You are viewing an old version of this page. View the current version.

Compare with Current View Page History

« Previous Version 16 Next »

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

    System:

    Interactions

    Model:

    Approach:

    Diagrammatic Representation

    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:

    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, Nx = 0 and Ny = +N. Gravity, however, is neither pure x or pure y in the tilted coordinates. Instead, we have mgx = mgsinθ and mgy = – mgcosθ. In the regular (untilted) coordinates, the situation is reversed. Gravity is simple, with mgx = 0 and mgy = -mg, but now the normal force is not aligned with either axis and has components Nx = Nsinθ and Ny = Ncosθ. Thus, it would seem that either choice results in the same level of simplification in the vector equations of Newton's 2nd Law.

    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 ay = 0 and ax = a. Thus, by using the tilted coordinates, we simplify two vectors, the normal force and the acceleration, while using untilted coordinates only simplifies gravity.

    Tilted Coordinates

    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?

    Mathematical Representation

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

      Method 1 – Regular (untilted) Coordinates

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

      We can then write the equations:

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [ \sum F_

      Unknown macro: {x}

      = N\sin\theta = ma_

      = ma\cos\theta ]
      [ \sum F_

      Unknown macro: {y}

      = N\cos\theta - mg = ma_

      = -ma\sin\theta ]\end

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

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

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [ ma \cot\theta \cos\theta - mg = -ma\sin\theta]\end

      Combining the acceleration terms by finding a common denominator and then using the trig identity

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [ \sin^

      Unknown macro: {2}

      \theta + \cos^

      \theta = 1]\end

      gives the result:

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [ g = \frac

      Unknown macro: {a}
      Unknown macro: {sintheta}

      ] \end

      Method 2 – Tilted Coordinates

      We have already drawn the relevant diagrams above, so we can simply write the equations:

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [ \sum F_

      Unknown macro: {x}

      = mg\sin\theta = ma_

      = ma ]
      [ \sum F_

      Unknown macro: {y}

      = N - mg\cos\theta = ma_

      = 0 ]\end

      The x-direction equation then immediately gives:

      Unknown macro: {latex}

      \begin

      Unknown macro: {large}

      [g\sin\theta = a]\end

      The same result as Method 1.

      deck: com.atlassian.confluence.macro.MacroExecutionException: java.lang.NullPointerException
      deck: com.atlassian.confluence.macro.MacroExecutionException: java.lang.NullPointerException
      • No labels