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Friction and Inclined Planes

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Friction and Inclined Planes - Complete Interactive Lesson

Part 1: Static vs Kinetic Friction

Static vs Kinetic Friction

Part 1 of 7 — Friction & Inclines

Friction is one of the most common forces you'll encounter on the AP Physics C exam. Understanding the distinction between static and kinetic friction — and when each applies — is essential.

The Nature of Friction

Friction arises from microscopic interactions between surfaces in contact. At the AP Physics C level, we model friction with two simple laws:

Type	Symbol	Condition	Formula
Static	$f_s$	Object at rest	$f_s \leq \mu_s N$
Kinetic	$f_k$	Object sliding	$f_k = \mu_k N$

Key Differences

Static friction is a variable force: it adjusts to match the applied force, up to a maximum $f_{s,\max} = \mu_s N$ .
Kinetic friction is a constant force (for a given normal force): $_{}$ .

The Normal Force

The normal force $N$ is perpendicular to the contact surface. On a flat surface with no vertical acceleration:

$N = mg - F\sin\theta$

where $F$ is an applied force at angle $\theta$ above the horizontal. If pushing downward at angle $\theta$ below horizontal:

$N = mg + F\sin\theta$

Free Body Diagrams with Friction

When drawing FBDs involving friction, always:

Identify the contact surface — friction acts along it.
Determine the normal force — perpendicular to the surface.
Check if the object moves — this determines static vs kinetic.
Direction of friction — opposes motion (kinetic) or opposes tendency of motion (static).

Worked Example

A 5 kg block is pushed along a floor by a force $F = 40$ N at $30°$ below the horizontal. $\mu_k = 0.25$ .

The Threshold of Motion

A critical AP Physics C skill is determining whether a block moves under a given force. The procedure:

Assume static: Calculate the force needed to keep the block stationary.
Compare to $f_{s,\max}$ : If the required friction $\leq \mu_s N$ , the block stays at rest.

Part 1 Summary

Concept	Key Formula
Static friction	$f_s \leq \mu_s N$
Kinetic friction	$_{}$

Part 2: Frictionless Inclines

Inclined Planes (No Friction)

Part 2 of 7 — Friction & Inclines

Before adding friction, let's master the frictionless incline — the foundation for all ramp problems.

Setting Up Coordinates

For an incline at angle $\theta$ :

Direction	Axis	Forces
Along the incline	$x$ -axis	$mg\sin\theta$ (down the ramp)
Perpendicular to incline

Part 3: Inclines with Friction

Inclined Planes with Friction

Part 3 of 7 — Friction & Inclines

Now we combine friction with inclined planes — the bread and butter of AP Physics C mechanics problems.

Forces on a Rough Incline

For a block on an incline of angle $\theta$ with friction:

Perpendicular to incline ( $y$ -direction): $N = mg\cos\theta$

Along the incline (-direction):

Part 4: Velocity-Dependent Friction

Friction with Calculus (Velocity-Dependent)

Part 4 of 7 — Friction & Inclines

In many real-world situations, the resistive force depends on velocity. This is the hallmark of AP Physics C — using differential equations to solve dynamics problems.

Linear Drag: $f = bv$

A common model for low-speed drag:

$m\frac{dv}{dt} = F_{\text{applied}} - bv$

Part 5: Systems on Inclines

Systems on Inclines

Part 5 of 7 — Friction & Inclines

Many AP Physics C problems involve multiple objects connected by strings, often on inclines. The key: apply Newton's second law to each object separately, then combine.

Atwood Machine on an Incline

A classic setup: mass $m_1$ on a rough incline (angle $\theta$ , coefficient $\mu_k$ ) connected by a massless string over a frictionless pulley to a hanging mass .

Part 6: Problem-Solving Workshop

Problem-Solving Workshop

Part 6 of 7 — Friction & Inclines

This workshop presents AP Physics C–style problems that integrate concepts from Parts 1–5. Practice the systematic approach:

Problem-Solving Framework

Step	Action
1	Draw a diagram and label all forces
2	Choose a coordinate system (tilted for inclines)
3	Write Newton's 2nd law for each object
4	Identify constraints (strings, contact)
5	Solve the system of equations
6	Check units and limiting cases

Problem 2: Block Launched Up a Ramp

A 3 kg block is launched up a $37°$ rough incline ( $\mu_k = 0.3$ ) with initial speed m/s.

