Power of Elevator: Calculating Motor Power Needed

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SUMMARY

The discussion focuses on calculating the motor power required for an elevator with a mass of 956 kg and a maximum load of 764 kg, factoring in a frictional force of 4150 N. Using the formula P = Fv, where F is the net force and v is the instantaneous speed of 4.69 m/s, the solution derives the necessary power output as approximately 107,144.62 W. The calculations incorporate gravitational acceleration (9.8 m/s²) and an upward acceleration of 1.07 m/s², confirming the logical progression of the problem-solving approach.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of basic physics concepts such as force, mass, and acceleration
  • Familiarity with power calculations in physics (P=Fv)
  • Ability to perform unit conversions and calculations in SI units
NEXT STEPS
  • Study the principles of elevator mechanics and design
  • Learn about frictional forces and their impact on motion
  • Explore advanced power calculations in mechanical systems
  • Investigate safety factors in elevator motor power requirements
USEFUL FOR

Mechanical engineers, physics students, and professionals involved in elevator design and maintenance will benefit from this discussion, particularly those focused on power calculations and system dynamics.

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Homework Statement


An elevator has a mass of 956 kg and car-
ries a maximum load of 764 kg. A constant
frictional force of 4150 N retards its motion
upward.
The acceleration of gravity is 9.8 m/s2 .
What power must the motor deliver at a
instantaneous speed of 4.69m/s if the elevator
is designed to provide an upward acceleration
of 1.07 m/s2? Answer in units of W.

The Attempt at a Solution



F=ma
N-mg=(me+ml)a
N=(me+ml)(a+g)
F=N-frictional force
P=Fv

I was wondering if this progression of the problem would be logical to solve for the power.
 
Last edited:
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I got it:

Fy=ma
N-mg-f=ma
N=(me+ml)(g+a) + friction

P= Nv

=107144.616
 

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