# An object with a mass of 9 kg is on a ramp at an incline of pi/12 . If the object is being pushed up the ramp with a force of  2 N, what is the minimum coefficient of static friction needed for the object to remain put?

Jan 14, 2018

To solve such problems always write down what is given and draw a diagram. Then use newtons laws to solve the problem.

#### Explanation:

Given:

The maximum value of static friction is:
$f = \setminus \mu m g \cos \left(\setminus \frac{\pi}{12}\right)$
$90 \sin \left(\setminus \frac{\pi}{12}\right) - 90 \setminus \mu g \cos \left(\setminus \frac{\pi}{12}\right) = 2 N$
$\frac{90}{4} - \frac{90}{4} \setminus \mu = 2 N$
45/2\mu=(45-4)/2
$\setminus \mu = \frac{41}{45}$

Jan 14, 2018

Minimum value of coefficient of static friction =0.26

#### Explanation:

Clearly,the component of weight which is acting along the inclined plane is $m g \sin \left(\frac{\pi}{12}\right)$
I.e $22.5 N$

But given only $2 N$of force is acting upward to keep it at its position which is not sufficient to overcome its component of weight acting downwards along the plane,hence frictional force must act upwards against the direction of ita component of weight acting downwards along the inclined plane.

Frictional force acting on the body$\left(f\right)$ = $\mu \cdot N$or $\mu \cdot m g \cos \left(\frac{\pi}{12}\right)$ or, $\mu \cdot 86.4 N$(where $\mu$ is the required minimum amount of static frictional coefficient)

So,from condition of linear equilibrium we can write,
$f + 2 = m g \sin \left(\frac{\pi}{12}\right)$ = 22.5
Solving we get, $\mu = 0.26$