Part 7: Review & Applications

Review & Applications

Part 7 of 7 — Friction & Inclines

Complete Topic Reference

Concept	Formula	Part
Static friction	$f_s \leq \mu_s N$

Typically

\mu_s > \mu_k

; it takes more force to start motion than to maintain it.

Step 1: Normal force (vertical equilibrium): $N = mg + F\sin 30° = 5(10) + 40(0.5) = 70 \text{ N}$

Step 2: Kinetic friction: $f_k = \mu_k N = 0.25 \times 70 = 17.5 \text{ N}$

Step 3: Horizontal acceleration: $F\cos 30° - f_k = ma$ $40(0.866) - 17.5 = 5a$ $a = \frac{34.64 - 17.5}{5} = 3.43 \text{ m/s}^2$

Key Insight: Pushing downward increases the normal force and thus the friction force. Pulling upward decreases it.

If exceeded: The block accelerates. Switch to kinetic friction for the dynamics.

The Transition: Calculus Perspective

At the instant motion begins, the friction force drops from $f_{s,\max} = \mu_s N$ to $f_k = \mu_k N$ . This creates an instantaneous jump in acceleration:

$a(t = 0^+) = \frac{F_{\text{applied}} - \mu_k N}{m}$

This discontinuity in friction is why objects often "jerk" when they start sliding.

Optimal Angle for Pulling

To minimize the pulling force $F$ needed to move a block, differentiate with respect to $\theta$ :

$F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$

Setting $\frac{dF}{d\theta} = 0$ :

$\sin\theta - \mu_s \cos\theta = 0 \implies \tan\theta = \mu_s$

$\theta_{\text{opt}} = \arctan(\mu_s)$

Next up: Part 2 — Inclined Planes (No Friction), where we analyze motion on smooth ramps using calculus.

Newton's Second Law on an Incline

Along the incline (taking down-the-ramp as positive):

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

This is constant acceleration, so all kinematics equations apply:

$v(t) = v_0 + (g\sin\theta)\,t$ $x(t) = x_0 + v_0 t + \frac{1}{2}(g\sin\theta)\,t^2$ $v^2 = v_0^2 + 2(g\sin\theta)\Delta x$

Energy Methods on Inclines

For a frictionless incline, mechanical energy is conserved:

$\frac{1}{2}mv_0^2 + mgh_0 = \frac{1}{2}mv^2 + mgh$

The height change on a ramp of length $L$ at angle $\theta$ :

$\Delta h = L\sin\theta$

Worked Example: Launched Up a Ramp

A block is launched up a frictionless $45°$ ramp with initial speed $v_0 = 10$ m/s.

How far up the ramp does it travel?

Using energy conservation: $\frac{1}{2}mv_0^2 = mg(L\sin\theta)$

Time to reach the top:

Decelerating at $a = -g\sin 45° = -7.07$ m/s²: $v = v_0 - (g\sin\theta)t = 0$

Velocity as a Function of Position

Using $v^2 = v_0^2 - 2g\sin\theta \cdot x$ :

$v(x) = \sqrt{v_0^2 - 2gx\sin\theta}$

This is valid until $v = 0$ at $x_{\max} = \frac{v_0^2}{2g\sin\theta}$ .

Calculus Approach: Variable Angle Ramps

What if the ramp angle changes? Consider a curved ramp where $\theta = \theta(x)$ .

For a small displacement $dx$ along the ramp, the height change is $dh = \sin\theta(x)\,dx$ .

