Posted by **Justin** on Tuesday, May 10, 2011 at 3:18pm.

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.

- Thermo engineering -
**bobpursley**, Tuesday, May 10, 2011 at 4:00pm
torque= force*radius

power= torque*angular speed where

angular speed= 300*2PI/60 rad/sec

this will give you power in watts.

- Thermo engineering -
**jackson**, Wednesday, September 10, 2014 at 1:05pm
torque=300Nm

power=31.41 Watts

- Thermo engineering -
**Maddy**, Friday, October 16, 2015 at 10:48pm
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

These are the answers in the back of the text.

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