Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 673

Fig. 33.3. Compound belt drive.

Let d 1 = Diameter of pulley 1,

N 1 = Speed of the pulley 1 in r.p.m. and

d 2 , d 3 , d 4 , N 2 , N 3 , N 4 = Corresponding values for pulleys 2, 3 and 4.

We know that the velocity ratio of the pulleys 1 and 2,

21

12

Nd

Nd

= ...(i)

Similarly, velocity ratio of the pulleys 3 and 4

4 3

24

N d

Nd

= ...(ii)

Multiplying equation (i) by equation (ii),

2413

1324

NN dd

NNdd

×=×

or

4 13

124

N dd

Ndd

×

=

× ...(Q^ N^2 = N^3 , being keyed to the same shaft)

A little consideration will show, that if there are six pulleys, then the velocity ratio,

6135

1246

Nddd

N ddd

××

=

××^

Product of diameters of drivers

Product of diameters of followers

=

Example 33.2. In a workshop, an engine drives a shaft by a belt. The diameters of the
engine pulley and the shaft pulley are 500 mm and 250 mm respectively. Another pulley of 700 mm
diameter on the same shaft drives a pulley 280 mm in diameter of the follower. If the engine runs at
180 revolutions per minute, find the speed of the follower.

Solution. Given: Diameter of the engine pulley (d 1 ) = 500 mm; Diameter of the shaft pulley
(d 2 ) = 250 mm; Diameter of pulley 2 (d 3 ) = 700 mm, Diameter of the follower (d 4 ) = 280 mm and
speed of the engine (N 1 ) = 180 r.p.m.
We know that speed of the shaft,

13

41

24

500 700

180

250 280

dd

NN

dd

× ×

=× = ×

××

= 900 r.p.m. Ans.

33.6.SLIP OF THE BELT

In the previous articles, we have discussed the motion of belts and shafts assuming a firm
frictional grip between the belts and the shafts. But sometimes the frictional grip becomes somewhat
loose. This may cause some forward motion of the driver without carrying the belt with it. This may