Free fall and air resistance

  • Thread starter eehiram
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  • #1
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Main Question or Discussion Point

My textbook source is:
Fundamentals of Physics, 6th edition, by Halliday, Resnick, Walker

According to Newton's well known 2nd Law of Motion:
Fnet = ma

In chapter 2, in the case of free fall, the Fgrav = mg,
where g = -9.8 m/s2, assuming that movement along the axis of y is positive going upward from the Earth's surface.

However, in order to make a slighter more elaborate calculation, let us attempt include to incorporate air resistance, as in chapter 6:

If we have the following values to insert into the Drag equation:

Mass density of air = ρ = 1.29 kg / m3

Object speed in m/s = v

Drag coefficient = Cdrag = needs to be looked up or calculated

Effective Cross-sectional Area in m2 = A



The drag equation in chapter 6, section 3 appears to be:

Fdrag = (1/2) ρv2CdragA


As air resistance increases with v2, the Fdrag reaches a value equal in magnitude and opposite in direction to Fgrav.

Then terminal velocity might be attained, and the object's free fall may cease to accelerate.

Terminal velocity vterminal can be solved for by calculating the case of Fnet = 0 = Fdrag + Fgrav

Fdrag = -Fgrav
(1/2) ρv2CdragA = -mg
etcetera...

Is this correct? This is not a homework assignment question.
 

Answers and Replies

  • #2
jtbell
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Notice that if you try to solve your last equation for v, you'll end up taking the square root of a negative number. You can get rid of that pesky minus sign by going back to your net force equation and recognizing that it's a vector equation:

$$\vec F_{net} = 0 \\
\vec F_{drag} + \vec F_{grav} = 0 \\
\vec F_{drag} = - \vec F_{grav}$$

The last equation says that the two forces are opposite in direction (minus sign) and equal in magnitude:

$$F_{drag} = F_{grav}$$

This equation is for the magnitudes, so we drop the minus sign.
 
  • #3
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Yes, you are right. I should have used vectors in my initial post. Thank you for correcting my mistake.
 

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