⛰️ Forces on an Incline

Part 1 of 7 — Inclined Planes

When an object sits on a tilted surface, gravity doesn't just pull it "down" — it has components along and perpendicular to the slope. Decomposing weight into these components is the foundation of every inclined plane problem.

The Tilted Coordinate System

On an incline, we tilt our x- and y-axes:

• x-axis: parallel to the surface (positive = down the slope)
• y-axis: perpendicular to the surface (positive = away from surface)

Why Tilt the Axes?

The normal force and friction are already along these tilted axes. The only force we need to decompose is gravity.

Weight Components

For an object on an incline of angle $\theta$:

$W_x = mg\sin\theta \quad \text{(along the slope, downhill)}$

$W_y = mg\cos\theta \quad \text{(into the slope)}$

Remembering Which Is Which

• At $\theta = 0°$: No component along the slope ($\sin 0° = 0$), full weight into surface ($\cos 0° = 1$). ✅
• At $\theta = 90°$: Full weight along the "slope" ($\sin 90° = 1$), nothing into surface ($\cos 90° = 0$). ✅

The sin component drives the object down the slope. The cos component is balanced by the normal force.

The Geometry Trick

Why is the angle of the incline equal to the angle between $\vec{W}$ and the perpendicular-to-surface direction?

Imagine the incline angle $\theta$ at the base. The surface is tilted $\theta$ from horizontal. The perpendicular-to-surface direction is tilted $\theta$ from vertical. Since gravity points straight down (vertical), the angle between gravity and the perpendicular direction is also $\theta$.

This means: $\text{Component perpendicular to surface} = mg\cos\theta$ $\text{Component parallel to surface} = mg\sin\theta$

The Normal Force

Since there's no acceleration perpendicular to the surface: $N = mg\cos\theta$

This is less than $mg$ — the steeper the incline, the smaller the normal force.

Check Your Understanding 🧠

Calculate the Components 🧮

A 20 kg block sits on a 37° incline ($g = 10$ m/s², $\sin 37° = 0.60$, $\cos 37° = 0.80$).

1. What is the component of weight parallel to the slope (in N)?

2. What is the normal force (in N)?

3. What is the ratio $W_{\parallel}/N$?

Round all answers to 3 significant figures.

Conceptual Checks 🔍

Exit Quiz ✅

🎿 Frictionless Inclines

Part 2 of 7 — Inclined Planes

Before adding friction, let's master the simpler case: objects sliding on perfectly smooth inclines. The only forces are gravity and the normal force — and since the normal force is perpendicular to motion, only gravity's component along the slope accelerates the object.

Acceleration on a Frictionless Incline

Free Body Diagram

On a frictionless incline of angle $\theta$:

• $N = mg\cos\theta$ (perpendicular — balanced)
• $mg\sin\theta$ (along the slope — unbalanced!)

Newton's Second Law (Along the Slope)

$mg\sin\theta = ma$ $\boxed{a = g\sin\theta}$

Key Insight

The acceleration is independent of mass! A 1 kg block and a 100 kg block slide at the same rate on the same frictionless incline.

Example Values

Notice: At 90° the "incline" is vertical, and the acceleration equals $g$.

Combining with Kinematics

Once you know $a = g\sin\theta$, use the standard kinematics equations:

$v = v_0 + at$ $\Delta x = v_0 t + \frac{1}{2}at^2$ $v^2 = v_0^2 + 2a\Delta x$

Example: Sliding Down from Rest

A block starts from rest and slides 2 m down a 30° frictionless incline. How fast is it going?

$a = g\sin 30° = 10 \times 0.50 = 5 \text{ m/s}^2$ $v^2 = 0 + 2(5)(2) = 20$ $v = \sqrt{20} \approx 4.5 \text{ m/s}$

Example: Sliding Up

A block is launched UP a 30° frictionless incline at 6 m/s. How far up does it go?

The deceleration is $a = -g\sin 30° = -5$ m/s² (opposing motion up the slope).

$0 = 36 + 2(-5)\Delta x$ $\Delta x = 36/10 = 3.6 \text{ m}$

Check Your Understanding 🧠

Frictionless Incline Problems 🧮

1. A block starts from rest on a 37° frictionless incline and slides for 2 seconds. What is its speed (m/s)? ($g = 10$ m/s², $\sin 37° = 0.60$)

2. How far has it traveled in those 2 seconds (in m)?

3. A block is launched at 10 m/s up a 30° frictionless incline. How far up the slope does it travel before stopping (in m)? ($g = 10$ m/s², $\sin 30° = 0.50$)

Conceptual Reasoning 🔍

Exit Quiz ✅

🧱 Inclines with Friction

Part 3 of 7 — Inclined Planes

Most real inclines have friction. Combining the incline weight components with friction forces is one of the most common AP Physics 1 problems. Let's master the approach.

