How Does Elevator Acceleration Affect Gold Trading Profits?

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Homework Help Overview

The discussion revolves around a scenario where olympic commissioners engage in buying and selling gold while in an elevator, with specific accelerations affecting the measurements of weight. The problem involves understanding how these accelerations influence the perceived weight of gold and consequently the profit calculations based on a constant price per ounce.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the gold nugget in an accelerating elevator and how this affects the weight readings on a spring scale. Questions arise regarding the implications of upward and downward accelerations on the perceived weight and the subsequent selling price.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about weight measurement under acceleration and clarifying the relationship between force and price. Some have provided insights into the equations governing the situation, while others are still grappling with the implications of these calculations for profit determination.

Contextual Notes

There is a focus on the specific accelerations during buying and selling, with participants noting the lack of explicit mass information and the need to consider how these factors influence the final selling price. The discussion also highlights the importance of distinguishing between force and price in their calculations.

  • #31
Yes! Now do all the cancellations that are possible in that expression.
 
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  • #32
Profit % = mp(adown+aup)/p(mg-ma) x 100%
Profit % = mp(adown+aup)/mp (g-a) x 100%
Profit % = (adown+aup)/(g-a) x 100%
Profit % = (2 m/s^2 + 2.5 m/s^2) / (9.8m/s^2 - 2 m/s^2) x 100%
Profit % = 4.5 m/s^2 / 7.8 m/s^2 x 100%
Profit % = 57.7 %
 
  • #33
Bingo.
 
  • #34
haruspex said:
Bingo.

I would like to thank you for all your help.
You have been extremely patient and once again thank you.
 
  • #35
Well i was thinking of the orientation shouldn't it be like -6.4% see:

Profit % = mp(adown(down)+aup(up))/p(mg(down)-ma(down)) x 100%
Profit % = mp(adown+aup)/mp (g-a) x 100%
Profit % = (adown+aup)/(g-a) x 100%
Profit % = (2 m/s^2(down) - 2.5 m/s^2(down)) / (9.8m/s^2 (down)- 2 m/s^2(down)) x 100%
Profit % = -0.5m/s^2/ 7.8 m/s^2 x 100%
Profit % = -6.4%
 

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