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Optimal Dragster Exhuast Pipe Angle

  1. Mar 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Top-fuel dragsters can accelerate from rest at a rate of about 5g. If this does not sound impressive, consider that for most automobiles the friction between the ground and the tires produces the acceleration. Since the coefficient of friction between regular treaded tires and pavement is typically less than or about equal to 1.0, a car relying on friction with normal force equal to its weight should accelerate at no more than about 1g.
    Dragsters use two methods to increase their acceleration. First, they greatly increase the friction between their tires and the road by eliminating tire tread (dragsters do not race in wet or snowy road conditions) and also by performing a “burnout” which lays down a patch of heated rubber on the track that the partially-melted tires will adhere to. The result of these measures is a coefficient of friction exceeding 2.0. The second important method for increasing acceleration is the use of engine exhaust to provide force. The high-speed exhaust gases exiting the dragster’s enormous 8000 hp engine produce a force with a size comparable to the car’s weight

    a.) Assuming a coefficient of friction equal to 2.5, at what angle θ measured with respect to the horizontal (bottom picture) should the exhaust pipes be oriented in order to achieve maximum acceleration? Assume the engine is capable of using the tires to produce a force equal to the maximum friction force (μN) regardless of how large the normal force becomes

    2. Relevant equations
    f=μN

    3. The attempt at a solution

    The answer I calculated is μ=1/tanθ but the correct answer is μ=tanθ, and the degree is 68. I'm really confuse about that.
     
  2. jcsd
  3. Mar 19, 2015 #2
    Can you argomentate with more details your conclusions?
    Are you sure that the text says that the angle should be measured with respect to the horizontal?

    warning: I'm not an expert but I would like become so ... in this life.
     
  4. Mar 19, 2015 #3
    Because engine is capable of using the tires to produce a force equal to the maximum friction force, therefore the frictional produced by gravity can be ignored. Just consider the exhaust force, Fcosθ=μFsinθ, so the result is μ=1/tanθ. But that's not correct answer.
    And the angle is measured with respect to the horizontal,
     
  5. Mar 20, 2015 #4
    I'm still thinking of it.
    However, let's try to think:
    Imagine that μ is unknown and we must measured it .
    Let's take a constant force F as a sample, applying it on a mass with weight force equal to F.
    If μ = 0.5 there is a few friction; F is double of the the maximum frictional force.
    If μ increases to 1, F becomes equal to the friction force.
    If μ increase to 2.5, there is a great friction; F now is smaller than the frictional force.
    So to move a mass with a great μ, it is necessary use a greater force respect of a small μ, because the same force F obtains greater frictional force.
    I think that it is right and I hope it is useful to you to solve the problem.
    However let's wait for an expert because I'm not so.
     
  6. Mar 20, 2015 #5
    Let F represent the magnitude of the force that the exhaust gases exert on the car in the mainly downward direction along the tailpipe. If the tailpipe is oriented at an angle of θ to the horizontal, what is the vertical component of this force? What is the horizontal component of this force? This force causes the normal force exerted by the ground on the tire to increase by the vertical component. How much is it capable of causing the frictional force exerted in the forward direction by the road on the tire (and car) to increase by? What is the total increase in the horizontal force on the car? At what angle θ is this increased horizontal force maximized?

    Chet
     
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