mcgooskie
May24-04, 07:26 PM
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 :yuck:
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 :yuck: