1. ## Power Question

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

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

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

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

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

2. Originally Posted by craig
Got a question involving power, but not sure how to start it.

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

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

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

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

let $\displaystyle F$ = the force exerted by the lorry against the resistive force.

$\displaystyle P = F \mu$ on the level road

$\displaystyle F = \frac{P}{\mu}$

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

$\displaystyle P = Fv - vW\sin{\alpha}$

$\displaystyle P = \frac{P}{\mu} v - vW\sin{\alpha}$

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

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

$\displaystyle P(\mu - v) = \mu v W\sin{\alpha}$

3. Thank you! Yet again you've saved me from another mechanics problem.

4. Sorry could you just explain something for me.

Originally Posted by skeeter
let $\displaystyle F$ = the force exerted by the lorry against the resistive force.

$\displaystyle P = F \mu$ on the level road

$\displaystyle F = \frac{P}{\mu}$

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

$\displaystyle P = Fv - vW\sin{\alpha}$

$\displaystyle P = \frac{P}{\mu} v - vW\sin{\alpha}$

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

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

$\displaystyle P(\mu - v) = \mu v W\sin{\alpha}$
Thanks again for the help.

5. $\displaystyle P = F \mu$ on the level road

$\displaystyle F = \frac{P}{\mu}$

$\displaystyle P = (F \textcolor{red}{+} W\sin{\alpha}) v$ 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)

$\displaystyle P = Fv + vW\sin{\alpha}$

$\displaystyle P = \frac{P}{\mu} v + vW\sin{\alpha}$

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

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

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

6. Originally Posted by skeeter
$\displaystyle P = F \mu$ on the level road

$\displaystyle F = \frac{P}{\mu}$

$\displaystyle P = (F \textcolor{red}{+} W\sin{\alpha}) v$ 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)

$\displaystyle P = Fv + vW\sin{\alpha}$

$\displaystyle P = \frac{P}{\mu} v + vW\sin{\alpha}$

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

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

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

Thanks for clearing that up.