219k views
1 vote
A 0.15-m-diameter pulley turns a belt rotating the drive- shaft of a power plant pump. The torque applied by the belt on the pulley is 200 N ⋅ m, and the power transmitted is 7 kW. Determine the net force applied by the belt on the pulley, in kN, and the rota- tional speed of the driveshaft, in RPM.

User Baumann
by
6.4k points

1 Answer

6 votes

Answer:

F= 2.66 KN

N=334.22 RPM

Step-by-step explanation:

Given that

d= 0.15 m

r= 0.075 m

T= 200 N .m

P = 7 KW = 7000 W

We know that

T= F .r

200 = F x 0.075


F=(200)/(0.075)\ N

F=2666.6 N

F= 2.66 KN

We know that


P=(2\pi NT)/(60)


N=(60* P)/(2* \pi T)


N=(60* 7000)/(2* \pi* 200)\ RPM

N=334.22 RPM

User Tylerjw
by
7.3k points