An object with a mass of 12 kg is on a plane with an incline of  - pi/6 . If it takes 6 N to start pushing the object down the plane and 4 N to keep pushing it, what are the coefficients of static and kinetic friction?

May 14, 2018

coefficient of static friction $\mu s = 0.366 . \sqrt{3}$
coefficient of kinetic friction $\mu k = 0.355 . \sqrt{3}$

Explanation:

the incline is making an angle of $- \frac{\pi}{6}$ = -30 degrees from the horizontal.

the forces operating on the body is the weight m.g vertically

down = $12 k g . 10 \frac{m}{s} ^ 2$ = 120 N

the normal reaction of the plane acting perpendicular to the plane. say R

breaking the weight in components along the plane and

perpendicular to the plane inclined at 30 degree to horizontal

$m g . \cos 30$ = $120 . \frac{\sqrt{3}}{2}$ = R and

mg. sin30 will act along the plane .

as a force of 6 N is needed to start the motion

so frictional force f s = 6 N + mg sin 30 = 6 N + 120/2 = 66 N

$f s = \mu s . R$ = 66N so, $\mu s = \left\{\frac{66}{60. \sqrt{3}}\right\}$

$\mu s$ = $0.366 \sqrt{3}$

however ,when the body starts moving a force of 4 N is sufficient to maintain the motion , therefore

$f k = \mu k . R$ = 64N so, $\mu k = \left\{\frac{64}{60. \sqrt{3}}\right\}$

$\mu k = 0.355 . \sqrt{3}$