Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 693

Resolving the reactions vertically to the groove.

R= R 1 sin α + R 1 sin α = 2R 1 sin α

or R 1 = 2sin

R

α

We know that the frictional force

=^221 2sin sin

R R
R
μ
μ=μ× =
αα= μR cosec α
Now consider a small portion of the belt as in Art. 33·14 subtending an angle δθ at the centre.
The tension on one side will be T and on the other side T + δT. Now proceeding as in Art. 33·14, we
get the frictional resistance equal to (μR cosec α) against μR. Thus the relation between T 1 and T 2 for
the rope drive will be:

1

2

2·3 log

T

T

⎛⎞

⎜⎟

⎝⎠

= μθ cosec α.

Example. 33·14. Find the power transmitted by a rope drive, from the following data:
Angle of contact = 180°
Pulley groove angle = 60°
Coefficient of friction = 0·2
Mass of rope = 0·4 kg/metre length
Permissible tension = 1·5 kN
Velocity of rope = 15 m/s
Solution. Given: Angle of contact (θ) = 180° = 3·142 rad; Pulley groove angle (2α) = 60°
or α = 30°; Coefficient of friction (μ) = 0·2; Mass of rope (m) = 0·4 kg/m; Permissible tension (T)
= 1·5 kN and velocity of rope = 15 m/s.

We know that the centrifugal tension,

TC=mv^2 = 0·4(15)^2 = 90 N

∴ T 1 =T – TC = 1500 – 90 = 1410 N

and^1
2

2·3 log

T

T

⎛⎞

⎜⎟

⎝⎠

=μθ cosec α =0·2 × 3·142 cosec 30°

= 0·2 × 3·142 × 2·0 = 1·257

∴

1

2

log

T

T

⎛⎞

⎜⎟

⎝⎠

=

1· 2 5 7

0·5465

2·3

=

or
2

1410

T

= 3·52 ...(Taking antilog of 0·5465)

∴ T 2 =

1410

400 N

3·52

=

We know that power transmitted by the rope drive,
P=(T 1 – T 2 ) v = (1410 – 400) × 15 N-m/s
= 15 150 W = 15·15 kW Ans.
Example 33.15. A rope drive is required to transmit 1 MW from a pulley of 1 meter diameter
running at 450 r.p.m. The safe pull in each rope is 2·25 kN and the rope has mass of 1 kg per meter.
The angle of lap and the groove angle is 150° and 45° respectively.
Find the number of ropes required for the drive, if the coefficient of friction between the rope
and the pulley is 0·3.