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Putting drag and frictional forces together on the open road

October 10, 2010

As my students finish up their work on the the BFPM challenge lab, I asked the following quesiton.

Draw me a free body diagram for a car driving at 20 mi/h and another car driving at 40 mi/h, both at constant velocity.

They quickly come away with FBDs that look like this:

student FBDs for two cars traveling at constant velocity.

A number of good points are visible in this FBD. Kids have realized that the net force must be zero, and many correctly identify that the frictional force of the road is making the car go. And if you ask, they realize that it is the tire pushing on the road that causes this force. But I still have questions. “How did you know how to draw the drag force in the 40 mi/hr case?”

Student: “I dunno, I just drew it bigger.”
Me: “but didn’t you find something about how the drag force is related to the velocity?”
Student: “Oh, yea. It is proportional to the square. So the force should be four times bigger! That’s a huge change. The engine has to push the tires to push on the road a lot harder at 40 mi/h”

Then I pose this question.

Many people don’t see a difference between driving at 65 and 80 mi/h. With your understanding of drag forces, what do you think? Is there only a slight difference between these two cases?”

And we set out to do some math. First, they see that the 80 mi/hr is

\frac{80\textrm{mi/h}}{65\textrm{mi/h}}=1.23 times larger.

And if the velocity is 1.23 times larger, this means the drag force, F_d must be 1.23^2 = 1.51 times larger. This means that the engine must push 50% harder on the wheels so that they push 50% harder on the road. Thus you must use gas at at 50% higher rate. Wow!

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