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The acceleration is not a force. When including a vector which is not a force in a free body diagram, you should indicate that distinction somehow. Here, we have used a dotted line for the acceleration.

What would the components of the acceleration be in the untilted coordinates?

{excerpt:hidden=true}Basic inclined plane problems illustrating the advantage of tilted coordinates.{excerpt}

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!incline.png!

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{card:label=Part A}

Find the magnitude of the acceleration of a box placed on a frictionless ramp that is inclined at θ = 30° above the horizontal.

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}Box as [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions* {cloak:id=inta} The box is subject to external influences from the earth (gravity) and the ramp (normal force).{cloak}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*
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{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

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We begin with a [free body diagram] and choice of a [coordinate system].  Although, as always, the choice of coordinates is arbitrary, a clever choice can help us reduce the work involved in solving the problem.  Consider the two possibilities below:

|!incline1a.jpg!|!incline2a.jpg!|
||Regular Coordinates||Tilted Coordinates||

Looking only at the forces in the free body diagrams, it is not clear that the tilted coordinates should be preferred.  If we use the tilted coordinates, then the normal force has very simple components, _N_~x~ = 0 and _N_~y~ = +{_}N{_}.  Gravity, however, is neither pure _x_ or pure _y_ in the tilted coordinates.  Instead, we have _mg_~x~ = _mg{_}sinθ and _mg_~y~ = -- _mg{_}cosθ.  In the regular (untilted) coordinates, the situation is reversed.  Gravity is simple, with _mg_~x~ = 0 and _mg_~y~ = -_mg_, but now the normal force is not aligned with either axis and has components _N_~x~ = _N{_}sinθ and _N_~y~ = _N{_}cosθ.  Thus, it would seem that either choice results in the same level of simplification in the vector equations of [Newton's 2nd Law|Newton's Second Law].  

|!incline1b.jpg!|!incline2b.jpg!|
||Regular Coordinates||Tilted Coordinates||

The tiebreaker in this case is only clear if you use your physical intution about the constraints in this problem.  By thinking about this problem, we know with certainty that the box will be sliding down the ramp.  It will not rise off the ramp or sink down through it.  Thus, the acceleration of the box will have to be parallel to the ramp.  It is important to remember that the acceleration will appear in the equations of Newton's 2nd Law as well.  Thus, the equations will be simpler if the acceleration has simple vector components.  Clearly, the acceleration will be simpler in the tilted coordinates, having _a_~y~ = 0 and _a_~x~ = _a_.  Thus, by using the tilted coordinates, we simplify _two_ vectors, the normal force and the acceleration, while using untilted coordinates only simplifies gravity.

|!incline2c.jpg!|
||Tilted Coordinates||

{note}The acceleration is _not_ a force.  When including a vector which is not a force in a free body diagram, you should indicate that distinction somehow.  Here, we have used a dotted line for the acceleration.{note}

{tip}What would the components of the acceleration be in the untilted coordinates?{tip}

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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

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To demonstrate the utility of the tilted coordinates, we will solve the problem using both options.

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{card:label=Method 1}

h4. Method 1 -- Regular (untilted) Coordinates

Before writing Newton's 2nd Law for each direction, we must break up the acceleration.

!incline1c.jpg!

We can then write the equations:

{latex}\begin{large}\[ \sum F_{x} = N\sin\theta = ma_{x} = ma\cos\theta \]
\[ \sum F_{y} = N\cos\theta - mg = ma_{y} = -ma\sin\theta \]\end{large}{latex}

{note}Be careful of the directions for the components of the acceleration.{note}

We can then solve the x-direction equation for the normal force and substitute into the y-equation, finding:

{latex}

Be careful of the directions for the components of the acceleration.

\begin{large}\[ ma \cot\theta \cos\theta - mg = -ma\sin\theta\]\end{large}{latex}

\begin{large}\[ \sum F_{x} = mg\sin\theta = ma_{x} = ma \]
\[ \sum F_{y} = N - mg\cos\theta = ma_{y} = 0 \]\end{large}{latex}

The same result as Method 1.