The elevator system shown has a car weight of 800 lb and carries passengers weig

Question

The elevator system shown has a car weight of 800 lb and carries passengers weighing 2500 lb.  The counterbalance weight is 2000 lb.  There are 3 cables connecting the car to the counterweights.  The system is designed to reach a maximum speed of 450 ft/min with equal acceleration and deceleration rates of 2.5 ft/s^2.  The distance between floors is 12 feet, but maximum speed is not reached when moving only one floor as the car accelerates in the first 6 ft and decelerates in the final 6 ft.

Asked:  Determine the horsepower input to the drive pulley while rotating at 50 rpm and producing upward acceleration of the elevator car.

Please show and explain all work for any credit.

The elevator system shown has a car weight of 800 lb and carries passengers weighing 2500 lb. The counterbalance weight is 2000 lb. There are 3 cables connecting the car to the counterweights. The system is designed to reach a maximum speed of 450 ft/min with equal acceleration and deceleration rates of 2.5 ft/s^2. The distance between floors is 12 feet, but maximum speed is not reached when moving only one floor as the car accelerates in the first 6 ft and decelerates in the final 6 ft. Asked: Determine the horsepower input to the drive pulley while rotating at 50 rpm and producing upward acceleration of the elevator car. Please show and explain all work for any credit.

Explanation / Answer

P =F*v

v = omega*R

omega =(50 rpm*2*pi)/60 =5.236 s^-1

R = 30 inch =30*2.54*10^-2 =0.762 m

v = 5.236*0.762 =3.99 m/s

a =2.5 ft/s^2 =2.5*0.305 m/s^2 =0.7625 m/s^2

m1 =800 +2500 =3300 lb =1496.85 kg

m2 =2000 lb =907.2 kg

F1 =m1*(g+a) = 1496.85(9.81+0.7625) =15825 N

F2 =m2(g-a) =907.2(9.81-0.7625) =8208 N

P = (F1-F2)*V = (15825-8208)*3.99 =30392 W =30392/745.7 =40.756 HP

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.