The equation of motion becomes: $m\frac{dv}{dt} = mg\sin\theta(x)$

Using the chain rule: $\frac{dv}{dt} = v\frac{dv}{dx}$ :

$mv\frac{dv}{dx} = mg\sin\theta(x)$

$\int_{v_0}^{v} v'\,dv' = g\int_0^x \sin\theta(x')\,dx'$

$\frac{v^2 - v_0^2}{2} = g\int_0^x \sin\theta(x')\,dx'$

Example: Parabolic Ramp

For a ramp shaped like $y = \frac{x^2}{2L}$ , the slope at position $x$ is:

$\tan\theta = \frac{dy}{dx} = \frac{x}{L}$

For small angles: $\sin\theta \approx \tan\theta = x/L$ .

$\frac{v^2}{2} = g\int_0^x \frac{x'}{L}\,dx' = \frac{gx^2}{2L}$

$v(x) = x\sqrt{\frac{g}{L}}$

The speed increases linearly with position — a result unique to this geometry.

Part 2 Summary

Concept	Formula
Acceleration on incline	$a = g\sin\theta$
Normal force	$N = mg\cos\theta$
Speed at bottom	$v = \sqrt{2gL\sin\theta} = \sqrt{2gh}$
Height change	$\Delta h = L\sin\theta$
Chain rule substitution	$a = v\frac{dv}{dx}$

Next up: Part 3 — Inclined Planes with Friction, combining everything from Parts 1 and 2.

Scenario	Net Force (down ramp +)
Sliding down	$mg\sin\theta - \mu_k mg\cos\theta = ma$
Sliding up	$-mg\sin\theta - \mu_k mg\cos\theta = ma$
At rest (checking)	$mg\sin\theta \leq \mu_s mg\cos\theta$ ?

A block starts sliding when the component of gravity along the ramp exceeds maximum static friction:

$mg\sin\theta_c = \mu_s mg\cos\theta_c$ $\tan\theta_c = \mu_s$ $\theta_c = \arctan(\mu_s)$

This gives a clean way to measure $\mu_s$ experimentally.

Dynamics on Rough Inclines

Sliding Down

Acceleration (taking down-ramp as positive): $a_{\text{down}} = g(\sin\theta - \mu_k \cos\theta)$

The block slides down only if $\tan\theta > \mu_s$ (i.e., $\theta > \theta_c$ ).

Sliding Up (Launched Upward)

When a block is launched up a rough incline, both gravity and friction decelerate it:

$a_{\text{up}} = -g(\sin\theta + \mu_k \cos\theta)$

Caution: After the block stops, it may or may not slide back down. It stays at rest if $\tan\theta \leq \mu_s$ .

Worked Example

A block is launched up a $37°$ rough incline ( $\mu_k = 0.3$ ) with $v_0 = 12$ m/s.

Deceleration going up: $a_{\text{up}} = -g(\sin 37° + 0.3\cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2$

Distance traveled up: $v^2 = v_0^2 + 2a\Delta x = 0$

Does it slide back? $\tan 37° = 0.75 > \mu_s$ (assuming $\mu_s \approx 0.35$ ), so yes.

Acceleration sliding back down: $a_{\text{down}} = g(\sin 37° - 0.3\cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2$

Speed at the bottom (returning): $v = \sqrt{2(3.6)(8.57)} = \sqrt{61.7} = 7.85 \text{ m/s}$

Notice: $v_{\text{return}} < v_0$ because friction dissipated energy.

Energy Methods with Friction

When friction is present, mechanical energy is not conserved. The work-energy theorem gives:

$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = W_{\text{net}} = -mgh - \mu_k N \cdot d$

For a ramp of length $d$ at angle $\theta$ :

$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + mgd\sin\theta - \mu_k mgd\cos\theta$

The energy dissipated by friction:

$\Delta E_{\text{thermal}} = \mu_k mgd\cos\theta = f_k \cdot d$

Round Trip Energy Loss

For a block launched up and sliding back to the start:

$\Delta E_{\text{total}} = 2\mu_k mgd\cos\theta$

Since $\frac{1}{2}mv_{\text{return}}^2 = \frac{1}{2}mv_0^2 - 2\mu_k mgd\cos\theta$ :

$v_{\text{return}} = \sqrt{v_0^2 - 4\mu_k gd\cos\theta}$

where $d = \frac{v_0^2}{2g(\sin\theta + \mu_k\cos\theta)}$ is the distance traveled up.

