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Math Help - Dimensions

  1. #1
    Senior Member slevvio's Avatar
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    Dimensions

    Hello everybody, I am having trouble with this question:

    Two common models for the resistive force experienced by a projectile in flight due to air resistance are:-

    (a) Force proportional to speed so that  F = -k_{1}v;
    (b) Force proportional to the square of speed so that  F = -k_{2}v^{2}.

    Investigate the dependence of k_1 on  \mu , the viscosity of air and  r , the effective radius of the body. Show that model (b) assumes that k_2 is independent of the viscosity of air.

    The bit in bold is the part I can't do - I got  k_1 \propto \mu r for the first bit. Thanks very much and any help is appreciated.
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  2. #2
    Senior Member slevvio's Avatar
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    I haven't received a solution yet so I will explain my working and maybe someone will see a mistake. Square brackets denote the dimensions of the quantity, and M = mass, T = time, L = length:

     F = -k_2v^2

     k_2 = -\frac{F}{v^2}

    So then  [k_2] = \frac{[F]}{[v]^2} = [M]\frac{[L]}{[T]^2}\cdot \frac{[T]^2}{[L]^2} = \frac{[M]}{[L]}

    Suppose k_2 \propto \mu^{\alpha}r^{\beta}

    Then [k_2] = [\mu]^{\alpha}[r]^{\beta}

    And \frac{[M]}{[L]} = [M]^{\alpha}[L]^{- \alpha}[T]^{-\alpha}[L]^{\beta}

    [M][L]^{-1} = [M]^{\alpha}[L]^{\beta - \alpha}[T]^{-\alpha}

    This gives us the equations  \alpha = 1, L = \beta - \alpha, - \alpha = 0 , which makes no sense, since how can \alpha be zero and one? Help please !
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