A wooden block slide directly down an inclined plane velocity of 6.0 m/s How large is the coefficient of kinetic friction, if the plane makes an angle of 25 deegre with the horizontal?

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- Jun 30th 2006, 12:19 PM #1

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- Jul 1st 2006, 01:29 AM #2

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Originally Posted by**Refujoi**

$\displaystyle N$, the Frictional force $\displaystyle F$, and the Gravitational force $\displaystyle mg$ on the block are in equilibrium.

Resolve the gravitational force into components down the plane and normal to

the plane and set these equal to the frictional force and the normal reaction

(allowing for the appropriate signs):

$\displaystyle F=mg \sin(\theta)$,

$\displaystyle N=mg \cos(\theta)$

where $\displaystyle \theta$ is the inclination of the plane.

Now $\displaystyle F=\mu N$, where $\displaystyle \mu$ is the coefficient of

kinetic/sliding ... friction. So:

$\displaystyle \mu\ mg \cos(\theta)=mg \sin(\theta)$

rearranging:

$\displaystyle \mu=\tan(\theta)$.

In this problem $\displaystyle \theat=25^{\circ}$, so:

$\displaystyle \mu=\tan(25^{\circ}) \approx 0.466$.

RonL