Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 681

Example 33.7. Find the necessary difference in tensions in N in the two sides of a belt drive,
when transmitting 120 W. at 30 m/sec.

Solution. Given: Transmitting power (p) = 120 W and speed of the belt (ν) = 30 m/s

Let (T 1 – T 2 ) = Necessary difference in tensions in the two sides of the belt.

We know that power transmitted by the belt (P)

12 12

12

(– ) (– )30

120 30 ( – )

4N 30

TTv TT

== =TT

∴ 12

120

(– ) 4N

30

TT== Ans.

33.14.RATIO OF TENSIONS

Consider a follower (i.e. driven) pulley rotating in the clockwise direction as shown in
Fig 33·9.

Fig. 33.9. Ratio of tensions.
Let T 1 = Tension in the belt on the tight side,
T 2 = Tension in the belt on the slack side, and
θ= Angle of contact in radians i.e., angle subtended by the arc
AB, along which the belt touches the pulley, at the centre.
Now consider a small portion of the belt PQ, subtending an angle δθ at the centre of the pulley
as shown in Fig. 33·9. The belt PQ is in equilibrium under the following forces :

1. Tension T in the belt at P,


2. Tension T + δT in the belt at Q,


3. Normal reaction R, and


4. Frictional force F = μ × R.
   where μ is coefficient of friction between belt and pulley.
   Resolving all the forces horizontally and equating the same,

()sin sin

22

RT T T

⎛⎞ ⎛⎞δθ δθ

=+δ ⎜⎟ ⎜⎟+

⎝⎠ ⎝⎠ ...(i)

Since δθ is very small, therefore substituting sin

22

⎛⎞δθ δθ

⎜⎟=

⎝⎠

in equation (i),

R= ()

22

TT T

δθ δθ

+δ +