Mining Methods & Machinery

Haulage & Rope Pull

The pull a rope haulage must exert to drag loaded cars up a gradient against gravity and track friction — and the power it costs.

PART 1

Topic Breakdown & Traps

On a rope-haulage incline, the rope must overcome two resistances: the gravity component of the load acting down the slope,

W sin θ

, and the track friction,

μ W cos θ

. Their sum is the rope pull

T

. Multiply by the haulage speed for the power, and by the number of cars in a trip for the full set.

Rope pull (hauling up an incline):

T = W (sin θ + μ cos θ)

W

= car weight,

θ

= gradient angle,

μ

= friction (track) coefficient.

Power:

P = T v

(

v

= haulage speed). For

N

cars:

T_{se t} = N T

P = T_{se t} v

⚠
**Gravity term is $sin θ$ , friction term is $cos θ$ .** Swapping them is wrong; for small gradients friction can rival gravity.
⚠
Hauling down reduces the pull. Going down-grade the gravity term *assists*, so $T = W (μ cos θ - sin θ)$ — sign flips.
⚠
Multiply by car count before power. Power is the whole trip's pull times speed, not a single car's.

PART 2

Q1BASIC1 Mark · MCQ

A loaded car weighing

20 kN

stands on an incline of

1 5^{\circ}

. The gravity component along the slope is:

Q2MEDIUM2 Marks · NAT

A car weighing

50 kN

is hauled up a

1 0^{\circ}

incline with a track-friction coefficient

μ = 0.02

. The rope pull required is ______ kN. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

A trip of

10

such cars (each needing

9.67 kN

) is hauled at

2 m/s

. The haulage power required is ______ kW. (Round off to one decimal place.)