Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 683

Let T 2 = Tension in the belt in slack side

We know that speed of the belt,

0·6 200

2rad/s

60 60

dN

v

ππ××

== =π

and 1
2

2·3 log 0·25 2·79 0·6975

T

T

⎛⎞

⎜⎟=μθ= × =

⎜⎟

⎝⎠

1

2

0·6975

log 0·3033

2·3

T

T

⎛⎞

⎜⎟⎜⎟==

⎝⎠

or

2

2·5

2·01

T

= ...(Taking antilog of 0·3033)

∴ 2

2·5

1·24 kN

2·01

T==

We know that power transmitted by the belt,
P=(T 1 – T 2 ) v = (2·5 – 1·24) 2π = 7·92 kW Ans.
Example 33.9. Two pulleys, one 450 mm diameter and the other 200 mm diameter are on
parallel shafts 1·95 m apart and are connected by a crossed belt. Find the length of the belt required
and the angle of contact between the belt and each pulley.
What power can be transmitted by the belt, when the larger pulley rotates at 200 rev/min, if
the maximum permissible tension in the belt is 1 kN, and the coefficient of friction between the belt
and pulley is 0·25?
Solution. Given: Diameter of larger pulley (d 1 ) = 450 mm = 0·45 m or radius (r 1 ) = 0·225 m;
Diameter of smaller pulley (d 2 ) = 200 mm = 0·2 m or radius (r 2 ) = 0·1 m; Distance between the
centres of the pulleys (l) = 1·95 m; Speed of the larger pulley (N) = 200 r.p.m; Maximum tension
in the belt (T 1 ) = 1 kN = 1000 N and coefficient of friction (μ) = 0·25.
Let T 2 = Tension in the belt in slack side

Length of belt
We know that length of the belt,

L=

2

12

12

()

()2

rr

rr l

l

+

π+ ++

=

(0·225 0·1)^2

(0·225 0·1) 2 1·95

1·9 5

• π++×+ m
  = 4·975 m Ans.
  Angle of contact between the belt and each pulley

Fig. 33.10.

Let θ= Angle of contact between the belt and each pulley.