A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit

Respuesta :

Answer:

T = 438.87 N.m

Explanation:

The power required to raise the 4961 N load in 10 meters for 2 minutes is:

[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]

P = Torque  × W

[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]

[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]

[tex]T = \dfrac{413.42}{0.942}[/tex]

T = 438.87 N.m