# An object with a mass of 8 kg is on a plane with an incline of pi/8 . If the object is being pushed up the plane with  8 N  of force, what is the net force on the object?

Jun 24, 2016

${\vec{F}}_{\text{total}} = 22 \hat{x} '$
where x'sin(π/8)=x

#### Explanation:

Assuming that:
- the plane is frictionless,
- the only forces at play are the weight of the mass ($\vec{W}$), the normal force (${\vec{F}}_{\text{N}}$) and the pushing force (${\vec{F}}_{\text{P}}$), and
- the direction of the pushing force is parallel to the plane.

For convinience, lets shift our coordinate system by $\frac{\pi}{8}$.
From $\left(x , y\right) \rightarrow \left(x ' , y '\right)$

Along the x' axis we have:
${\vec{F}}_{\text{P}} = - 8 N \hat{x} '$

${\vec{W}}_{\text{x'}} = \left(8 k g\right) \left(9.8 \frac{m}{{s}^{2}}\right) \left(\sin \left(\frac{\pi}{8}\right)\right) \hat{x} ' = 30.00 N \hat{x} '$

And along y' we have:
${\vec{W}}_{\text{y'}} = - \left(8 k g\right) \left(9.8 \frac{m}{{s}^{2}}\right) \left(\cos \left(\frac{\pi}{8}\right)\right) \hat{y} ' = - 72.43 N \hat{y} '$

${\vec{F}}_{\text{N"=-vecW_"y}} = 72.43 \hat{y} '$

Then lets just add the vectors per axis.

For x'
${\vec{F}}_{\text{x'"=vecF_"P"+vecW_"x'}}$
${\vec{F}}_{\text{x'}} = - 8 N \hat{x} ' + 30.00 N \hat{x} '$
${\vec{F}}_{\text{x'}} = + 22.00 N \hat{x} '$

And for y'
${\vec{F}}_{\text{y'"=vecF_"N"+vecW_"y'}}$
${\vec{F}}_{\text{y'}} = 72.43 N \hat{y} ' - 72.43 \hat{y} '$
${\vec{F}}_{\text{y'}} = 0 N \hat{y} '$

Thus, the net force is just
${\vec{F}}_{\text{total"=vecF_"x'}} = 22.00 N \hat{x} '$

From this we can tell that the net force is parallel to the plane and that the net force points against the direction of the pushing force.

*Note: Its best to use a free body diagram if you're confused about the direction of forces.

Hope this helps!