Part 3 Summary

Concept	Formula
Critical angle	$\theta_c = \arctan(\mu_s)$
Accel. sliding down	$a = g(\sin\theta - \mu_k\cos\theta)$
Decel. sliding up	$a = -g(\sin\theta + \mu_k\cos\theta)$
Energy loss (one way)	$\Delta E = \mu_k mgd\cos\theta$

Next up: Part 4 — Velocity-Dependent Friction, where we use differential equations to handle drag and other calculus-based friction models.

At terminal velocity, $a = 0$ :

$v_T = \frac{F_{\text{applied}}}{b}$

For an object falling under gravity with linear drag:

$v_T = \frac{mg}{b}$

$m\frac{dv}{dt} = mg - bv$

Separation of variables:

$\frac{dv}{mg - bv} = \frac{dt}{m}$

$-\frac{1}{b}\ln\left(\frac{mg - bv}{mg}\right) = \frac{t}{m}$

$v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right) = v_T\left(1 - e^{-t/\tau}\right)$

where $\tau = m/b$ is the time constant.

Quadratic Drag: $f = cv^2$

At higher speeds, drag is proportional to $v^2$ :

$m\frac{dv}{dt} = mg - cv^2$

Terminal Velocity

$v_T = \sqrt{\frac{mg}{c}}$

Solving the ODE

$\frac{dv}{g - (c/m)v^2} = dt$

Let $\alpha = c/m$ . Then:

$\frac{dv}{g - \alpha v^2} = dt$

Using partial fractions or the substitution $v = v_T \tanh(u)$ :

$v(t) = v_T \tanh\left(\frac{gt}{v_T}\right)$

where $v_T = \sqrt{g/\alpha} = \sqrt{mg/c}$ .

Position by Integration

$x(t) = \int_0^t v_T \tanh\left(\frac{gt'}{v_T}\right)dt' = \frac{v_T^2}{g}\ln\cosh\left(\frac{gt}{v_T}\right)$

Key Behavior

Time	Speed	Acceleration
$t = 0$	$0$	$g$
$t \to \infty$

Velocity-Dependent Friction on Inclines

Consider a block sliding down an incline with velocity-dependent friction $f = bv$ :

$m\frac{dv}{dt} = mg\sin\theta - bv$

This has the same form as free fall with linear drag. The solution:

$v(t) = \frac{mg\sin\theta}{b}\left(1 - e^{-bt/m}\right)$

Terminal velocity on the incline:

$v_T = \frac{mg\sin\theta}{b}$

Worked Example

A 0.5 kg block slides down a $30°$ incline with velocity-dependent friction $f = 2v$ (in SI units). Find:

Terminal velocity: $v_T = \frac{mg\sin\theta}{b} = \frac{0.5 \times 10 \times 0.5}{2} = 1.25 \text{ m/s}$

Time constant: $\tau = \frac{m}{b} = \frac{0.5}{2} = 0.25 \text{ s}$

Speed at $t = 0.5$ s: $v(0.5) = 1.25\left(1 - e^{-0.5/0.25}\right) = 1.25(1 - e^{-2}) = 1.25(0.865) = 1.08 \text{ m/s}$

Acceleration at $t = 0.5$ s: $a(t) = g\sin\theta \cdot e^{-bt/m} = 5 \cdot e^{-2} = 0.677 \text{ m/s}^2$

Part 4 Summary

Model	ODE	Solution	$v_T$
Linear drag ( $f=bv$ )	$m\dot{v} = mg - bv$	$v_T(1-e^{-t/\tau})$	$mg/b$
Quadratic drag ( $f=cv^2$ )	$m\dot{v} = mg - cv^2$
Linear drag, no driving	$m\dot{v} = -bv$	$v_0 e^{-t/\tau}$

AP Tip: The AP Physics C exam frequently tests your ability to set up and solve first-order ODEs with separation of variables.

Next up: Part 5 — Systems on Inclines (Atwood machines, connected blocks).

