Asked by Justin

The driveshaft of a building’s air-handling fan is turned at 300 RPM by a belt running on a 0.3-m-diameter pulley. The net force applied by the belt on the pulley is 2000 N. determine the torque applied by the belt on the pulley, in N.m, and the power transmitted, in kW.

Answers

Answered by bobpursley
torque= force*radius

power= torque*angular speed where
angular speed= 300*2PI/60 rad/sec

this will give you power in watts.
Answered by jackson
torque=300Nm
power=31.41 Watts
Answered by Maddy
torque=F*D/2=2000n(0.15m)=300 N*m
Power = Torque*angular speed
=(300 N*m)(300 RPM) (2*PI/60)
=9424 W = 9.42 kW

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