How Does an Elevator's Motor Calculate Force and Work?

[zoner7]

Homework Statement

An elevator consists of an elevator cage and a counterweight attached to the ends of a cable that runs over a pulley. The mass of the cage (with its load) is 1200kg, and the mass of the counterweight is 1000kg. The elevator is driven by an electric motor attached to the pulley. Suppose the elevator is initially at rest on the first floor of the building and the motor makes the elevator accelerate upward at the rate of 1.5 m/s.

Mass of elevator = 1200kg
Mass of counterweight = 1000kg
Acceleration = 1.5 m/sa) What is the tension in the part of the cable attached to the elevator cage? What is the tension in the part of the cable attached to the counterweight.

b) The acceleration lasts exactly 1.0s. How much work has the electric motor done in this interval? Ignore frictional forces and the mass of the pulley.

Time = 1.0s
Interval = 1.5m

c) After the acceleration interval of 1.0s, the motor pulls the elevator at a constant speed until it reaches the third floor, exactly 10.0m above the first floor. What is the total amount of work that the motor has done up to this point.

Distance = 10.0m

Homework Equations

[tex]\Sigma[/tex]F = m*a
Work = F*D

The Attempt at a Solution

a)
Tension on the cable connected to elevator:
[tex]\Sigma[/tex]F = m*a
T-mg = ma
T = mg + ma
T = 1200kg(9.8m/s^2 + 1.5m/s)
T = 13,560 N

Tension on the cable connected to counterweight:
[tex]\Sigma[/tex]F = m*a
-T-mg = ma
-T = ma + mg
T = -m(a + g)
T = 8,300N

This problem confused me. I set the tension to be negative because the motor is pushing down on the cable. This really dosen't make sense. Maybe some force is negative, but it sure isn't tension. Also, since the elevator is going down, why isn't acceleration negative? Keep in mind, I did get the right answer.

b)
W = F * D
W = (13,560N - 8,300N) * 1.5m
W = 7890 J

I know this answer isn't right. I don't understand what I am supposed to do for the force.

c)
[tex]\Sigma[/tex]F = ma
[tex]\Sigma[/tex]F = 0
F - mg = 0

This problem is similar to the last one. Once again, I'm not sure how to calculate the force. Do I need to examine just the force on the elevator and the motor or that caused by the counterweight as well? How are the elevator and the counterweight related? Obviously the counterweight reduces the amount of force that the motor needs to apply in order to overcome the inertia of the elevator.

 

[jbsteele]

I would really appreciate it if someone could explain this problem to me. I'm really confused about how the tension and forces are related. Any help is greatly appreciated. Thank you!

 

[baba89]

But how do I account for that in my calculations?

a) The tension in the cable attached to the elevator cage can be calculated by using the equation \SigmaF = m*a. The elevator is accelerating upward, so the acceleration should be positive. The equation would then be T-mg = ma. Solving for T, we get T = ma + mg. Plugging in the given values, we get T = 1200kg(9.8m/s^2 + 1.5m/s) = 13,560 N. This is the correct answer.

The tension in the cable attached to the counterweight can be calculated in a similar manner. The counterweight is accelerating downward, so the acceleration should be negative. The equation would then be -T-mg = ma. Solving for T, we get T = -ma - mg. Plugging in the given values, we get T = -1000kg(-1.5m/s^2 + 9.8m/s^2) = 13,300 N. This is the correct answer.

b) The work done by the electric motor can be calculated using the equation W = F*D. In this case, the force is the net force on the elevator, which is the difference between the tension on the cable attached to the elevator cage and the tension on the cable attached to the counterweight. So, the force is (13,560 N - 13,300 N) = 260 N. The distance is the acceleration interval of 1.0s, so D = 1.0s * 1.5m/s = 1.5m. Plugging in the values, we get W = 260N * 1.5m = 390 J. This is the correct answer.

c) To calculate the total amount of work done by the motor, we need to consider both the acceleration interval and the constant speed interval. During the constant speed interval, the elevator is not accelerating, so the net force on it is 0. The only force acting on the elevator is the tension in the cable attached to it, which is equal to its weight. So, the work done during the constant speed interval is W = F*D = mg*D = 1200kg * 9.8m/s^2 * 10.0m = 117,600 J.

Adding this to the work done during the acceleration