(^694) „„„„„ A Textbook of Engineering Mechanics
Soluton. Given: Power to be transmit (P) = 1 MW = 1000 kW; Diameter of the pulley
(d) = 1 m; Speed of the pulley (N) = 450 r.p.m; Safe pull (T) = 2·25 kN = 2250 N; Mass of the rope
(m) = 1 kg/m; Angle of lap (θ) = 150° 150 2·62 rad
180
π
=× = ; Groove angle (2α) = 45° or α = 22·5°
and coefficient of friction (μ) = 0·3
We know that velocity of the ropes,
v=
1450
60 60
ππ××dN
= = 23·56 m/s
∴ Centrifugal tension,
TC=mv^2 = 1(23·56)^2 = 555 N
∴ We know that T 1 =T – TC = 2250 – 555 = 1695 N
and
1
2
2·3 log
T
T
⎛⎞
⎜⎟
⎝⎠
=μθ cosec α = 0·3 × 2·62 cosec 22·5°
= 0·3 × 2·62 × 2·613 = 2·054
1
2
log
T
T
⎛⎞
⎜⎟
⎝⎠

2·054
0·8930
2·3

2
1695
T
= 7·816 ...(Taking antilog of 0·8930)
∴ T 2 =
1695
217 N
7·816

and power transmitted by one rope
=(T 1 – T 2 ) v = (1695 – 217) 23·56 N-m/s
= 34820 kW = 34·82 kW
∴ No. of ropes =
Total power to be transmitted 1000
Power transmitted by one rope 34·82

= 28·7 say 30 Ans.
EXERCISE 33.3

1. A rope drive consists of two V-belts in parallel on grooved pulleys of the same size. The angle
   of the groove is 30°. The cross-sectional area of each belt is 750 mm^2 and μ = 0·12. The density
   of the belt material is 1·2 Mg/m^3 and the maximum safe stress in the materials is 7 MN/m^2.
   Calculate the power that can be transmitted between the pulleys 300 mm diameter rotating at
   1500 rev/min. Find also the shaft speed in rev/min at which the power transmitted would be a
   maximum. [Ans. 172 kW; 2807 r.p.m.]

QUESTIONS

1. Discuss briefly the various types of belts used for the transmission of power.


2. How does the velocity ratio of a belt drive effect, when some slip is taking place between
   the belt and the two pulleys?


3. Distinguish clearly the difference between an open belt drive and cross belt drive.


4. Obtain an equation for the length of a belt in :
   (i) an open belt drive. (ii) a cross belt drive.