Block on incline ( $m_1$ ): $T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a$

Hanging block ( $m_2$ , assuming it descends): $m_2 g - T = m_2 a$

Solving for $a$ and $T$

Adding the equations: $m_2 g - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = (m_1 + m_2)a$

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

$T = m_2(g - a) = \frac{m_1 m_2 g(1 + \sin\theta + \mu_k\cos\theta)}{m_1 + m_2}$

Two Blocks on Different Inclines

Consider masses $m_1$ and $m_2$ connected by a string over a pulley, each on different inclines at angles $\theta_1$ and $\theta_2$ .

Assuming $m_1$ moves up its incline and $m_2$ moves down its incline:

For $m_1$ (moves up, friction opposes): $T - m_1 g\sin\theta_1 - \mu_1 m_1 g\cos\theta_1 = m_1 a$

For $m_2$ (moves down, friction opposes): $m_2 g\sin\theta_2 - T - \mu_2 m_2 g\cos\theta_2 = m_2 a$

Adding: $a = \frac{m_2 g(\sin\theta_2 - \mu_2\cos\theta_2) - m_1 g(\sin\theta_1 + \mu_1\cos\theta_1)}{m_1 + m_2}$

Checking Direction

Before solving, determine which way the system tends to move:

Compare $m_1 g\sin\theta_1$ vs $m_2 g\sin\theta_2$ (gravitational components).

Stacked Blocks on an Incline

For a block of mass $m_1$ on top of a block $m_2$ on a ramp:

The friction between blocks is what accelerates/decelerates $m_1$ .
If the blocks move together: treat as one system to find $a$ , then isolate one block to find the friction between them.
If they slide relative to each other: apply kinetic friction at the interface.

Constraint Equations

When blocks are connected by strings through pulleys, the constraint equation relates their accelerations.

Simple Constraint (Single String)

If a single string connects two blocks, their speeds are equal: $v_1 = v_2 \implies a_1 = a_2 = a$

Pulley Ratio Constraint

If a string wraps around a movable pulley attached to block 2: $\Delta x_2 = \frac{\Delta x_1}{2} \implies a_2 = \frac{a_1}{2}$

And the tensions relate as: $T_2 = 2T_1$

Worked Example: Pulley System

A 5 kg block on a $37°$ rough incline ( $\mu_k = 0.25$ ) is connected to a 3 kg hanging mass through a pulley attached to the incline block.

With constraint $a_{\text{hang}} = 2a_{\text{incline}}$ :

Let $a$ be the incline block's acceleration (up the incline).

Incline block: $2T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a$

Hanging block: $m_2 g - T = m_2(2a)$

From the hanging block: $T = m_2 g - 2m_2 a = 3(10) - 6a = 30 - 6a$

Substituting: $2(30 - 6a) - 50\sin 37° - 0.25(50)\cos 37° = 5a$

$60 - 12a - 30 - 10 = 5a$ $20 = 17a \implies a = 1.18 \text{ m/s}^2$

Part 5 Summary

System Type	Key Approach
Atwood on incline	Separate FBDs, same $a$ and $T$
Two inclines	Compare gravitational components to find direction
Stacked blocks	Check if friction is sufficient; if not, blocks separate
Pulley constraints	Relate $a_1$ and $a_2$ via string length

Next up: Part 6 — Problem-Solving Workshop with multi-step AP-style problems.

Phase 1: Going up

$a_{\text{up}} = -g(\sin 37° + \mu_k \cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2$

Distance traveled up: $0 = v_0^2 + 2a_{\text{up}}d \implies d = \frac{v_0^2}{2 \times 8.4} = \frac{100}{16.8} = 5.95 \text{ m}$

Time to reach top: $t_1 = \frac{v_0}{|a_{\text{up}}|} = \frac{10}{8.4} = 1.19 \text{ s}$

Phase 2: Check if it slides back

$\tan 37° = 0.75$ . If $\mu_s \approx 0.35 < 0.75$ , it slides back.