FBD on an Incline with Friction

Forces on an object on an incline with friction:

Perpendicular to surface (y-axis):

• $N$ (away from surface)
• $mg\cos\theta$ (into surface)
• These balance: $N = mg\cos\theta$

Parallel to surface (x-axis):

• $mg\sin\theta$ (down the slope)
• $f$ (friction — direction depends on motion!)

Friction Direction Rules

• Sliding down: Friction acts up the slope
• Sliding up: Friction acts down the slope
• Stationary: Friction opposes the tendency of motion

Case 1: Sliding Down the Incline

$ma = mg\sin\theta - f_k = mg\sin\theta - \mu_k mg\cos\theta$ $\boxed{a = g(\sin\theta - \mu_k \cos\theta)}$

Case 2: Sliding Up the Incline

Both gravity and friction oppose motion: $ma = mg\sin\theta + f_k = mg\sin\theta + \mu_k mg\cos\theta$ $\boxed{a_{\text{decel}} = g(\sin\theta + \mu_k \cos\theta)}$

Note: The deceleration going up is greater than the acceleration going down!

Worked Example

A 5 kg block slides down a 37° incline with $\mu_k = 0.25$ ($g = 10$ m/s², $\sin 37° = 0.60$, $\cos 37° = 0.80$).

Step 1: Normal force: $N = mg\cos 37° = 50 \times 0.80 = 40$ N

Step 2: Friction: $f_k = \mu_k N = 0.25 \times 40 = 10$ N (up the slope)

Step 3: Net force along slope: $F_{\text{net}} = mg\sin 37° - f_k = 30 - 10 = 20$ N

Step 4: Acceleration: $a = 20/5 = 4$ m/s²

Compare to frictionless: $a = g\sin 37° = 6$ m/s²

Friction reduced the acceleration by $\mu_k g\cos 37° = 2$ m/s².

Check Your Understanding 🧠

Incline + Friction Problems 🧮

A 10 kg block slides down a 53° incline with $\mu_k = 0.30$ ($g = 10$ m/s², $\sin 53° = 0.80$, $\cos 53° = 0.60$).

1. What is the normal force (in N)?

2. What is the kinetic friction force (in N)?

3. What is the acceleration down the slope (in m/s²)?

Round all answers to 3 significant figures.

Special Case: Constant Velocity on an Incline

If a block slides at constant velocity down an incline: $mg\sin\theta = \mu_k mg\cos\theta$ $\mu_k = \tan\theta$

This is identical to the critical angle for static friction! Measuring the angle at which a block slides at constant velocity gives you $\mu_k$ directly.

Will It Slide?

A block is placed on a rough incline. It will slide if: $mg\sin\theta > f_{s,\max} = \mu_s mg\cos\theta$ $\tan\theta > \mu_s$

If $\mu_s = 0.577$, the critical angle is $\tan^{-1}(0.577) = 30°$. Any steeper and it slides.

Exit Quiz ✅

🔗 Connected Systems on Inclines

Part 4 of 7 — Inclined Planes

Many AP Physics problems involve objects connected by strings over pulleys, with one or both on inclines. These are classic Atwood-on-incline problems. The key: all connected objects share the same magnitude of acceleration.

The Classic Setup

Object on Incline + Hanging Mass

A block of mass $m_1$ sits on an incline of angle $\theta$, connected by a string over a frictionless pulley to a hanging mass $m_2$.

For $m_1$ (on the incline, taking up-the-slope as positive): $T - m_1 g\sin\theta = m_1 a$

For $m_2$ (hanging, taking downward as positive): $m_2 g - T = m_2 a$

Solving for Acceleration

Add the equations to eliminate $T$: $m_2 g - m_1 g\sin\theta = (m_1 + m_2)a$

$\boxed{a = \frac{m_2 g - m_1 g\sin\theta}{m_1 + m_2}}$

Solving for Tension

Substitute $a$ back into either equation: $T = m_2(g - a) = \frac{m_1 m_2 g(1 + \sin\theta)}{m_1 + m_2}$

Direction Check

• If $m_2 g > m_1 g\sin\theta$: the hanging mass falls and $m_1$ goes up the incline
• If $m_2 g < m_1 g\sin\theta$: $m_1$ slides down and $m_2$ rises

Worked Example

$m_1 = 4$ kg on a frictionless 30° incline, connected to $m_2 = 3$ kg hanging ($g = 10$ m/s²).

