Given:- x= 1.2m ; $N_{2}$ = 350 rpm, $N_{2}$ = 140 rpm.; $d_{2}$ = 400 mm =0.4m;

t =5 mm; 0.005m; b = 80 mm= 0.08m; $\mu$ = 0.3, $\sigma$ = 1.4 MN/m$^{2}$ = 1.4 × 106 N/m$^{2}$

1.Diameter of the driving pulley

Let $d_{1}$ = Diameter of the driving pulley,

We know that $\frac{N_{2}}{N_{1}}$ =$\frac{d_{1}}{d_{2}}$or $d_{1}$ = $\frac{N_{2}d_{2}}{N_{1}}$ =(140 ×0.4)/350 =0.16m.

2.Maximum power transmitted by the belting

First of all, let us find the angle of contact of the belt on the smaller pulley (or driving pulley).

Let $\theta$ = Angle of contact of the belt on the driving pulley.

From fig. we find that

$\sin\alpha=\frac{O_{2}M}{O_{1}O_{2}}=\frac{(r_{2}-r_{1})}{x}=\frac{(d_{2}-d_{1})}{2x}=\frac{0.4 - 0.16}{2\times 1.2}=0.1$

$\alpha$ = 5.740

$\theta$ = 1800 -2 $\alpha$ = 1800 -2$\times$5.740 = 168.520

$T_{1}$ = Tension in the tight side of the belt, and

$T_{2}$= Tension in the slack side of the belt.

Let

We know that

$2.3log\frac{(T_{1}}{T_{2})}=\mu\theta$

$log\frac{(T_{1}}{T_{2})}$ =0.882/2.3 =0.3835 or $\frac{(T_{1}}{T_{2})}$ = 2.42 ….(Taking antilog of 0.38)

We know that maximum tension to which the belt can be subjected,

$T_{1}$ = $\sigma$ × b × t = 1.4×106×0.08×0.005 = 560N

$T_{2}$ =$T_{1}$/2.42 =560/2.42 =231.4N.

Velocity of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times0.16\times 350}{60}$ = 2.93m/s

Power transmitted, P = $(T_{1} - T_{2})$ν =(560-231.4)2.93 = 963W = 0.963 kW

**3. Required initial belt tension**

We know that the initial belt tension,

$T_{0}=\frac{(T_{1}+T_{2})}{2}$ = (560+231.4)/2 =395.7N.