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Math Help - Power Question

  1. #1
    Super Member craig's Avatar
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    Power Question

    Got a question involving power, but not sure how to start it.

    Lorry of weight W generates power P. The lorry has max. speed of \mu on a level road, but v < \mu on a slope inclined at angle \alpha to the horizontal.

    If the power and resistance of the lorry stay the same and are constant, show that:

    \mu vW \sin{\alpha} = P(\mu - v)

    Guessing this involves resolving the forces, not sure where to start though.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by craig View Post
    Got a question involving power, but not sure how to start it.

    Lorry of weight W generates power P. The lorry has max. speed of \mu on a level road, but v < \mu on a slope inclined at angle \alpha to the horizontal.

    If the power and resistance of the lorry stay the same and are constant, show that:

    \mu vW \sin{\alpha} = P(\mu - v)

    Guessing this involves resolving the forces, not sure where to start though.

    Thanks in advance
    let F = the force exerted by the lorry against the resistive force.

    P = F \mu on the level road

    F = \frac{P}{\mu}


    P = (F - W\sin{\alpha}) v on the incline

    P = Fv - vW\sin{\alpha}<br />

    P = \frac{P}{\mu} v - vW\sin{\alpha}<br />

    P \mu = Pv - \mu v W\sin{\alpha}

    P \mu - Pv = \mu v W\sin{\alpha}<br />

    P(\mu - v) = \mu v W\sin{\alpha}
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  3. #3
    Super Member craig's Avatar
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    Thank you! Yet again you've saved me from another mechanics problem.
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  4. #4
    Super Member craig's Avatar
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    Sorry could you just explain something for me.

    Quote Originally Posted by skeeter View Post
    let F = the force exerted by the lorry against the resistive force.

    P = F \mu on the level road

    F = \frac{P}{\mu}


    P = (F - W\sin{\alpha}) v on the incline

    P = Fv - vW\sin{\alpha}<br />

    P = \frac{P}{\mu} v - vW\sin{\alpha}<br />

    P \mu = Pv - \mu v W\sin{\alpha}

    P \mu - Pv = \mu v W\sin{\alpha}<br />
Here you have subtracted Pv from both sides, but shouldn't the right hand side be - \mu v W\sin{\alpha}<br />
?

    P(\mu - v) = \mu v W\sin{\alpha}
    Thanks again for the help.
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  5. #5
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    P = F \mu on the level road

    F = \frac{P}{\mu}


    P = (F \textcolor{red}{+} W\sin{\alpha}) v on the incline

    (mea culpa ... edit ... should be a + since the force the lorry exerts up the incline has to equal the sum of the resistive force and parallel weight component)

    P = Fv + vW\sin{\alpha}<br />

    P = \frac{P}{\mu} v + vW\sin{\alpha}<br />

    P \mu = Pv + \mu v W\sin{\alpha}

    P \mu - Pv = \mu v W\sin{\alpha}<br />

    P(\mu - v) = \mu v W\sin{\alpha}[/quote]
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  6. #6
    Super Member craig's Avatar
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    Quote Originally Posted by skeeter View Post
    P = F \mu on the level road

    F = \frac{P}{\mu}


    P = (F \textcolor{red}{+} W\sin{\alpha}) v on the incline

    (mea culpa ... edit ... should be a + since the force the lorry exerts up the incline has to equal the sum of the resistive force and parallel weight component)

    P = Fv + vW\sin{\alpha}<br />

    P = \frac{P}{\mu} v + vW\sin{\alpha}<br />

    P \mu = Pv + \mu v W\sin{\alpha}

    P \mu - Pv = \mu v W\sin{\alpha}<br />

    P(\mu - v) = \mu v W\sin{\alpha}
    [/QUOTE]

    Thanks for clearing that up.
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