# An object with a mass of 2 kg is on a ramp at an incline of pi/12 . If the object is being pushed up the ramp with a force of  8 N, what is the minimum coefficient of static friction needed for the object to remain put?

Jun 3, 2018

$\mu = 0.155$

#### Explanation:

we can write $\frac{\pi}{12}$ as 15° in degree.

Now please consider the image and look at the free body diagram and the forces acting on it.

$8 N$ force acting upwards and $m g \sin \theta$, $\mu m g \cos \theta$ downwards. where the $\mu$ is coefficient of static friction, $m$ is mass and $g$ is accelaration due to gravity.

Thus for the object to be put,
$\Rightarrow 8 N = m g \sin \theta + \mu m g \cos \theta$

rArr (8 - 2*9.8*sin15°)/(2*9.8*cos15°) = mu

as sin15°=0.256 and cos15°=0.998

after calculation $\mu = 0.155$

Hence coefficient of static friction is $\mu = 0.155$