Weight Decomposition on a Ramp (FGX and FGY)

Setup and coordinate choice

• Box on Earth sitting on a ramp; weight force is $F_G$, directed downward.

• Use rotated axes: $x$ along the ramp, $y$ perpendicular to the ramp; this often makes the sum of forces in the $y$ direction zero, simplifying problems.

• Weight has two components: parallel to ramp (along $x$) and perpendicular to ramp (along $y$). Name them $FGX$ (parallel) and $FGY$ (perpendicular).

• The components have directions set by the chosen axes (often both are negative in the example).

Components of weight

• Ramp angle relative to ground: $\theta$.

• Magnitudes:

  • $FGX = |F_G| \sin \theta$

  • $FGY = |F_G| \cos \theta$

• Signs depend on axis orientation; the magnitudes above give the resolved parts along the chosen directions.

Determining sine vs cosine

• Visual/geometry reasoning: the component adjacent to $\theta$ in the right triangle is the perpendicular component (FGY); the opposite side is the parallel component (FGX).

• Therefore:

  • $FGY = |F_G| \cos \theta$

  • $FGX = |F_G| \sin \theta$

• This matches the alternate geometry approach: the angle involved is consistent with identifying adjacent vs opposite sides.

Summary

• For a ramp with angle $\theta$ to the ground:

  • $FGY = |F_G| \cos \theta$ (perpendicular to ramp)

  • $FGX = |F_G| \sin \theta$ (parallel to ramp)

• The actual signs depend on the chosen +x/+y directions.

Friction: Static and Kinetic (Key Points)

Static friction

• Occurs when surfaces are not moving relative to each other.

• Maximum static friction: $Fs(max) = \mu s \cdot N$

• Static friction can match the applied force up to the maximum: $Fs = F(applied)$ for $F(applied) \le Fs(max)$

• If the applied force exceeds the maximum, slipping occurs at the threshold: $F(applied) = Fs(max) = \mu s \cdot N$

Kinetic friction

• Friction when surfaces slide relative to each other.

• Approximately constant: $Fk = \mu k \cdot N$, independent of applied force and velocity.

• Direction opposite the motion.

Coefficients and normal force

• Normal force: $N$; friction scales with $N$.

• Coefficients: $\mu s$, $\mu k$ depend on surfaces (experimental).

• Bounds: $Fs \le \mu s \cdot N$; $Fk = \mu k \cdot N$

Free-body diagram approach

• Determine whether in static or kinetic regime based on motion.

• If static: $Fs = F(applied)$ with $F(applied) \le \mu s \cdot N$; at threshold, static friction is overcome.

• If kinetic: $Fk = \mu k \cdot N$

• Solve using Newton's second law: $\sum F = ma$

Quick recap

• Static friction matches applied force up to the maximum: $Fs(max) = \mu s \cdot N$.

• Slip occurs when $F(applied) = Fs(max)$.

• After slip: kinetic friction dominates: $Fk = \mu k \cdot N$.