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

In summary, the conversation discusses a problem where 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 and the minimum reading is 400 Newtons. The conversation also includes a discussion on drawing a free-body-diagram of the passenger and using it to determine the elevator's acceleration when the scale readings are at the maximum and minimum. The conclusion is that the elevator's acceleration is 3.52 m/s^2 when the scale reads 800 Newtons and -3.14 m/s^2 when the scale reads 400 Newtons. Both the passenger and the elevator experience the same accelerations.
  • #1
Cicima
20
0
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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  • #2
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.
 
  • #3
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...
 
  • #4
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!
 
  • #5
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?
 
  • #6
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.
 

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

What is elevator acceleration?

Elevator acceleration refers to the rate of change in the elevator's velocity over a certain period of time. It is measured in meters per second squared (m/s²) or feet per second squared (ft/s²).

How is elevator acceleration calculated?

Elevator acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = Δv / Δt, where a is the acceleration, Δv is the change in velocity, and Δt is the change in time.

What factors affect elevator acceleration?

Several factors can affect elevator acceleration, including the weight of the elevator and its load, the power and efficiency of the motor, and the friction and resistance of the elevator's components.

How does acceleration impact elevator safety?

Acceleration is an important factor in elevator safety as it affects the smoothness of the ride and the comfort of passengers. Too much acceleration can cause discomfort and even injuries, while too little acceleration can result in longer wait times and inefficient operation.

How can elevator acceleration be controlled?

Elevator acceleration can be controlled by adjusting the motor's power and speed, as well as by maintaining the elevator's components to reduce friction and resistance. Elevator manufacturers also have safety regulations and standards in place to ensure proper acceleration levels for passenger safety.

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