE· Reviewed Jun 2026

Mining Methods & Machinery

Belt-conveyor capacity and drive power, Euler belt friction, gradient haulage resistance and balanced-winder hoisting power — the materials-handling numerics GATE leans on hardest.

Section 1

Recent Trend Analysis (2017–2026)

Materials handling carries 4–7 marks and has migrated almost entirely from descriptive method questions to machinery numericals.

The decade-long shift in question style:
- 2017–2019 — method recall: compare bord-and-pillar vs longwall, list haulage types, define belt-conveyor components. Largely 1-mark theory.
- 2020–2022 — single-formula compute: a belt capacity in

t / h

, a drive power from effective tension, or an Euler belt-friction tension ratio.
- 2023–2026 — full multi-variable NAT chains: balanced-winder hoisting power (steady + acceleration peak), gradient haulage resistance with rolling friction, and capacity-to-power coupling on a single conveyor.

Exact recurring themes you must own:
- Belt capacity

Q = 3.6 A v ρ

and its density-unit trap.
- Drive power

P = T_{e} v

and the Euler ratio

T_{1} / T_{2} = e^{μ θ}

(wrap angle in radians).
- Hoisting power for a balanced (tail-rope) winder — out-of-balance = payload only — plus the inertia peak during acceleration.
- Gradient resistance

sin α

vs rolling resistance

μ cos α

.
- A rising count of MSQs on method selection / machine suitability — one wrong tick zeroes the mark.

Section 2

Master Formula Matrix & Derivations

Belt Conveyor Capacity

Throughput is the load cross-section swept past a point at the belt speed, weighted by bulk density.

\overset{m}{˙} = A v ρ (k g / s) \Rightarrow Q = 3.6 A v ρ (t / h, ρ in k g / m^{3})

⚡ Exam shortcut ·

ρ

is given in

t / m^{3}

use

Q = 3600 A v ρ

. The factor

3.6

already folds in

\times 3600 s / h \div 1000 k g / t

Variable Index (SI units)

Symbol	Meaning	SI unit
$Q$	Mass throughput	$t / h$
$\overset{m}{˙}$	Mass flow rate	$k g / s$
$A$	Load cross-sectional area on belt	$m^{2}$
$v$	Belt speed	$m / s$
$ρ$	Bulk density of material	$k g / m^{3}$

Drive Power & Euler Belt Friction

The motor supplies the effective belt tension at belt speed; the drive pulley can only develop tension up to the Euler capstan limit before slipping.

P = T_{e} v, \frac{T _{1}}{T _{2}} = e^{μ θ}

⚡ Exam shortcut ·

T_{e} = T_{1} - T_{2}

is the effective (driving) tension. The wrap angle

θ

must be in radians (

18 0^{\circ} = π

). Input (motor) power

= P / η

Variable Index (SI units)

Symbol	Meaning	SI unit
$P$	Drive (output) power	$W$
$T_{e}$	Effective tension ( $T_{1} - T_{2}$ )	$N$
$T_{1}, T_{2}$	Tight-side / slack-side tension	$N$
$μ$	Pulley–belt friction coefficient	dimensionless
$θ$	Angle of wrap on drive pulley	$r a d$

Haulage Tractive Resistance & Power

A load on a gradient must overcome the component of gravity along the slope plus rolling friction across it.

R = W g (sin α + μ cos α), P = \frac{R v}{η}

⚡ Exam shortcut ·

For a grade quoted as '1 in

n

sin α = 1/ n

and

cos α \approx 1

. Use + when hauling up, − for the friction term when the grade aids descent.

Variable Index (SI units)

Symbol	Meaning	SI unit
$R$	Total tractive resistance (rope pull)	$N$
$W$	Mass hauled (cars + load)	$k g$
$g$	Gravitational acceleration ( $9.81$ )	$m / s^{2}$
$α$	Gradient angle	degrees
$μ$	Rolling / track resistance coefficient	dimensionless
$v$	Haulage speed	$m / s$

Hoisting / Winding Power (Steady + Peak)

A balanced (tail-rope) winder lifts only the net out-of-balance load at steady speed; during acceleration the effective inertia adds a transient force.

P_{s t e a d y} = \frac{W _{p} g v}{η}, P_{p e ak} = \frac{( W _{p} g + m _{e f f} a ) v}{η}

⚡ Exam shortcut ·

Balanced winding ⇒ out-of-balance

=

payload only (skip + counterweight cancel). Without a tail rope, add the unbalanced rope weight.

m_{e f f}

lumps cage, load, ropes and the rotational-inertia equivalent.

Variable Index (SI units)

Symbol	Meaning	SI unit
$P$	Motor (input) power	$W$
$W_{p}$	Payload mass	$k g$
$v$	Hoisting (rope) speed	$m / s$
$a$	Acceleration during ramp-up	$m / s^{2}$
$m_{e f f}$	Effective accelerated mass	$k g$
$η$	Drive efficiency (fraction)	dimensionless

Rope Factor of Safety

Winding ropes are sized so the breaking strength comfortably exceeds the maximum static-plus-dynamic load.

