Area under the curve of a mass time graph

Click For Summary
SUMMARY

The discussion centers on calculating the area under a mass-time graph to determine changes in velocity using Newton's second law. The user blueblast suggests converting mass readings in kilograms to force in Newtons and emphasizes the importance of deducting the resting weight (800 N) from the total force measured. This adjustment allows for accurate momentum calculations by integrating the area under the curve. The user confirms that their method aligns with results obtained through a more complex approach, validating the effectiveness of this simplified technique.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with force and mass units (Newtons and kilograms)
  • Basic knowledge of graphing and integration concepts
  • Experience with momentum calculations
NEXT STEPS
  • Study the principles of momentum and its relation to force and time
  • Learn about graphing techniques for visualizing force vs. time data
  • Explore advanced integration methods for calculating areas under curves
  • Investigate real-world applications of Newton's laws in elevator systems
USEFUL FOR

Students studying physics, particularly those preparing for exams like F=MA, as well as educators and anyone interested in practical applications of Newton's laws in dynamics.

blueblast

Homework Statement



Screen Shot 2017-10-05 at 7.21.32 PM.png


Homework Equations



Force * time = mass * change in velocity[/B]

The Attempt at a Solution



What I did was I converted the y-axis from kilograms to Newtons, since the "mass" reading is the force that the scale experiences. Then, the area under the curve will be the change in momentum, and if you divide that by the mass(80 kg, it is given that the elevator is stationary at t = 0), you can get the change in velocity. How would you use this to solve this problem? I know there are other ways to solve this, I'm just curious if my method works.

This is from the 2013 F=MA exam.

Thanks,

blueblast
 
Physics news on Phys.org
You need to deduct the student's resting weight from the curve before measuring the area under it, as most of the force measured by the scale is being used to resist that weight.

Having done that, if you make a graph of the area under the curve against time (ie the integral of the graphed function), that will show momentum, and you'll be able to see where the period of maximum momentum is.
 
  • Like
Likes   Reactions: blueblast
andrewkirk said:
You need to deduct the student's resting weight from the curve before measuring the area under it, as most of the force measured by the scale is being used to resist that weight.

Having done that, if you make a graph of the area under the curve against time (ie the integral of the graphed function), that will show momentum, and you'll be able to see where the period of maximum momentum is.

Could you possibly give me an example for one of the time periods?
 
Rather than go to all the trouble of calculating area under the graph, why not look at it and work out in what time period the lift is accelerating downwards, when it is accelerating upwards (ie slowing down), and when it is traveling at a constant rate. Think about yourself traveling in a lift. If it is going down, at what point in the ride do you feel lighter than usual, and at what point do you feel heavier than usual?

I love to jump in the air at these two times to feel the sensation of altered weight. is it just me or does anybody else like doing that?
 
andrewkirk said:
Rather than go to all the trouble of calculating area under the graph, why not look at it and work out in what time period the lift is accelerating downwards, when it is accelerating upwards (ie slowing down), and when it is traveling at a constant rate. Think about yourself traveling in a lift. If it is going down, at what point in the ride do you feel lighter than usual, and at what point do you feel heavier than usual?

I love to jump in the air at these two times to feel the sensation of altered weight. is it just me or does anybody else like doing that?

The only reason I actually want to get the values is because I solved it using a different way involving Newton's second law. That way seemed much more complicated than it needed to be, so I wanted to see if this method's values matched up with the more complicated way. I got the same results by subtracting the weight (800 N) from whatever value the y-axis displayed at the time(like you told me to do), then found the area under the curve. After that, I divided the area by the actual mass (80 kg), and got the same results as my other method :) Is this what you were suggesting initially?

Also, regarding your comment on jumping, I have always been afraid that the elevator cable will snap under my weight. A strange phobia of mine.
 
blueblast said:
Is this what you were suggesting initially?
Yes, that's right.
 
  • Like
Likes   Reactions: blueblast

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
2K
Replies
7
Views
1K
Replies
1
Views
3K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K