Driving force: $m_2 g - m_1 g\sin 30° = 30 - 20 = 10$ N

Total mass: $m_1 + m_2 = 7$ kg

Acceleration: $a = 10/7 \approx 1.43$ m/s²

Tension: $T = m_2(g - a) = 3(10 - 1.43) = 25.7$ N

Check: For $m_1$: $T - m_1 g\sin 30° = 25.7 - 20 = 5.7$ N. $m_1 a = 4 \times 1.43 = 5.7$ N. ✅

With Friction

If the incline has friction ($\mu_k$), add friction to $m_1$'s equation:

$m_1$ sliding up: $T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a$

$a = \frac{m_2 g - m_1 g\sin\theta - \mu_k m_1 g\cos\theta}{m_1 + m_2}$

Check Your Understanding 🧠

Connected System Calculations 🧮

$m_1 = 6$ kg on a frictionless 30° incline, connected over a pulley to $m_2 = 5$ kg hanging freely ($g = 10$ m/s², $\sin 30° = 0.50$).

1. What is the net driving force of the system (in N)?

2. What is the acceleration (in m/s², to one decimal)?

3. What is the tension in the string (in N, to one decimal)?

Round all answers to 3 significant figures.

System Analysis 🔍

Exit Quiz ✅

📐 Acceleration and Velocity on Inclines

Part 5 of 7 — Inclined Planes

This lesson focuses on using kinematics on inclines — calculating how fast objects move, how far they travel, and how long it takes. We'll combine the incline acceleration formula with the kinematic equations you already know.

Kinematics on Inclines — Review

Once you find $a$ on an incline, the kinematics equations work exactly the same:

$v = v_0 + at$ $\Delta x = v_0 t + \frac{1}{2}at^2$ $v^2 = v_0^2 + 2a\Delta x$

Key Accelerations

Worked Example: Up and Back Down

A block is launched at $v_0 = 8$ m/s up a 30° incline with $\mu_k = 0.20$ ($g = 10$ m/s²).

Going up: $a_{\text{up}} = g(\sin 30° + \mu_k\cos 30°) = 10(0.50 + 0.173) = 6.73$ m/s²

Distance up: $v^2 = v_0^2 - 2a_{\text{up}}d$ $0 = 64 - 2(6.73)d \Rightarrow d = 64/13.46 = 4.75 \text{ m}$

Going down: $a_{\text{down}} = g(\sin 30° - \mu_k\cos 30°) = 10(0.50 - 0.173) = 3.27$ m/s²

Speed at bottom: $v^2 = 2(3.27)(4.75) = 31.1$ $v = \sqrt{31.1} \approx 5.6 \text{ m/s}$

The block returns slower than it left! Friction removed energy in both directions.

Time Problems

How long to slide down?

A block starts from rest and slides $d$ meters down an incline with acceleration $a$: $d = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2d}{a}}$

Example

A block slides from rest down a 5 m frictionless 45° incline ($g = 10$ m/s²).

$a = g\sin 45° = 7.07 \text{ m/s}^2$ $t = \sqrt{\frac{2 \times 5}{7.07}} = \sqrt{1.414} = 1.19 \text{ s}$

How long to stop going up?

$v_0 = 10$ m/s up a 30° rough incline ($\mu_k = 0.25$).

$a = g(\sin 30° + 0.25\cos 30°) = 10(0.50 + 0.217) = 7.17 \text{ m/s}^2$ $t = v_0/a = 10/7.17 = 1.39 \text{ s}$

Check Your Understanding 🧠

Kinematics on Inclines 🧮

A block starts from rest and slides 4 m down a 37° incline with $\mu_k = 0.25$ ($g = 10$ m/s², $\sin 37° = 0.60$, $\cos 37° = 0.80$).

1. What is the acceleration down the slope (in m/s²)?

2. What is the speed at the bottom (in m/s)?

3. How long does it take to reach the bottom (in s, to one decimal)?

Round all answers to 3 significant figures.

Reasoning Checks 🔍

Exit Quiz ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Inclined Planes

This workshop combines all incline concepts: weight components, friction on inclines, connected systems, and kinematics. Practice the complete problem-solving approach for AP-level incline problems.

