Question

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?

Answer 1

`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!