# 5.2 Drag forces  (Page 3/6)

 Page 3 / 6

## 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 $v$ versus mass. Also plot ${v}^{2}$ 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, ${F}_{\text{net}}=0$ . 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 $\text{mg}=\frac{1}{2}{\text{ρCAv}}^{2}$ .

Thus the terminal velocity ${v}_{t}$ can be written as

${v}_{\text{t}}=\sqrt{\frac{2\text{mg}}{\text{ρCA}}}.$

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

$A=\left(2 m\right)\left(0\text{.}\text{35 m}\right)=0\text{.}\text{70}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}.$

Using our equation for ${v}_{\text{t}}$ , we find that

$\begin{array}{lll}{v}_{\text{t}}& =& \sqrt{\frac{2\left(\text{85}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)}{\left(1.21\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left(1.0\right)\left(0.70\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)}}\\ & =& \text{44 m/s.}\end{array}$

Discussion

This result is consistent with the value for ${v}_{\text{t}}$ mentioned earlier. The 75-kg skydiver going feet first had a $v=\text{98 m}/\text{s}$ . 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

${F}_{\text{s}}=6\mathrm{\pi r\eta v},$

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

how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
haider
topic of question?
haider
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melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
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