Chapter 5 Lecture: Inclined plane (Frictionless)

Rotated Coordinate Systems on an Inclined Plane

• Motivation for rotating coordinate axes:

  • Standard conventional Cartesian coordinate systems define the positive $x$-axis horizontally and the positive $y$-axis vertically.

  • When analyzing an inclined plane, using a conventional horizontal and vertical coordinate system complicates the equations because both the normal force and the acceleration have simultaneous non-zero horizontal and vertical components.

  • Rotating the coordinate axes simplifies calculations and provides a direct, intuitive interpretation of the physical system.

• Definition of the rotated coordinate axes:

  • The $x$-axis is aligned parallel to the surface of the incline. Down the incline is defined as the positive $x$-direction, while up the incline is defined as the negative $x$-direction.

  • The $y$-axis is aligned perpendicular to the surface of the incline. Upward and away from the surface is defined as the positive $y$-direction.

• Analytical advantages of coordinate rotation:

  • Normal force alignment: Because the normal force is always strictly perpendicular to the contact surface, it lies entirely along the rotated $y$-axis ($N_x = 0$ and $N_y = N$).

  • Acceleration alignment: Because the block is constrained to slide along the surface of the ramp, the linear acceleration occurs strictly along the rotated $x$-axis ($a_x = a$ and $a_y = 0$).

  • Decoupling: Motion is divided into a one-dimensional equilibrium problem along the perpendicular $y$-axis and a one-dimensional acceleration problem along the parallel $x$-axis.

Physical Setup and Free Body Diagram Analysis

• System configuration:

  • A block of mass $m$ sits on a ramp or incline elevated at an angle $\theta$ above the horizontal.

  • The contact surface between the block and the ramp is assumed to be completely frictionless.

  • Initially at rest, the block accelerates downward along the incline surface due to gravity.

• Simplification of subscript notation:

  • When analyzing a single isolated object (a single mass $m$), double-subscript interaction notation (such as $\mathbf{F}_{\text{Earth on mass}}$ or $\mathbf{F}_{\text{ramp on mass}}$) is omitted without loss of clarity.

• Contact force (Normal Force, $N$):

  • The normal force is a contact force exerted by the ramp surface pushing upward on the block.

  • The defining characteristic of a normal force is that it is always directed perpendicular ($90^\circ$) to the contact surface.

  • In the rotated coordinate frame, the normal force vector acts exclusively in the positive $y$-direction:     $N_x = 0$     $N_y = N$

• Non-contact force (Gravitational Force / Weight, $w$):

  • Weight represents the gravitational force exerted by Earth on the mass.

  • Weight is calculated as the product of mass and gravitational acceleration:     $w = mg$

  • Weight points vertically downward, perpendicular to the flat horizontal ground.

  • Because the coordinate axes are rotated by an angle $\theta$, the downward weight vector does not align with either the rotated $x$-axis or the rotated $y$-axis, requiring vector decomposition into rotated components $w_x$ and $w_y$.

Geometric Derivation of the Weight Vector Angle

• Determination of the internal angle between weight and the rotated axes:

  • The incline forms an angle $\theta$ with the horizontal ground.

  • A vertical line dropped from the incline to the horizontal floor forms a right-angled triangle with the ground, where the angle between the vertical line and the horizontal base is $90^\circ$.

  • The third angle of this right triangle, formed between the incline surface and the vertical gravity line, equals:     $90^\circ - \theta$

  • The rotated negative $y$-axis is defined to be perpendicular ($90^\circ$) to the surface of the incline.

  • The angle between the vertical weight vector ($w$) and the rotated negative $y$-axis is the complement of $(90^\circ - \theta)$:     $90^\circ - (90^\circ - \theta) = \theta$

  • Therefore, the angle between the downward vertical weight vector ($w$) and the rotated negative $y$-axis is equal to the incline angle $\theta$.

Vector Decomposition of the Weight Force

• Projection analogy:

  • Decomposing a vector onto coordinate axes corresponds to the physical shadow cast by a vector onto the perpendicular axes when illuminated by an overhead or lateral light source.

  • Completing the perpendicular projections produces a rectangle/parallelogram containing right-angled triangles.

• Trigonometric relationships (SOH CAH TOA):

  • The weight vector magnitude $w = mg$ forms the hypotenuse of the component right triangle.

  • The component adjacent to the angle $\theta$ lies along the negative $y$-axis ($w_y$).

  • The component opposite to the angle $\theta$ lies parallel to the ramp surface along the positive $x$-axis ($w_x$).

  • Definition of sine:     $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{w_x}{w}$     $w_x = w \sin(\theta) = mg \sin(\theta)$

  • Definition of cosine:     $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{w_y}{w}$     $w_y = w \cos(\theta) = mg \cos(\theta)$

  • Definition of tangent:     $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{w_x}{w_y}$

• Inversion of typical component conventions:

  • Unlike conventional horizontal/vertical systems where the $x$-component uses cosine and the $y$-component uses sine, measuring the incline angle $\theta$ relative to the negative $y$-axis causes the parallel $x$-component to depend on $\sin(\theta)$ and the perpendicular $y$-component to depend on $\cos(\theta)$.

