Answer to Question #218683 in Mechanical Engineering for Kevine

Question

An open belt 100mm wide connects two pulleys mounted on parallel shaft with their centres 2.4m apart. The diameter of the larger pulley is 450mm and that of the smaller pulley 300mm. The coefficient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to 14N/mm width. If the larger pulley rotates at 120 rpm. Find the maximum power that can be transmitted

Answer 1

Max stress in the belt = $T_1=14N/mm =14 N *100 mm/mm=1400N$

$v=\frac{\pi D_1N_1}{60}=\frac{\pi*0.45*120}{60}=2.827 m/s$

$\alpha =sin^{-1}(\frac{0.45-0.3}{2*2.4})=1.79^0$

$\therefore 2 \alpha =3.58 ^0$

$\theta = (180-2 \alpha)*\frac{\pi}{180}=3.079 rad$

$\frac{T_1}{T_2}=e^{\mu \theta}=e^{0.3*3.079}=2.518 \implies T_2=\frac{1400}{2.518}=555.996 N$

Maximum power transmitted, $P=(T_1-T_2)*V=(1400-555.996)*2.827=2386W=2.39kW$