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How much force to drag an object?

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    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 exaclty 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!
     
  2. jcsd
  3. Jan 29, 2013 #2

    CWatters

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    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.
     
    Last edited: Jan 29, 2013
  4. Jan 29, 2013 #3

    mfb

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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.
     
  5. Jan 29, 2013 #4
    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)
     
  6. Jan 29, 2013 #5

    mfb

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    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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