Engineering Mechanics

(^680) „„„„„ A Textbook of Engineering Mechanics

3. An engine, running at 150 r.p.m. drives a line shaft by means of a belt. The engine pulley is 750
   mm diameter and the pulley on the line shaft being 450 mm diameter. The 900 mm pulley on
   the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Find the speed of the
   dynamo shaft, when (i) there is no slip; and (ii) there is a slip of 2% at each drive.
   [Ans. 1500 r.p.m.; 1380 r.p.m.]


4. Two parallel shafts 6 m apart are provided with 900 mm and 300 mm diameter pulleys and are
   connected by means of a cross belt. The direction of rotation of the follower pulley is to be
   reversed by changing over to an open belt drive. How much length of the belt has to be reduced?
   [Ans. 40 mm]

33.13.POWER TRANSMITTED BY A BELT

Fig. 33.8. Power transmitted by a belt.

In Fig. 33·8, is shown the driving pulley (i.e. driver) A and the follower B. We know that the

driving pulley pulls the belt from one side, and delivers the same to the other. It is thus obvious, that

the tension in the former side (i.e., tight side) will be more than that in the latter side (i.e., slack side)

as shown in Fig. 33·8.

Let T 1 = Tension in the tight side

T 2 = Tension in the slack side and

v = Velocity of the belt

We know that the effective turning (i.e. driving) force at the circumference of the follower is

the difference between the two tensions (i.e., T 1 – T 2 ).

∴ Work done = Force × Distance = (T 1 – T 2 ) × v

∴ Power = (T 1 – T 2 )v J/s = (T 1 – T 2 )v

Notes:1.The torque exerted on the driving pulley (T 1 – T 2 ) r 1

2.Similarly, the torque exerted on the follower = (T 1 – T 2 ) r 2 where r 1 and r 2 are in metres.

Example 33·6. The tensions in the two sides of the belt are 1000 and 800 newtons respec-

tively. If the speed of the belt is 75 metres per second, find the power transmitted by the belt.

Solution. Given: Tension in the tight side (T 1 ) = 1000 N; Tension in the slack side (T 2 ) = 800

N and speed of the belt (ν) = 75 m/s

We know that power transmitted by the belt,

P = (T 1 – T 2 ) v = (1000 – 800) × 75 = 15 000 N-m/s

= 15 000 W = 15 kW Ans.