$a_{\text{down}} = g(\sin 37° - \mu_k \cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2$

Speed at bottom: $v = \sqrt{2(3.6)(5.95)} = \sqrt{42.84} = 6.55 \text{ m/s}$

Energy check: Energy dissipated = $\mu_k mg\cos\theta \times 2d = 0.3(30)(0.8)(11.9) = 85.7$ J.

Initial KE = $\frac{1}{2}(3)(100) = 150$ J. Final KE = $150 - 85.7 = 64.3$ J. $v = \sqrt{2(64.3)/3} = 6.55$ m/s. ✓

Problem 3: Velocity-Dependent Friction on a Ramp

A 1 kg block starts from rest at the top of a $45°$ incline with velocity-dependent friction $f = 4v$ (SI units). Find:

(a) Terminal velocity: $v_T = \frac{mg\sin\theta}{b} = \frac{1(10)(\frac{\sqrt{2}}{2})}{4} = \frac{7.07}{4} = 1.77 \text{ m/s}$

(b) Velocity as a function of time: $v(t) = v_T(1 - e^{-bt/m}) = 1.77(1 - e^{-4t})$

(c) Distance traveled in 1 second: $x(t) = \int_0^t v(t')\,dt' = v_T\left[t + \frac{m}{b}e^{-bt/m}\right]_0^t = v_T\left(t + \frac{1}{4}e^{-4t} - \frac{1}{4}\right)$

$x(1) = 1.77\left(1 + 0.25e^{-4} - 0.25\right) = 1.77(1 + 0.00458 - 0.25) = 1.77(0.755) = 1.34 \text{ m}$

(d) Acceleration at $t = 0.5$ s: $a(t) = g\sin\theta \cdot e^{-bt/m} = 7.07 e^{-2} = 0.957 \text{ m/s}^2$

Workshop Summary

Common AP Pitfalls

Mistake	Correction
Wrong friction direction	Always opposes motion (or tendency of motion)
Forgetting $N \neq mg$ on inclines	$N = mg\cos\theta$ on an incline
Using $\mu_s$ when sliding	Use $\mu_k$ once the object is in motion
Ignoring the "does it move?" check	Always compare applied force to $f_{s,\max}$ first
Wrong sign on $a$ when going up vs down	Friction always opposes velocity direction

Next up: Part 7 — Review & Applications, consolidating everything with real-world contexts.

Application: Braking on a Hill

A car of mass $m$ is traveling at speed $v_0$ down a hill of angle $\theta$ . The brakes provide a constant friction force $F_b$ .

Equation of motion: $ma = mg\sin\theta - F_b$

Stopping distance: $d = \frac{mv_0^2}{2(F_b - mg\sin\theta)}$

This is valid only if $F_b > mg\sin\theta$ (brakes can actually slow the car).

Critical Braking Condition

The minimum braking force to stop on a hill: $F_b > mg\sin\theta$

If the tires provide braking through friction: $F_b = \mu_k N = \mu_k mg\cos\theta$

The car can stop only if: $\mu_k mg\cos\theta > mg\sin\theta \implies \mu_k > \tan\theta$

This is why steep icy hills ( $\mu_k$ small, $\theta$ large) are so dangerous.

With ABS (Maintaining Static Friction)

Modern ABS brakes prevent wheel lock, using $\mu_s$ instead of $\mu_k$ :

$d_{\text{ABS}} = \frac{v_0^2}{2g(\mu_s\cos\theta - \sin\theta)}$

Since $\mu_s > \mu_k$ , ABS gives shorter stopping distances.

Application: Drag on a Falling Object

A skydiver ( $m = 80$ kg) falls with quadratic drag $f = 0.25v^2$ .

Terminal velocity: $v_T = \sqrt{\frac{mg}{c}} = \sqrt{\frac{800}{0.25}} = \sqrt{3200} = 56.6 \text{ m/s} \approx 127 \text{ mph}$

Velocity as a function of time: $v(t) = 56.6\tanh\left(\frac{10t}{56.6}\right) = 56.6\tanh(0.177t)$

At $t = 5$ s: $v(5) = 56.6\tanh(0.883) = 56.6(0.708) = 40.1 \text{ m/s}$

At $t = 15$ s: $v(15) = 56.6\tanh(2.65) = 56.6(0.990) = 56.0 \text{ m/s}$

The skydiver is essentially at terminal velocity after about 15 seconds.

