What is the maximum acceleration for an elevator during normal operation?

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Homework Help Overview

The discussion revolves around determining the maximum acceleration of an elevator during normal operation, specifically focusing on the relationship between apparent weight and acceleration as experienced by a person standing on a scale inside the elevator.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate maximum acceleration using the apparent weight readings from a scale, questioning whether to add the calculated acceleration to gravitational acceleration.
  • Some participants question the necessity of assuming a specific weight, considering how varying weights might influence the calculated acceleration.

Discussion Status

The discussion includes attempts to clarify the calculations and assumptions involved in determining maximum acceleration. Some guidance has been offered regarding the interpretation of gravitational acceleration in the context of the problem.

Contextual Notes

Participants are exploring the implications of using different weights in the calculations and how that affects the resulting acceleration, indicating a need for clarity on the relationship between force, mass, and acceleration in this scenario.

mcgooskie
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This should be a simple problem, but I think I'm making it harder than it is.

In this problem, I am interested in maximum acceleration for an elevator during normal operation. While the elevator is at rest, on the ground floor, I get in, put down my bathroom scale and stand on it. I continue standing while the elevator is going up. During my trip to the 45th floor, the scale reading increases by a max of 25lbs.

Assumptions:
It doesn't matter what floor you go to.
When the elevator accelerated upward, the apparent weight is greater than mg by the amount ma. It's as if gravity were increased from g to g+a.

According to an equation I found in my textbook...
Fn-mg=ma (where a is in the y direction)
Fn=mg+ma (where Fn is the reading on the scale, the apparent weight)

Since the reading is given in lbs, I am going to use 32.2ft/s^2 for gravity.
I am going to start with an arbitrary weight of 100lbs, which would give a max weight of 125lbs.
And W=mg so...100lbs=(m)(32.2ft/s^2)=3.11slugs

Fn=mg+ma
125lbs=(3.11slugs)(32.2ft/s^2)+(3.11slugs)(a ft/s^2)
125lbs-100.14lbs=(3.11slugs)(a ft/s^2)
24.86lbs=(3.11slugs)(a ft/s^2)
7.99ft/s^2=a

If I did the problem right...
Is 7.99ft/s^2 my answer for max acceleration? Or do i add that to 32.2ft/s^2 for a max acceleration of 40.19ft/s^2?

Thanks! Kelli
 
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mcgooskie said:
If I did the problem right...
Is 7.99ft/s^2 my answer for max acceleration? Or do i add that to 32.2ft/s^2 for a max acceleration of 40.19ft/s^2?
You did the problem correctly. (But see my steps below.) You calculated the max acceleration (assuming a 100 lb weight). Why would you add it to g? (g is the acceleration of a freely falling body--the elevator isn't falling, is it?)

Here's how I'd do it:

Fn - mg = ma
(mg + 25) - mg = ma
25 = ma

a = 25/m = 25 (32.2/100) = 8.05 ft/s^2 (upwards)
 
Do I need to even assume a 100lb weight? What if i used a weight different than 100lbs, wouldn't that affect the acceleration?
 
mcgooskie said:
Do I need to even assume a 100lb weight? What if i used a weight different than 100lbs, wouldn't that affect the acceleration?
Yes it would. The same net force (25 lbs) acting on a different mass would yield a different acceleration.
 

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