Calculating Elevator Acceleration: Free-Body Diagram and Problem Solving Tips

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Homework Help Overview

The discussion revolves around a physics problem involving an elevator and a passenger standing on a scale. The passenger's weight is given, and the scale readings during upward and downward acceleration are provided. Participants are tasked with drawing a free-body diagram and determining the elevator's acceleration based on the scale's readings.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations related to the passenger's weight and the net forces acting on the passenger when the scale reads maximum and minimum values. Questions arise about the correctness of the calculations and the interpretation of acceleration directions.

Discussion Status

Some participants have provided guidance on interpreting the net forces and calculating the elevator's acceleration. There is an ongoing exploration of the second part of the problem, with multiple interpretations of the results being discussed. Participants are clarifying their understanding of the relationship between the passenger's acceleration and the elevator's acceleration.

Contextual Notes

Participants are working under the constraints of an exam preparation context, with a focus on understanding the principles of forces and acceleration in a physics scenario. There is an emphasis on ensuring clarity in the calculations and the assumptions made during the problem-solving process.

Cicima
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I got a list of problems that will be on some of our exams soon, and i don't know how to do a few of them, here's one.

a 60kg passenger standing on a scale in an elevator notes that the scale's maximum reading is 800 Newtons when the elevator accelerates upward. the scales minimum reading is 400 Newtons.
You must draw a free-body-diagram of the passenger. You must use the free-body-diagram to determine the elevators acceleration when the scales reading is at the maximum and the minimum.


I know how to draw the diagram, just now sure how to solve the problem.
My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
800-588.6=211.4
a=F/m = 211.4 N/60 2Kg = 3.52 m/s^2 down
gravity is expierienced, so total acceleration= 3.52m/s^2 +9.81 m/s^2 = 13.33 m/s^2 down...Thats the total acceleration of the elevator when the scale is its max. Does that look anywhere near correct? Or am i doing this completely wrong? Please help, i have a few more problems too, i really need to know these for the exam i will be taking.
 
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Cicima said:
I got a list of problems that will be on some of our exams soon, and i don't know how to do a few of them, here's one.

a 60kg passenger standing on a scale in an elevator notes that the scale's maximum reading is 800 Newtons when the elevator accelerates upward. the scales minimum reading is 400 Newtons.
You must draw a free-body-diagram of the passenger. You must use the free-body-diagram to determine the elevators acceleration when the scales reading is at the maximum and the minimum.


I know how to draw the diagram, just now sure how to solve the problem.
My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
800-588.6=211.4
a=F/m = 211.4 N/60 2Kg = 3.52 m/s^2 down
gravity is expierienced, so total acceleration= 3.52m/s^2 +9.81 m/s^2 = 13.33 m/s^2 down...Thats the total acceleration of the elevator when the scale is its max. Does that look anywhere near correct? Or am i doing this completely wrong? Please help, i have a few more problems too, i really need to know these for the exam i will be taking.
When the scale reads 800, the net force is UP (800 up -588down = 211 UP) So the accelration is up. Forget about total aceleration! The net force is 211 up, so the acceleration works out to 3.52m/s/s UP. That's Newton's law. Now try the 2nd part when the scale reads lower.
 
PhanthomJay said:
When the scale reads 800, the net force is UP (800 up -588down = 211 UP) So the accelration is up. Forget about total aceleration! The net force is 211 up, so the acceleration works out to 3.52m/s/s UP. That's Newton's law. Now try the 2nd part when the scale reads lower.
Before i try to find the 2nd part, when the elevator is going down, let me make sure i understand what your saying about my answer...

When the scale is reading 800 Newtons, the elevators acceleration is 3.52 M/S^2 up. Is that correct? It seems like too little work was done to prove that...
 
Cicima said:
Before i try to find the 2nd part, when the elevator is going down, let me make sure i understand what your saying about my answer...

When the scale is reading 800 Newtons, the elevators acceleration is 3.52 M/S^2 up. Is that correct? It seems like too little work was done to prove that...
Yes, you are right. That is the person's acceleration with respect to the ground. But since the person stays in contact with the floor, I should have noted that the acceleration of the elevator with respect to the ground must be the same!
 
PhanthomJay said:
When the scale reads 800, the net force is UP (800 up -588down = 211 UP) So the accelration is up. Forget about total aceleration! The net force is 211 up, so the acceleration works out to 3.52m/s/s UP. That's Newton's law. Now try the 2nd part when the scale reads lower.
My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
588.6-400=188.6
a=F/m = 188.6 N/60 2Kg = 3.14 m/s^2 down, so it the answer would be -3.14 m/s/s. Right? So the total answer would be that when the scale is at its MAX at 800 N, the elevators acceleration is 3.52M/S/S and when the scale is at its min. the elevators acceleration is -3.145M/S/S?
 
Cicima said:
My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
588.6-400=188.6
a=F/m = 188.6 N/60 2Kg = 3.14 m/s^2 down, so it the answer would be -3.14 m/s/s. Right? So the total answer would be that when the scale is at its MAX at 800 N, the elevators acceleration is 3.52M/S/S and when the scale is at its min. the elevators acceleration is -3.145M/S/S?
Yes, that is correct. The person accelerates upward at 3.52m/s/s and feels heavier, and then decelerates (accelerates down) at 3.14m/s/s when coming to a stop, and feels lighter during this deceleration phase. These are also the same magnitudes and directions for the elevator's accelerations.
 

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