🎉 Topic Complete: Friction & Inclines

You've mastered the full AP Physics C treatment of friction and inclines:

Part	Topic	Status
1	Static vs kinetic friction	✅
2	Inclined planes (no friction)	✅
3	Inclined planes with friction	✅
4	Velocity-dependent friction (calculus)	✅
5	Systems on inclines	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Takeaway: On the AP exam, friction problems test your ability to (1) set up correct FBDs, (2) handle the static/kinetic transition, and (3) solve differential equations for velocity-dependent forces.

L = \frac{v_0^2}{2g\sin\theta} = \frac{100}{2(10)(\sin 45°)} = \frac{100}{14.14} = 7.07 \text{ m}

t = \frac{v_0}{g\sin\theta} = \frac{10}{7.07} = 1.41 \text{ s}

g∫0xsinθ(x′)dx′

\Delta x = \frac{v_0^2}{2|a|} = \frac{144}{16.8} = 8.57 \text{ m}

Δ x = \frac{v _{0}^{2}}{2∣ a ∣} = \frac{144}{16.8} = 8.57 m

m2g(sinθ2−μ2cosθ2)−m1g(sinθ1+μ1cosθ1)

The side with greater gravitational component tends to slide down.

a < 0

from your assumption, the system moves the other way (flip the friction directions and resolve).

Friction and Inclined Planes

Friction and Inclined Planes - Complete Interactive Lesson

Part 1: Static vs Kinetic Friction

Static vs Kinetic Friction

The Nature of Friction

Key Differences

The Normal Force

Free Body Diagrams with Friction

Worked Example

The Threshold of Motion

Part 1 Summary

Part 2: Frictionless Inclines

Inclined Planes (No Friction)

Setting Up Coordinates

Part 3: Inclines with Friction

Inclined Planes with Friction

Forces on a Rough Incline

Part 4: Velocity-Dependent Friction

Friction with Calculus (Velocity-Dependent)

Linear Drag: f=bvf = bvf=bv

Part 5: Systems on Inclines

Systems on Inclines

Atwood Machine on an Incline

Part 6: Problem-Solving Workshop

Problem-Solving Workshop

Problem-Solving Framework

Problem 2: Block Launched Up a Ramp

Part 7: Review & Applications

Review & Applications

Complete Topic Reference

The Transition: Calculus Perspective

Optimal Angle for Pulling

Newton's Second Law on an Incline

Energy Methods on Inclines

Worked Example: Launched Up a Ramp

Velocity as a Function of Position

Calculus Approach: Variable Angle Ramps

Example: Parabolic Ramp

Part 2 Summary

Critical Angle

Dynamics on Rough Inclines

Sliding Down

Sliding Up (Launched Upward)

Worked Example

Energy Methods with Friction

Round Trip Energy Loss

Part 3 Summary

Terminal Velocity

Solving the ODE

Quadratic Drag: f=cv2f = cv^2f=cv2

Terminal Velocity

Solving the ODE

Position by Integration

Key Behavior

Velocity-Dependent Friction on Inclines

Worked Example

Part 4 Summary

Free Body Diagrams

Solving for aaa and TTT

Two Blocks on Different Inclines

Checking Direction

Stacked Blocks on an Incline

Constraint Equations

Simple Constraint (Single String)

Pulley Ratio Constraint

Worked Example: Pulley System

Part 5 Summary

Solution

Problem 3: Velocity-Dependent Friction on a Ramp

Workshop Summary

Common AP Pitfalls

Application: Braking on a Hill

Critical Braking Condition

With ABS (Maintaining Static Friction)

Application: Drag on a Falling Object

🎉 Topic Complete: Friction & Inclines

Linear Drag: $f = bv$

Quadratic Drag: $f = cv^2$

Solving for $a$ and $T$