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?

1 Answer
Jun 24, 2016

vecF_"total"=22hatx'
where x'sin(π/8)=x

Explanation:

Assuming that:
- the plane is frictionless,
- the only forces at play are the weight of the mass (vecW), the normal force (vecF_"N") and the pushing force (vecF_"P"), and
- the direction of the pushing force is parallel to the plane.

For convinience, lets shift our coordinate system by pi/8.
From (x,y) rarr(x',y')

Along the x' axis we have:
vecF_"P"=-8N hatx'

vecW_"x'"=(8kg)(9.8m/(s^2))(sin(pi/8))hatx'=30.00Nhatx'

And along y' we have:
vecW_"y'"=-(8kg)(9.8m/(s^2))(cos(pi/8))haty'=-72.43Nhaty'

vecF_"N"=-vecW_"y"=72.43haty'

Then lets just add the vectors per axis.

For x'
vecF_"x'"=vecF_"P"+vecW_"x'"
vecF_"x'"=-8Nhatx' + 30.00Nhatx'
vecF_"x'"=+22.00Nhatx'

And for y'
vecF_"y'"=vecF_"N"+vecW_"y'"
vecF_"y'"=72.43Nhaty' -72.43haty'
vecF_"y'"=0Nhaty'

Thus, the net force is just
vecF_"total"=vecF_"x'"=22.00Nhatx'

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!