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

  • Thread starter Thread starter mcgooskie
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion focuses on determining the maximum acceleration of an elevator during normal operation, based on the apparent weight increase experienced by a person standing on a scale inside the elevator. The initial calculation yielded a maximum acceleration of 7.99 ft/s², but there was confusion about whether to add this to the acceleration due to gravity. Clarifications indicate that the maximum acceleration should not be combined with gravity since the elevator is not in free fall. The correct calculation suggests that the maximum acceleration is approximately 8.05 ft/s², derived from the net force acting on the mass. The impact of varying weights on acceleration was also acknowledged, emphasizing that different weights would yield different acceleration results.
mcgooskie
Messages
4
Reaction score
0
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
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top