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Air Resistance

We have neglected air resistance thus far mainly for one reason, and that is that air resistance makes problems more difficult. Air resistance, or drag, depends on the velocity that an object is going, and not always in a simple way. Thus it is difficult to work problems without a computer.

There are a few things we can say about drag. Consider an object moving through a substance, such as air or water. It is reasonable that a larger object would experience more drag than a smaller one. We might also expect that the drag would be larger for a faster moving object. Finally it is reasonable to expect that a dense fluid would exert more drag than a “thin” fluid. It will not be a surprise, then, that the formula for drag is

Fdrag = CAv2


where C is a constant equal to about 0.2, ρ is the density of the fluid, A is the cross-sectional area of the moving object, and v is its velocity. This equation is reasonably accurate if the fluid is “roughed up” a bit by the object’s passing through. For cars driving and people walking through air and for fish swimming through water, this is a reasonable assumption.

You do not have to memorize this equation, but do know how to use it. The force is placed in the force diagram like all the other forces, except it points necessarily opposite the direction of motion.
A dog Nikki (7 kg) falls five stories (3.3 meters each) down from a roof with his stomach pointing down. He is 0.2 meters tall, and 0.15 meters wide and 0.4 meters from nose to tail. The density of air is 1.29 kg/m3.
  1. If we ignore air resistance, what is the terminal velocity of the dog?
  2. What is the air resistance by the time the dog gets to the bottom?
  3. Is air resistance a small or large effect in this problem?
  1. First, we DRAW A DIAGRAM (Figure 6-11). We are hoping that the force of air resistance is small. We have

Fnet = Fgrav = –mg

a = –g = –10 m/s2

v1 = 0

y = –16.5 m


We use the kinematic equation that does not involve ∆t, so we have
  1. The cross-sectional area presented by the dog is (0.4 m)(0.15 m) = 0.06 m2. It does not matter how tall Nikki is. Thus
  2. Air resistance is about 10% of the dog’s weight (70 N). This implies that the dog’s weight is the dominant force all the way down, and we are justified in ignoring air resistance. Our answer is good with no more than about 10% error.

Nikki is a stunt dog. He was unharmed.


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Figure 6-11



A rubber ball (radius 2 cm, mass 5 grams) is dropped from a height of 50 meters. What is its velocity when it reaches the ground?



If we work out the problem as in Example 1a above, we run into trouble. Ignoring air resistance yields v2 = 32 m/s. (You should work this out.) The cross-sectional area is the area of a circle πr2, not the total surface area of a sphere. If we calculate the force due to air resistance at the bottom of flight, we obtain



whereas the weight of the ball is Fgrav = mg = 0.05 N. The calculated force Fair is not small at all in comparison. In fact, it is much larger than the force of gravity. Any assumption that air resistance is negligible is not valid. We need a new idea in order to solve this.


The new idea follows. If the ball has fallen so far that it has stopped accelerating then the gravity force down and the air resistance force up are balanced. Then the force diagram would look like the one in Figure 6-12. The force equation becomes


Fair – mg = (Fnet)y = 0


CρAv2 – mg = 0



As the ball falls, its velocity increases until it begins to get close to 12 m/s. The air resistance increases until it balances the force of gravity. Force balance occurs when v = 12 m/s. This velocity is called terminal velocity.

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Figure 6-12




A man (60 kg) falls from a very tall building (200 m). The cross-sectional area for his fall is 0.3 m2. 

  1. What equation governs the fall?
  2. What is his velocity when he reaches the bottom?
..\art 6 jpg\figure 6-to.jpg
Figure 6-13
  1. First, we DRAW A DIAGRAM (Figure 6-13). The force equation becomes
    Fair – mg = ma
    CρAv2 – mg = ma
  2. An analysis similar to the one in Example 1 shows that ignoring air resistance yields v2 = 63 m/s. If we calculate Fair, as in Example 1, we obtain 310 N, about half of the man’s weight. This shows that ignoring air resistance is wrong since 310 N is not small compared to Fgrav = 600 N. But Fair is not large compared to Fgrav either. Our calculations show that air resistance is too large to be ignored, but not so large as to assure force balance. This problem is just too hard.

In this chapter we looked at friction and air resistance. Friction is a force which opposes the slipping of surfaces, and it always acts parallel to the surface. Situations involving friction fall into two categories: static and kinetic. If the surfaces are not slipping, then static friction is the force which maintains the status. We solve for the frictional force using force balance. The static friction cannot be larger than the maximum, given by Fs,max = μsN.


If the surfaces are slipping, then kinetic friction opposes the slipping. Its magnitude is given by Fk = μkN.


The MCAT does not have many problems involving air resistance. Generally you just need to know that air resistance is a retarding force that depends on the surface area, density of the medium, and velocity of the object.


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