Why are there no force resolutions at ##30^0##?

[Benjamin_harsh]

Homework Statement

A 10 kg box rest on a 30 degree incline and begins to slide down. What is the acceleration if no friction is present?

Relevant Equations

$m.a_{x} = mg.sinθ$ , $a_{x} = 10.sin(30)$

There is a component the weight force that accelerates the box downwards, $F_{g}$, which is equal to mg.sinθ of below triangle.

$F_{g}= mg.sinθ$

$F_{N}= mg.cosθ$

$\sum F_{X} = F_{g}$

$m.a_{x} = mg.sinθ$

$a_{x} = 10.sin(30)$

$a_{x} = \large \frac {1}{2}$

$a_{x} = 5 \large \frac {m}{sec^2}$

Why there is no force resolutions at $30^0$? I mean why resolutions taken at below the box?

 

[PeroK]

Why there is no force resolutions at $30^0$? I mean why resolutions taken at below the box?

I don't understand your question.

 

[Benjamin_harsh]

I don't understand your question.

Why did we resolve forces at $θ$ instead of $30^0$?

 

[PeroK]

Why did we resolve forces at $θ$ instead of $30^0$?

$\theta = 30°$

 

[Benjamin_harsh]

$\theta = 30°$

Why did resolve forces near body instead of corner of the triangle?

 

[PeroK]

Why did resolve forces near body instead of corner of the triangle?

It's usual to show the resolution of forces where they apply. I.e. on the block itself. There are no relevant forces acting at the corner of the triangle. There's nothing going on there.

 

[Benjamin_harsh]

It's usual to show the resolution of forces where they apply. I.e. on the block itself. There are no relevant forces acting at the corner of the triangle. There's nothing going on there.

Thank you