D_{1} = Diameter of the driver = 120cm = 1.2m

R_{1} = Radius of the driver = 0.6m

N_{1} = Speed of the driver in R.P.M. = 200RPM

D_{2} = Diameter of the driven or Follower = 50cm = 0.5m

R_{2} = Radius of the driven or Follower = 0.25m

X = Distance between the centers of two pulleys = 4m

µ = Coefficient of friction = 0.3

T_{1} = Tension in the tight side of the belt = 1855.3N

Calculation for power transmitting:

**Let **

P = Maximum power transmitted by belt drive

= (T_{1}–T_{2}).V/1000 KW ...(i)

Where, T_{2} = Tension in the slack side of the belt

V = Velocity of the belt in m/sec.

= πDN/60 m/sec, D is in meter and N is in RPM ...(ii)

For T_{2},

We use the relation Ratio of belt tension = T_{1}/T_{2 }= eµθ ...(iii)

But angle of contact is not given, let

θ = Angle of contact and,

θ = Angle of lap for open belt, Angle of contact (θ) = Π – 2α ...(iv)

Sinα = (r_{1} – r_{2})/X = (0.6 – 0.25)/4 α = 5.02° ...(v)

Using the relation (iii), θ = Π – 2α = 180 – 2 X 5.02

= 169.96° = 169.96° X Π/180 = 2.97 rad ...(iv)

Now using the relation (iii)1855.3/T_{2} = e(0.3)(2.967)

T_{2} = 761.8N ...(vii)

For finding the velocity, using the relation (ii) V

= (3.14 X 1.2 X 200)/60 = 12.56 m/sec ...(viii)

For finding the Power, using the relation (i)

P = (1855.3 – 761.8) X 12.56 P = 13.73 KW

We know that, 1. Torque exerted on the driving pulley = (T_{1} – T_{2}).R_{1 }

= (1855.3 – 761.8) X 0.6

**= 656.1Nm**

2. Torque exerted on the driven pulley

= (T_{1} – T_{2}).R_{2} = (1855.3 – 761.8) X 0.25

**= 273.4.1Nm**