Find force (N) of atmosphere on area

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SUMMARY

The discussion focuses on calculating the force exerted on the eardrum due to atmospheric pressure variations. The correct approach involves using the formula F = PA, where P represents the net pressure difference across the eardrum. Given that the area of the eardrum is 53 mm² (or 53 x 10⁻⁶ m²) and the pressure variation is ±200 Pa, the net force exerted on the eardrum is calculated as F = 200 Pa * 53 x 10⁻⁶ m², resulting in a force of 0.0106 N, not 5353 N as initially calculated.

PREREQUISITES
  • Understanding of pressure and force relationships in physics
  • Familiarity with the formula F = PA
  • Knowledge of unit conversions, specifically mm² to m²
  • Basic concepts of sound pressure levels and thresholds of pain
NEXT STEPS
  • Review the principles of pressure calculations in fluid mechanics
  • Study unit conversion techniques, particularly for area measurements
  • Explore sound pressure levels and their physiological effects on human hearing
  • Learn about the relationship between pressure variations and force in different contexts
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Students studying physics, particularly those focusing on mechanics and sound, as well as educators looking for practical examples of pressure calculations in real-world scenarios.

Calhoun295
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Homework Statement



Your eardrum has an area of about 53 mm2. Sounds become painful to your ear when the pressure variations that are involved in sound reach 200 Pa above and below normal air pressure. At this level, called the threshold of pain, how much force is exerted on your eardrum?

Homework Equations



Pressure=[tex]\frac{Force}{Area}[/tex]

The Attempt at a Solution



F=PA
F= 101000N/m^2 (.053m^2)
F= 5353 N

I've tried normal atmosphere above and below 200Pa. The answer isn't correct according to wileyplus. I'm obviously doing it wrong, so what's wrong with my equation?
 
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Hi Calhoun295! :smile:

(try using the X2 tag just above the Reply box :wink:)
Calhoun295 said:
F=PA
F= 101000N/m^2 (.053m^2)
F= 5353 N

I've tried normal atmosphere above and below 200Pa.

No, you'll have Patm on one side of your eardrum, and Patm ± 200 on the other side … net pressure = ±200. :wink:
 
Also note that [tex]53 mm^2= 53 \cdot ((10^{-3} m)^2) = 53\cdot 10^{-6} m^2[/tex]
 

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