Mining Methods & Machinery

Mine Hoisting

Static rope load and motor power for a winding system — including the extra force demanded during acceleration.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A hoist raises a skip or cage (its own dead weight plus the payload) up the shaft on a rope wound over a drum. At steady speed the rope carries just the static weight of the suspended load; while accelerating, it must also supply the inertia force

m a

, so the effective force rises to

m (g + a)

. The motor power is this force times the rope speed, divided by the drive efficiency.

The Core Formula Matrix

Static rope load:

F = (m_{p a y l o a d} + m_{s k i p}) g

.

Effective force while accelerating at

a

F = m (g + a)

.

Motor power:

P = \frac{F v}{η}

v

= rope (hoisting) speed,

η

= efficiency.

The ‘IIT Traps’

⚠
Include the skip/cage dead weight. The rope lifts payload *and* conveyance; using payload alone under-sizes the motor.
⚠
**Acceleration adds $m a$ , giving $m (g + a)$ **, not $m g$ . Forgetting it under-estimates the peak power.
⚠
Divide by efficiency for motor power. $P = F v / η$ ; the output $F v$ is less than the motor's input.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A skip of

4000 kg

carries a payload of

8000 kg

. The static rope load (at steady speed) is approximately:

Q2MEDIUM2 Marks · NAT

The same load (

117.72 kN

) is hoisted at

10 m/s

through a drive of efficiency

0.90

at steady speed. The motor power is ______ kW. (Round off to the nearest whole number.)

Q3HARD2 Marks · NAT

During start-up the

12000 kg

load is accelerated at

1 m/s^{2}

while moving at

10 m/s

(

η = 0.90

). The peak motor power is ______ kW. (Round off to two decimal places.)