Inclined Planes

Introduction

An inclined plane is a flat surface tilted at an angle to the horizontal.

Why they're important:

• Reduces force needed to lift objects (mechanical advantage)
• Common in real world: ramps, hills, roofs
• Requires careful component analysis

Coordinate System

Standard choice: Rotate axes so one axis is parallel to the incline.

• x-axis: Parallel to incline (positive = up the incline)
• y-axis: Perpendicular to incline (positive = away from surface)
• Angle: $\theta$ = angle incline makes with horizontal

Why this choice?

• Motion is along the incline (parallel to x-axis)
• No motion perpendicular to incline (y-direction equilibrium)
• Simplifies math!

Forces on an Incline

Weight Components

Weight $W = mg$ points straight down (not perpendicular to incline!).

Break weight into components:

Parallel to incline (down the slope): $W_x = mg\sin\theta$

Perpendicular to incline (into the surface): $W_y = mg\cos\theta$

Derivation from geometry:

• Angle between weight and perpendicular to incline = $\theta$
• Therefore: parallel component uses $\sin\theta$, perpendicular uses $\cos\theta$

Normal Force

Normal force $N$ acts perpendicular to the surface (along y-axis).

In y-direction (no acceleration perpendicular to incline): $N - mg\cos\theta = 0$ $N = mg\cos\theta$

Key point: Normal force is NOT equal to $mg$ on an incline!

Friction

Friction acts parallel to incline, opposing motion or attempted motion.

• If object slides down: friction points up the incline
• If object is pushed up: friction points down the incline

Maximum static friction: $f_{s,max} = \mu_s N = \mu_s mg\cos\theta$

Kinetic friction: $f_k = \mu_k N = \mu_k mg\cos\theta$

Motion Down a Frictionless Incline

Free body diagram:

• Weight component parallel: $mg\sin\theta$ (down incline)
• Weight component perpendicular: $mg\cos\theta$ (into surface)
• Normal force: $N = mg\cos\theta$ (out from surface)
• No friction

Apply Newton's Second Law (x-direction, parallel to incline):

$\sum F_x = ma$ $mg\sin\theta = ma$ $a = g\sin\theta$

Key results:

• Acceleration is independent of mass!
• Acceleration increases with steeper incline ($\theta$ larger)
• At $\theta = 90°$: $a = g$ (free fall)
• At $\theta = 0°$: $a = 0$ (horizontal surface)

Motion with Friction

Object at Rest on Incline

Question: What coefficient of friction is needed to prevent sliding?

Analysis:

Parallel to incline (equilibrium): $f_s = mg\sin\theta$

Maximum static friction: $f_{s,max} = \mu_s N = \mu_s mg\cos\theta$

Condition for no sliding: $mg\sin\theta \leq \mu_s mg\cos\theta$ $\sin\theta \leq \mu_s \cos\theta$ $\tan\theta \leq \mu_s$

Critical angle: $\theta_c = \tan^{-1}(\mu_s)$

• If $\theta < \theta_c$: object stays at rest
• If $\theta = \theta_c$: object on verge of sliding
• If $\theta > \theta_c$: object slides down

Object Sliding Down Incline

Free body diagram:

• Weight component parallel: $mg\sin\theta$ (down)
• Kinetic friction: $f_k = \mu_k mg\cos\theta$ (up)
• Normal force: $N = mg\cos\theta$

Apply Newton's Second Law:

$\sum F_x = ma$ $mg\sin\theta - \mu_k mg\cos\theta = ma$ $a = g(\sin\theta - \mu_k\cos\theta)$

Special cases:

• If $\sin\theta = \mu_k\cos\theta$: $a = 0$ (slides at constant velocity)
• If $\sin\theta > \mu_k\cos\theta$: $a > 0$ (speeds up)
• If $\sin\theta < \mu_k\cos\theta$: Can't slide down (would need initial motion)

Object Pushed Up Incline

Forces parallel to incline:

• Applied force $F_{app}$ (up incline)
• Weight component: $mg\sin\theta$ (down incline)
• Friction: $f_k = \mu_k mg\cos\theta$ (down incline, opposes motion)

Newton's Second Law: $F_{app} - mg\sin\theta - \mu_k mg\cos\theta = ma$

To push at constant velocity ($a = 0$): $F_{app} = mg\sin\theta + \mu_k mg\cos\theta$ $F_{app} = mg(\sin\theta + \mu_k\cos\theta)$

Problem-Solving Strategy

1. Draw the incline and identify angle $\theta$
2. Rotate coordinate system (x parallel, y perpendicular to incline)
3. Draw free body diagram with rotated axes
4. Break weight into components:
   • Parallel: $mg\sin\theta$
   • Perpendicular: $mg\cos\theta$
5. Find normal force from y-direction equilibrium: $N = mg\cos\theta$
6. Determine friction (static or kinetic)
7. Apply Newton's Second Law in x-direction
8. Solve for unknowns

Common Angles and Trig Values

| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ | |---------|-------------|-------------|-------------| | $0°$ | $0$ | $1$ | $0$ | | $30°$ | $0.5$ | $0.866$ | $0.577$ | | $37°$ | $0.6$ | $0.8$ | $0.75$ | | $45°$ | $0.707$ | $0.707$ | $1$ | | $53°$ | $0.8$ | $0.6$ | $1.33$ | | $90°$ | $1$ | $0$ | undefined |

Real-World Applications

• Ramps: Reduce force needed to lift heavy objects
• Roads on hills: Banking prevents cars from sliding
• Ski slopes: Steeper = faster (larger $g\sin\theta$)
• Conveyor belts: Angle and friction determine maximum load