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  4. Drag forces

6.2 Drag forces  (Page 3/6)

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Take-home experiment

This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity v size 12{v} {} versus mass. Also plot v 2 size 12{v rSup { size 8{2} } } {} versus mass. Which of these relationships is more linear? What can you conclude from these graphs?

A terminal velocity

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

Strategy

At terminal velocity, F net = 0 size 12{F rSub { size 8{"net"} } =0} {} . Thus the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find mg = 1 2 ρCAv 2 size 12{ ital "mg"=0 "." 5ρ ital "CAv" rSup { size 8{2} } } {} .

Thus the terminal velocity v t size 12{v rSub { size 8{t} } } {} can be written as

v t = 2 mg ρCA . size 12{v= sqrt { { {2 ital "mg"} over {ρ ital "CA"} } } } {}

Solution

All quantities are known except the person’s projected area. This is an adult (82 kg) falling spread eagle. We can estimate the frontal area as

A = ( 2 m ) ( 0 . 35 m ) = 0 . 70 m 2 . size 12{2`m times `0 "." "35"`m=0 "." "70"`m rSup { size 8{2} } } {}

Using our equation for v t size 12{v rSub { size 8{t} } } {} , we find that

v t = 2 ( 85 kg ) ( 9.80 m/s 2 ) ( 1.21 kg/m 3 ) ( 1.0 ) ( 0.70 m 2 ) = 44 m/s.

Discussion

This result is consistent with the value for v t size 12{v rSub { size 8{t} } } {} mentioned earlier. The 75-kg skydiver going feet first had a v = 98 m / s size 12{v="94"`m/s} {} . He weighed less but had a smaller frontal area and so a smaller drag due to the air.

The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don’t reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled “On Being the Right Size.”

To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law    , which states that

F s = 6 πrηv , size 12{F rSub { size 8{s} } =6πrηv} {}

where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.

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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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