Scale Readings in Accelerating Elevators: Calculations and Analysis

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Homework Help Overview

The discussion revolves around a physics problem involving a man standing on a scale in an accelerating elevator. The scenario includes calculating the scale reading when the elevator accelerates downward at 1.5 m/s² and exploring the implications of maximum acceleration during free fall.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational force and the scale reading, questioning the impact of downward acceleration on the scale's measurement. There are attempts to derive the scale reading using force equations, and some participants express confusion about the concept of maximum acceleration.

Discussion Status

Participants are actively engaging with the problem, offering various interpretations and calculations. Some guidance has been provided regarding the direction of forces and the expected behavior of the scale reading under different conditions, but there is no explicit consensus on the correct approach or final answer.

Contextual Notes

There is uncertainty regarding the definition of maximum acceleration in the context of the problem, and some participants question the assumptions made about the forces acting on the man in the elevator.

Soaring Crane
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A 910 N man stands on a scale in an elevator that is accelerating downward 1.5 m/s^2. What does the scale read in N? If the elevator moves down with max. acceleration, what will the scale read?

I started to do this:

F = ma
910 N = m(9.8 m/s^2)
m = 92.86 kg

mg - F_s = ma
910 N - F_s = 92.86 kg*(-1.5 m/s^2)
F_s = 139.29 N + 910 N = 1049.29 N

But if the acc. is downward, wouldn't the scale have a reading that is < 910?

I also don't understand the second part. What is the maximum acc.?

Please correct me on my mistakes and help me on the second part. Thanks.
 
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The scale should have a reading of less than 910N.

Weight of person - Resultant Contact Force = mass of person x acceleration of lift.

I don't know what maximum acceleration is. Perhaps they are asking you to find the person's scale reading in the lift in free fall. Try using the formula I gave above.
 
You need to remember that your formula must reduce to what you know when the acceleration of the lift is zero.
That is, the force on the scales will be mg (when a=0). Also don't forget that this force is directed downwards (in the direction of the acceleration of the elevator).

I believe the formula you need is

[tex]F_s = m(g-a)[/tex]

Using this you should get something like 770N
 
I think what I gave you was exactly the same as what recon gave...
sorry.
 
Soaring Crane said:
mg - F_s = ma
910 N - F_s = 92.86 kg*(-1.5 m/s^2)
F_s = 139.29 N + 910 N = 1049.29 N

But if the acc. is downward, wouldn't the scale have a reading that is < 910?

I also don't understand the second part. What is the maximum acc.?

Please correct me on my mistakes and help me on the second part. Thanks.

That was good to realize your answer couldn't be right. If you can visulize the problem, common sense catches lots of mistakes.

The mistake in your calculation is the direction gravity is pulling the object. The gravitational force is -910 N (down), the same direction that the elevator is accelerating.

Unless the elevator has a motor to push it downward, the maximum acceleration would be 9.8 m/s^2 since the elevator will be accelerated by gravity. (My brother-in-law took a job as a coal miner and the initiation for his first day on the job was for his new 'buddies' to turn the lights off in the elevator right before it went into free-fall down the shaft. In my brother-in-law's case, his first day was also his last. He decided to take a job as a prison guard, instead - it paid less, but you didn't have to buy new underwear so often.)
 

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