F . S . = \frac{T _{b r e ak}}{( W _{p} + W _{c a g e} + W _{r o p e} ) g}

⚡ Exam shortcut ·

Statutory winding-rope F.S. is typically

\geq 5

and decreases with depth as rope self-weight grows. Add the inertia term

m_{e f f} a

for the dynamic check.

Variable Index (SI units)

Symbol	Meaning	SI unit
$F . S .$	Factor of safety	dimensionless
$T_{b r e ak}$	Rope breaking strength	$N$
$W_{r o p e}$	Suspended rope mass	$k g$

Section 3

The "IIT Trap" Warning System

Bulk-density units. $Q = 3.6 A v ρ$ needs $ρ$ in $k g / m^{3}$ ; with $ρ$ in $t / m^{3}$ the factor is $3600$ . Mixing the two slips the capacity by $1000 \times$ .
**Belt speed $m / s$ vs $m / min$ .** Capacity and power both use $v$ in $m / s$ . A speed given in $m / min$ must be divided by 60 first.
Euler wrap angle in radians. $T_{1} / T_{2} = e^{μ θ}$ with $θ$ in radians — feeding degrees makes the exponent ~57× too large.
Effective vs tight-side tension. Drive power uses $T_{e} = T_{1} - T_{2}$ , not $T_{1}$ . Using the tight-side tension alone over-states the power.
**Gradient: $sin α$ , not $tan α$ .** Grade resistance is $W g sin α$ ; rolling resistance uses $cos α (\approx 1)$ . For a '1 in $n$ ' grade, $sin α = 1/ n$ .
Balanced vs unbalanced winding. With a tail rope the out-of-balance is the payload only. Adding cage/counterweight masses (which cancel) double-counts the load.
Weight vs inertia in the peak. Peak force $= W_{p} g + m_{e f f} a$ — the $g$ term is weight, the $a$ term is inertia; dropping either (or using $g$ for both) is a classic NAT error.
Efficiency placement. Motor input power is $P_{o u t} / η$ (divide). Multiplying by $η$ understates the required motor rating.

Section 4

High-Fidelity Core Examples

Example 12-mark complexity

A troughed belt conveyor carries coal of bulk density

ρ = 900 k g / m^{3}

at a belt speed

v = 2.5 m / s

with a load cross-sectional area

A = 0.05 m^{2}

. The effective belt tension is

T_{e} = 18 k N

. Determine (a) the conveyor capacity in

t / h

and (b) the drive (output) power.

Given Parameters Matrix (clean SI)

Bulk density ( $ρ$ )	$900 k g / m^{3}$
Belt speed ( $v$ )	$2.5 m / s$
Load area ( $A$ )	$0.05 m^{2}$
Effective tension ( $T_{e}$ )	$18 k N = 18000 N$

Algebraic Derivation Track

Step 1 — Capacity (density in

k g / m^{3}

⇒ factor

3.6

Q = 3.6 A v ρ = 3.6 \times 0.05 \times 2.5 \times 900

Q = 3.6 \times (0.05 \times 2.5 \times 900) = 3.6 \times 112.5 = 405 t / h

Step 2 — Drive power at belt speed:

P = T_{e} v = 18000 \times 2.5 = 45000 W = 45 k W

🎯 Final target & accepted range ·

Capacity

Q = 405 t/h

(accept

400 - 410

); drive power

P = 45 kW

. Motor input would be

45/ η

— e.g.

\approx 50 k W

η = 0.9

Example 22-mark complexity

A balanced (tail-rope) winder hoists a payload

W_{p} = 8000 k g

at a steady rope speed

v = 12 m / s

. The effective accelerated mass (cage, load, ropes and rotating parts) is

m_{e f f} = 20000 k g

and the acceleration during ramp-up is

a = 0.8 m / s^{2}

. Drive efficiency is

η = 0.90

. Find (a) the steady motor power and (b) the peak motor power at the end of acceleration. Take

g = 9.81 m / s^{2}

Given Parameters Matrix (clean SI)

Payload ( $W_{p}$ )	$8000 k g$
Rope speed ( $v$ )	$12 m / s$
Effective mass ( $m_{e f f}$ )	$20000 k g$
Acceleration ( $a$ )	$0.8 m / s^{2}$
Efficiency ( $η$ )	$0.90$

Algebraic Derivation Track

Step 1 — Out-of-balance force (balanced winder ⇒ payload only):

F_{oo b} = W_{p} g = 8000 \times 9.81 = 78480 N

Step 2 — Steady motor power:

P_{s t e a d y} = \frac{F _{oo b} v}{η} = \frac{78 480 \times 12}{0.90} = \frac{941 760}{0.90} = 1046400 W \approx 1046 k W

Step 3 — Peak force (weight + inertia at end of acceleration):

F_{p e ak} = W_{p} g + m_{e f f} a = 78480 + 20000 \times 0.8 = 78480 + 16000 = 94480 N

Step 4 — Peak motor power:

P_{p e ak} = \frac{F _{p e ak} v}{η} = \frac{94 480 \times 12}{0.90} = \frac{1 133 760}{0.90} \approx 1260 k W

🎯 Final target & accepted range ·

Steady power

\approx 1046 kW

(accept

1040 - 1050 kW

); peak power

\approx 1260 kW

(accept

1255 - 1265 kW

). The motor must be rated for the peak, not the steady, demand.