Incline Problem-Solving Strategy

Step 1: Draw the FBD with Tilted Axes

• x-axis along the slope, y-axis perpendicular
• Weight components: $mg\sin\theta$ (along), $mg\cos\theta$ (perp)

Step 2: Find the Normal Force

$N = mg\cos\theta \pm F_{\text{applied, perp}}$

Step 3: Determine Friction (if any)

• Is it frictionless? → No friction term
• Sliding? → $f_k = \mu_k N$
• Stationary? → $f_s \leq \mu_s N$

Step 4: Apply Newton's Second Law Along the Slope

$\sum F_{\parallel} = ma$

Step 5: Solve for Unknowns, Then Use Kinematics if Needed

Worked Example

A 12 kg block is pulled up a 30° rough incline ($\mu_k = 0.20$) by a rope parallel to the surface with 100 N ($g = 10$ m/s²).

Normal force: $N = mg\cos 30° = 120 \times 0.866 = 103.9$ N

Friction (opposes motion up): $f_k = 0.20 \times 103.9 = 20.8$ N

Along slope: $100 - mg\sin 30° - f_k = ma$ $100 - 60 - 20.8 = 12a$ $19.2 = 12a$ $a = 1.6 \text{ m/s}^2$

Workshop Multiple Choice 🎯

Workshop Calculations 🧮

1. A 5 kg block slides from rest down a 53° incline with $\mu_k = 0.30$ for 2 seconds. What is its final speed (in m/s)? ($g = 10$ m/s², $\sin 53° = 0.80$, $\cos 53° = 0.60$)

2. A 10 kg block on a frictionless 30° incline is connected to a 4 kg hanging mass. What is the acceleration (in m/s²)? ($\sin 30° = 0.50$)

3. What minimum $\mu_s$ is needed to keep a block stationary on a 60° incline? ($\tan 60° = 1.73$)

Round all answers to 3 significant figures.

Reasoning Workshop 🔍

Exit Quiz — Incline Workshop ✅

🎯 Synthesis & AP Review

Part 7 of 7 — Inclined Planes

Congratulations on completing the Inclined Planes unit — and the entire Dynamics section! This final lesson reviews all incline concepts and connects them to the broader AP Physics 1 framework.

Complete Inclined Planes Reference

Weight Decomposition

$W_{\parallel} = mg\sin\theta \quad \text{(drives motion along slope)}$ $W_{\perp} = mg\cos\theta \quad \text{(balanced by normal force)}$

Normal Force

$N = mg\cos\theta \quad \text{(no extra perpendicular forces)}$

Acceleration Formulas

Connected System (Incline + Hanging Mass)

$a = \frac{m_{\text{hang}}g - m_{\text{incline}}g\sin\theta}{m_{\text{hang}} + m_{\text{incline}}}$

Key Principles

• Acceleration on a frictionless incline is mass-independent
• Normal force on an incline is less than $mg$
• Steeper incline = more acceleration, less normal force
• Going up a rough incline: deceleration > acceleration going down
• A block returns slower after going up and coming back on a rough incline

Conceptual Review 🧠

AP-Style Calculations 📝

1. A block slides from rest down a frictionless 30° incline that is 10 m long. What is its speed at the bottom (in m/s)? ($g = 10$ m/s², $\sin 30° = 0.50$)

2. A 5 kg block on a 37° incline ($\mu_k = 0.25$, $g = 10$ m/s²) slides down. What is the acceleration (in m/s²)? ($\sin 37° = 0.60$, $\cos 37° = 0.80$)

3. What angle gives a frictionless incline acceleration of $5$ m/s²? ($g = 10$ m/s², $\sin 30° = 0.50$)

AP Reasoning 🎯

Connecting to the Full Dynamics Framework

You've now completed all four dynamics topics:

1. Newton's First & Second Laws

$F_{\text{net}} = ma$ The foundation of all dynamics. Every incline problem uses this.

2. Newton's Third Law

Action-reaction pairs in connected systems (string tension, normal force pairs).

3. Friction

$f_s \leq \mu_s N$ and $f_k = \mu_k N$. Essential for realistic incline problems.

4. Inclined Planes

Weight decomposition + friction + kinematics. The ultimate application of all dynamics concepts combined.

AP Exam Tips

• Always draw a FBD — it's required for free-response credit
• Tilt your axes on incline problems
• Check your direction — what's positive?
• Verify answers — does the acceleration make sense? Is it less than $g$?
• Watch for "constant velocity" — that means $a = 0$!

Final Exit Quiz — Inclined Planes Unit ✅