Elevator and unknown passenger mass

In summary, an elevator motor with 60% efficiency lifts a 2000kg elevator 30m using 80kW of power in 15 seconds. With five passengers of equal mass, the mass of each passenger is determined to be 76.822kg. The motor's efficiency is used to calculate the total mass, assuming that the power measured is for the loaded elevator.
  • #1
retracell
22
0

Homework Statement


An elevator motor which is 60% efficient uses 80kW of power while lifting a 2000kg elevator 30m. If the elevator contains five passengers of the same mass and it takes 15 seconds to accomplish this task, then determine the mass of each passenger.

Homework Equations


d1=d0+v0t+1/2at2
Fnet=ma
W=FDcos[tex]\theta[/tex]
[tex]\epsilon[/tex]=Eoutput/Einputx100%

The Attempt at a Solution


d1=d0+v0t+1/2at2
30=0+0+1/2(152)a
30=112.5a
a=0.2666666

Let x rep. each individual mass
Fapplied-mg=ma
F=(2000+5x)(0.26666)+(2000+5x)(9.8)
F=(2000+5x)(10.06666)

W=(2000+5x)(10.06666)(30)(1)
W=(2000+5x)(301.9996)
W=603999.2+1509.998x

P=(603999.2+1509.998x)/15
P=40266.61333+100.6665333x

0.6=40266.61333+100.6665333x/80000
0.6(80000)=40266.61333+100.6665333x
7733.38667=100.6665333x
x=76.822kg

Is this correct?
 
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  • #2
Welcome to PF.

I think it is a little easier to figure than your approach.

Your motor runs at 60% efficiency for 15 s. That means that it delivered 80K*(.6)*15 Joules.

The motor lifted the car m*g*h.

Doesn't that mean total M = 80K*(.6)*15/(30*9.8) ?
 
  • #3
LowlyPion said:
Welcome to PF.

I think it is a little easier to figure than your approach.

Your motor runs at 60% efficiency for 15 s. That means that it delivered 80K*(.6)*15 Joules.

The motor lifted the car m*g*h.

Doesn't that mean total M = 80K*(.6)*15/(30*9.8) ?

Thank you for your reply.

From what I understand, you're applying E=m*g*h, then rearranging to m=E/g*h right?
Does this mean that time is proportional to mass in this situation because I thought the elevator motor's time is constant.

I keep thinking that it is on two separate occasions, one where there are no passengers and one with five passengers.

When calculating problems such as these, do you always use the output power/energy or is there a way to solve using input and the above efficiency equation.
 
  • #4
I'm not sure what efficiency equation you would use, since the efficiency of the power conversion to mechanical work is apparently given.

The presumption is that the motor only delivers 60% of its 80kw to mechanical power. The mechanical power available to work then is given by the time it took to accomplish the task.

Power times time is work and work in this case would be increasing Potential Energy.
 
  • #5
Suppose it asks us to find the work done without the passengers. Would the 15 seconds still apply or will it change? Just need to clarify on what determines it an empty elevator or a loaded one.
 
  • #6
In practice the motor would draw more current and create more torque within a range of loads and the added mass would be simply accounted for by just the increased power.

But the statement of the problem only provides the single rate of delivering work. Hence you would expect that the time for an empty elevator would be less than a full one. I think the statement of the problem leaves something to be desired. I think the problem would have been better served to indicate that the power measured was for the loaded elevator.
 
  • #7
Yes! That's exactly what got me confused and made me think that there are two separate systems.

Thank you very much for your clarification.
 

What is the effect of a passenger's mass on an elevator's movement?

The mass of a passenger does not have a significant impact on an elevator's movement. Elevators are designed to carry a certain weight capacity, and as long as this capacity is not exceeded, the movement of the elevator will not be affected by the passenger's mass.

How does an elevator determine the weight of its passengers?

Most elevators use a sensor system that measures the force exerted on the floor of the elevator. This force is then converted into weight, which is used to determine the total weight of the passengers in the elevator.

Can an elevator malfunction due to an unknown passenger's mass?

No, elevators are designed to handle a wide range of weights and can function properly even if the exact weight of a passenger is unknown. The weight capacity of the elevator is the most important factor in ensuring safe operation.

What is the average weight of a passenger used in elevator design?

The average weight used in elevator design is typically around 150-200 pounds per person. This is a standard weight used to calculate the weight capacity of the elevator and ensure safe operation.

How does an elevator adjust for varying passenger weights?

Elevators are equipped with a weight sensor system that constantly measures the weight of the passengers. If the weight capacity is reached, the elevator will not allow any more passengers to enter until some have exited. This ensures that the weight limit is not exceeded and the elevator can operate safely.

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