Answer to Question #202464 in Mechanical Engineering for Brian

Question

 An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centers 2.4 m apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coefficient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to 14 N/mm width. If the larger pulley rotates at 120 r.p.m., 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, $P=(T_1-T_2)*V=(1400-555.996)*2.827=2386W=2.39kW$