How much force to drag an object?

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Aerodynamic drag can be equated to the force required to pull a weight, with 48 pounds of drag at 70 mph equivalent to dragging a 300-pound steel chunk down the road. To calculate the horsepower needed to overcome this drag, it was determined that approximately 9 horsepower is required, which is a significant portion of the 30-35 horsepower typically needed for cars at that speed. The coefficient of friction between steel and the road surface is crucial for accurately determining how much weight would produce equivalent drag. This coefficient is a dimensionless value that relates the contact force to friction force and remains relatively constant regardless of velocity or surface area. Understanding these principles can help illustrate the impact of aerodynamic drag on vehicle performance.
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Hi. I'm not a student, and I'm old, so I don't remember physics class in high school. So I apologize for my question.

I'm trying to figure out how you would translate aerodynamic drag to someone like me- a layman. I have a wing on my car that, if I were to tilt it to 25 degrees angle of attack, would produce 120 pounds of downforce at 70mph, but would also produce 48 pounds of drag. I was trying to explain the problem of drag to a good friend, and I'm having trouble equating that to something he can appreciate.

If I'm thinking correctly, this 48 pounds of drag would be like attaching a rope on the back of the car and dragging a 300 pound chunk of steel down the road since the drag from friction would be about 48 pounds. But how do I determine exactly how much a chunk of steel must weigh to equal 48 pounds of friction drag? I assume that contact surface would be a determining factor aslo.

Thanks for your help!
 
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would produce 120 pounds of downforce at 70mph, but would also produce 48 pounds of drag

How about working out the horsepower wasted pushing the wing through the air?...

Sorry but it's fastsest for me to work in SI/metric so..

48lbs = 22kg
Force in Newtons = 22 * 9.8 = 216N

Velocity = 70mph = 31m/s

Power in watts = force * velocity = 216 * 31 = 6700 W

Then ..

6700/750 = about 9 Horsepower.

PS Google suggests 30-35hp is needed for most cars to do 70mpg so that 9hp is a significant percentage.
 
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You need the coefficient of friction between steel and a road. This is a dimensionless value, relating the contact force (300 pounds * gravitational acceleration) to the force of friction. As a good approximation, the coefficient does not depend on the velocity or the surface area.
 
Thanks for the replies. Yes, I figured I'd need to know the coefficient of friction. How do I figure that out? :D

Thanks for the hp figure. That is helpful to be sure. But I was hoping we could further illustrate it by saying it takes 9hp to drag X pounds of iron down a paved road. Or, in other words, 48 pounds of aerodynamic drag is equal to pulling a steel anchor weighing X pounds down the road. (Yes, it's silly, but we who put wings on our cars are by definition, silly... lol)
 
c131frdave said:
How do I figure that out? :D
Test it (not on a public road please :D) or look it up in a table.

The additional force downwards will add friction as well - the wheels of your car are probably better than steel sliding on a road, but it is not negligible.
 
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