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Math Help - dimensional homogeneity

  1. #1
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    dimensional homogeneity

    Hi, How to show that Newton's second law F=\frac{d\left(mv\right) }{dt} for dimensional homogeneity.

    thanks
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    Quote Originally Posted by hazeleyes View Post
    Hi, How to show that Newton's second law F=\frac{d\left(mv\right) }{dt} for dimensional homogeneity.

    thanks
    You need to show that the units match on both sides of the equation.

    Force has units

    F=\frac{(\text{kg})(\text{m})}{\text{s}^2}

    m=\text{kg}

    v=\frac{\text{m}}{\text{s}}

    Now we need to take the derivative on the right hand side

    \frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)

    So now analyze the units

    \frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}

    So the units on both sides are the same
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    Quote Originally Posted by TheEmptySet View Post
    You need to show that the units match on both sides of the equation.

    Force has units

    F=\frac{(\text{kg})(\text{m})}{\text{s}^2}

    m=\text{kg}

    v=\frac{\text{m}}{\text{s}}

    Now we need to take the derivative on the right hand side

    \frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)

    So now analyze the units

    \frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}

    So the units on both sides are the same
    May I ask how is mass a function of time?
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    Quote Originally Posted by bugatti79 View Post
    May I ask how is mass a function of time?
    OK, it could be the a mass flow rate like density by the cross sectional area by velocity
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    Quote Originally Posted by bugatti79 View Post
    May I ask how is mass a function of time?
    With rockets, you are ejecting mass to propel yourself forward. Mass is most definitely a function of time then. The most general form of Newton's Second Law is the version in his Principia:

    \mathbf{F}=\dot{\mathbf{p}}.
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    Quote Originally Posted by TheEmptySet View Post
    You need to show that the units match on both sides of the equation.

    Force has units

    F=\frac{(\text{kg})(\text{m})}{\text{s}^2}

    m=\text{kg}

    v=\frac{\text{m}}{\text{s}}

    Now we need to take the derivative on the right hand side

    \frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)

    So now analyze the units

    \frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}

    So the units on both sides are the same
    but isn't Volume flow rate(dv/dt)= m^3s^{-1}
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  7. #7
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    In Newton's Second Law, v is velocity, not volume. So it would have units of m/s. Usually, volume is an upper-case V (math is case-sensitive!).
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    but why dm/dt is kg/s?
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    Quote Originally Posted by hazeleyes View Post
    but why dm/dt is kg/s?
    It is the change is mass with respect to time so the units are mass over time or kg/s
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    Thanks so much!!
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  11. #11
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    Quote Originally Posted by TheEmptySet View Post
    It is the change is mass with respect to time so the units are mass over time or kg/s
    More generally, if I have a function y=f(x), and the units of y are [y], and the units of x are [x], then the units of

    \frac{dy}{dx} are \frac{[y]}{[x]}.

    Similarly, the units of

    \int y(x)\,dx are [y][x].
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