The belt is 5 mm thick and 80 mm wide. The coefficient of friction between the belt and pulley is 0.3 and the maximum permissible tension in the belting is 1.4MN/m2.

Determine:

1.diameter of the driving pulley,

2.maximum power that may be transmitted by the belting, and

3.required initial belt tension.

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 …

a Flat belt type is use it For a belt drive two pulleys 1.2m apart e driver pulley with a diameter 40 cm is routing with speed 350 rpm while diameter of driven pulley is 100 cm coefficient of friction of the contact surface between belt and pulley is 0.3 …