Does air resistance have a max value?

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SUMMARY

Air resistance, or drag force, does not have a theoretical maximum value; however, it reaches a practical limit when an object in free fall attains terminal velocity. This occurs when the gravitational force acting on the object is balanced by the drag force, which is dependent on the object's weight and the drag coefficient. The drag coefficient varies based on the shape and orientation of the object. Understanding these principles is crucial for analyzing the behavior of free-falling objects in atmospheric conditions.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of terminal velocity
  • Knowledge of drag force and its calculation
  • Basic grasp of the drag coefficient and its influencing factors
NEXT STEPS
  • Research the drag equation and its applications in physics
  • Study the factors affecting drag coefficients for various shapes
  • Explore the concept of terminal velocity in different mediums
  • Investigate real-world examples of free-fall scenarios and their outcomes
USEFUL FOR

Physics students, engineers, and anyone interested in understanding the dynamics of free-falling objects and the effects of air resistance on motion.

r4g3r4hk
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I know that friction on a surface has a maximum value. When an object is pushed with a force greater than the maximum friction value, the object will move.
But when a free-falling object falls from a height, does air resistance have a maximum value as well?
 
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r4g3r4hk said:
I know that friction on a surface has a maximum value. When an object is pushed with a force greater than the maximum friction value, the object will move.
That's an odd way to view it. Static friction is a function of the normal force. There is no theoretical limit, only practical/materials limits.
But when a free-falling object falls from a height, does air resistance have a maximum value as well?
By constraining the scenario to [atmospheric, I assume] free fall, you are defining the drag limit to be equal to the weight of the object.
 
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Thanks for the replies! I believe I understand now.
 
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