Questions & Discussion: Coordinate System Angle Conventions

• Question / Prompt regarding standard angle measurement:

  • In standard vector mathematics, angles are conventionally measured counterclockwise from the positive horizontal $x$-axis, which automatically assigns $\cos$ to the $x$-component and $\sin$ to the $y$-component. Should that convention be maintained here?

• Explanation and resolution:

  • If an angle is measured strictly from the positive $x$-axis counterclockwise, the automatic convention ($A_x = A \cos(\theta)$ and $A_y = A \sin(\theta)$) holds without requiring geometric visualization.

  • However, in a rotated incline problem, using the standard counterclockwise angle requires calculating a new, non-standard reference angle rather than using the provided ramp angle $\theta$.

  • To directly utilize the given physical angle $\theta$ of the incline, the convention is adapted: the geometry must be evaluated explicitly, yielding:     $w_x = mg \sin(\theta)$     $w_y = mg \cos(\theta)$

  • Both methods are mathematically valid, but using the explicit geometry of the incline angle $\theta$ avoids recomputing the angle of the coordinate system.

Application of Newton's Second Law of Motion

• General equation of Newton's Second Law:   $\sum \mathbf{F} = m \mathbf{a}$

• Analysis along the rotated $x$-axis (parallel to incline):

  • Once the weight vector is resolved into orthogonal components, the original vector $w$ is discarded from the summation.

  • The only force acting along the rotated $x$-axis is the parallel component of gravity, $w_x$, acting down the incline:     $\sum F_x = w_x = m a_x$

  • Substituting $w_x = mg \sin(\theta)$:     $mg \sin(\theta) = m a_x$

  • Mass $m$ cancels from both sides:     $a_x = g \sin(\theta)$

  • Conclusion: The acceleration down a frictionless inclined plane is entirely independent of the mass of the object and depends solely on the acceleration due to gravity $g$ and the incline angle $\theta$.

• Analysis along the rotated $y$-axis (perpendicular to incline):

  • The forces acting along the rotated $y$-axis are the normal force $N$ in the positive $y$-direction and the perpendicular component of gravity $w_y$ in the negative $y$-direction:     $\sum F_y = N - w_y = m a_y$

  • Because the block remains in continuous contact with the plane and does not accelerate off or through the ramp, motion perpendicular to the plane is zero:     $a_y = 0$

  • Substituting $a_y = 0$:     $N - w_y = 0$     $N = w_y$

  • Substituting $w_y = mg \cos(\theta)$:     $N = mg \cos(\theta)$

  • Conclusion: The normal force exerted by the ramp on the block equals $mg \cos(\theta)$.

Trigonometric Bounds and Limiting Case Analysis

• Mathematical bounds of trigonometric functions:

  • Both $\sin(\theta)$ and $\cos(\theta)$ oscillate between the values of $-1$ and $+1$.

  • Within the physical domain of an inclined plane, the incline angle is bounded between horizontal and vertical:     $0^\circ \le \theta \le 90^\circ$

  • Within this quadrant ($[0^\circ, 90^\circ]$):

  • Both $\sin(\theta)$ and $\cos(\theta)$ remain non-negative, taking values in the interval $[0, 1]$.

  • The value of $\cos(\theta)$ decreases monotonically from $1$ at $\theta = 0^\circ$ to $0$ at $\theta = 90^\circ$.

  • The value of $\sin(\theta)$ increases monotonically from $0$ at $\theta = 0^\circ$ to $1$ at $\theta = 90^\circ$.

• Physical consequences of function bounds:

  • Maximum normal force: Because $\cos(\theta) \le 1$, the normal force $N = mg \cos(\theta)$ can never exceed the total weight of the object ($mg$).

  • Maximum acceleration: Because $\sin(\theta) \le 1$, the acceleration down the incline $a = g \sin(\theta)$ can never exceed the free-fall acceleration due to gravity ($g$).

• Limiting Case 1: Horizontal Flat Surface ($\theta = 0^\circ$):

  • Incline geometry: The surface is completely flat and horizontal.

  • Normal force evaluation:     $N = mg \cos(0^\circ) = mg(1) = mg$     The normal force matches the full gravitational weight of the block, meaning the surface supports the entire load.

  • Acceleration evaluation:     $a = g \sin(0^\circ) = g(0) = 0$     The block experiences no lateral acceleration along the surface.

• Limiting Case 2: Perfectly Vertical Plane ($\theta = 90^\circ$):

  • Incline geometry: The ramp is rotated to a completely vertical wall.

  • Normal force evaluation:     $N = mg \cos(90^\circ) = mg(0) = 0$     The block loses contact force with the vertical wall; the normal force vanishes.

  • Acceleration evaluation:     $a = g \sin(90^\circ) = g(1) = g$     The block falls freely under gravity, accelerating at $g$.

• Validation of derived models:

  • The extreme-case behaviors at $\theta = 0^\circ$ and $\theta = 90^\circ$ match known physical realities (flat resting object vs. vertical free fall), confirming the correctness of $a_x = g \sin(\theta)$ and $N = mg \cos(\theta)$.