How do I Find Average Force in a Force vs Time graph?

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Homework Help Overview

The discussion revolves around finding the average force from a Force vs Time graph, focusing on the relationship between force, time, and impulse in the context of physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of average force and its calculation from the graph, with some attempting to relate the area under the curve to impulse. Questions arise regarding the interpretation of different sections of the graph and how to average the forces represented.

Discussion Status

There is an ongoing exploration of the relationship between impulse and average force, with some participants providing insights into the concept of time-averaged force. However, there is no explicit consensus on the method of calculation, and various interpretations are being discussed.

Contextual Notes

Participants are navigating through the definitions and implications of impulse and average force, with some uncertainty about the specific calculations involved. The original poster expresses confusion about the process, indicating a need for clarification on the steps to take.

haroldham
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This is really simple, but I can not remember how to find the average force in a Force vs TIme graph.
 
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Think about the height of a "[certain] rectangle." Think area.
 
robphy said:
Think about the height of a "[certain] rectangle." Think area.
Ok, you completely lost me, but I think I figured it out. This certain graph increases, remains constant, then decreases. Do I take the final force of each section, add it to the initial force of each section, divide that value by the time of that section, then average those together?
 
Presumably, you mean "time-averaged force".
The area under your graph is equal to the impulse (ie, the change in momentum).
The "time-averaged force" you seek is the constant force you would apply in the same amount of time in order to obtain the same change in momentum. Translate the last sentence into facts about the graph.
 
Oh, ok I get it. It is impulse divided by time.
 

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