# An object with a mass of 6 kg is on a plane with an incline of pi/12 . If the object is being pushed up the plane with  6 N  of force, what is the net force on the object?

Jul 17, 2017

$\sum F = 9.23$ $\text{N}$ directed down the plane

#### Explanation:

I'll assume the surface is frictionless, as there is no information regarding friction given.

Referring to the image below:

I'll call the positive $x$-axis going down the plane.

We know the normal force $n$ and quantity $m g \cos \theta$ are equal in magnitude, so we need not resolve the vertical direction.

Since the surface is (supposedly) frictionless, the only forces being applied are

• gravitation force, equal to $m g \sin \theta$ (directed down the incline)

• applied force, equal to $6$ $\text{N}$ (directed up the incline)

We need to find the quantity $m g \sin \theta$ by knowing

• $m = 6$ $\text{kg}$

• $g = 9.81$ ${\text{m/s}}^{2}$

• $\theta = \frac{\pi}{12}$

Therefore,

$m g \sin \theta = \left(6 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) \sin \left(\frac{\pi}{12}\right) = 15.2$ $\text{N}$

The net horizontal force $\sum {F}_{x}$ is given by

$\sum {F}_{x} = m g \sin \theta - {F}_{\text{applied}} = 15.2$ $\text{N}$ $- 6$ $\text{N}$

= color(red)(9.23 color(red)("N"

directed down the plane