Calculating Acceleration in Elevators and Balloons

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Homework Help Overview

The discussion revolves around calculating the acceleration of an elevator and a weather balloon. The first problem involves a student on a scale in an elevator, where the scale readings change as the elevator moves. The second problem concerns a weather balloon and the forces acting on it as it ascends.

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Approaches and Questions Raised

  • Participants explore the implications of changing scale readings in the elevator scenario, questioning the meaning of the increase in weight. Some suggest using Newton's second law (F=ma) to analyze the forces involved. Others discuss proportionality as a method to find acceleration.

Discussion Status

Participants have shared various approaches to the problems, with some providing calculations and others expressing confusion about specific terms. There is no explicit consensus on a single method, but multiple interpretations and calculations are being explored.

Contextual Notes

Some participants indicate uncertainty about the assumptions underlying the problems, such as the interpretation of scale readings and the conditions of the balloon's ascent.

Moe_the_Genius
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1. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. As the elevator moves up, the scale reading increases to 936 N, then decreases back to 836 N. Find the acceleration of the elevator.

2. The instruments attached to a weather balloon have a mass of 5.0 kg. The balloon is released and exerts an upward force of 98 N on the instruments. What is the acceleration of the balloon and the instruments? After the balloon has accelerated for 10 s, the instruments are released. What is the velocity of the instruments at the moment of their release?

For the first one, I don't understand what it means by "increases to 936 N". Thanks for any help.
 
Last edited:
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The scale's display reads 936 N rather than 836 N.

I'm assuming this is homework so please show us an attempt at doing the problem.
 
Okay, for your first problem, the increase in weight, in Newtons, occurs becaus the elevator moves up. For instance, when you're in an elevator, you feel heavier when going up, vice versa for going down, you feel lighter. For the problem itself, you might be able to use F=Ma, where your force is the difference between 936 and 836, which is 100, and the mass is just the original divided by 9.8=>kgs. I tried the problem out, and I got a decent acceleration. I may be wrong because I'm used to having time and distance thrown in.
 
Last edited:
Moe_the_Genius said:
1. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. As the elevator moves up, the scale reading increases to 936 N, then decreases back to 836 N. Find the acceleration of the elevator.

For the first one, I don't understand what it means by "increases to 936 N". Thanks for any help.

The scale reads 936N instead of 836N.

The question can be solved using proportionality.

836N/9.8m/s^2 = 936N/xm/s^2

x = (936*9.8)/836

x = 10.97

So that is your "gravity". Subtract actual gravity from that number and you get the elevator's acceleration of 1.17 m/s^2.
 
Both ways work, I took a different approach, but my answer also comes out to be 1.17.
 

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