This should be a simple problem, but I think I'm making it harder than it is.(adsbygoogle = window.adsbygoogle || []).push({});

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:

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# Homework Help: Elevators and Acceleration

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