Inclined Planes and Atwood Machines

AP Physics 1· AP Physics 1 CED — Dynamics· 14 min read

1. Frictionless Inclined Planes★★☆☆☆⏱ 3 min

An inclined plane constrains motion to only be parallel to its sloped surface. To simplify calculations, we rotate the coordinate system so that the positive $x$ -axis is parallel to the incline (default pointing down) and the positive $y$ -axis is perpendicular to the surface, pointing outward.

Weight $m g$ acts straight down, so we decompose it into parallel and perpendicular components. Using similar triangles, the angle between the weight vector and the negative $y$ -axis equals the incline angle $θ$ , giving the components below:

Parallel (x-direction, down incline): $m g sin θ$
Perpendicular (y-direction, into incline): $m g cos θ$

Since there is no acceleration perpendicular to the incline, net force in $y$ is zero, so normal force $N = m g cos θ$ . For frictionless inclines, net parallel force gives $a = g sin θ$ .

📐 Worked Example

A 5 kg crate slides down a frictionless incline that makes a 30° angle with the horizontal. What is the magnitude of the normal force on the crate, and what is the crate's acceleration?

1
Draw a free-body diagram with weight $m g$ downward, normal force $N$ perpendicular to the incline. Rotate coordinates to align with the incline.
2
Decompose weight into components:
3
$F_{\parallel} = mg \sin\theta, \quad F_{\perp} = mg \cos\theta$
4
Sum forces in the $y$ -direction (net force = 0):
5
$\sum F_y = N - mg \cos\theta = 0 \implies N = mg \cos\theta$
6
Plug in values: $m = 5 kg$ , $g = 9.8 m/s^{2}$ , $cos 3 0^{\circ} \approx 0.866$ :
7
$N = (5)(9.8)(0.866) \approx 42.4\ \text{N}$
8
Sum forces in the $x$ -direction to find acceleration:
9
$\sum F_x = mg \sin\theta = ma \implies a = g \sin\theta = 9.8(0.5) = 4.9\ \text{m/s}^2$
10
Final answer: Normal force = ~42.4 N, acceleration = 4.9 m/s² down the incline.

2. Inclined Planes with Friction★★★☆☆⏱ 4 min

Friction always opposes motion (or the tendency for motion). Static friction acts on stationary objects with maximum magnitude $f_{s} \leq μ_{s} N$ , while kinetic friction acts on sliding objects with fixed magnitude $f_{k} = μ_{k} N$ , where $μ_{k} < μ_{s}$ .

Normal force does not change with friction: $N = m g cos θ$ still holds because there is no acceleration perpendicular to the incline. For an object sliding down, friction acts up the incline, giving acceleration $a = g (sin θ - μ_{k} cos θ)$ . An object will start sliding if $tan θ > μ_{s}$ , the critical condition for sliding.

📐 Worked Example

A 2 kg box is at rest on an incline angled 25° to the horizontal. The coefficient of static friction is 0.5, and coefficient of kinetic friction is 0.4. Does the box slide, and if so, what is its acceleration?

1
Calculate normal force first:
2
$N = mg \cos\theta = (2)(9.8)(\cos 25^\circ) \approx 17.76\ \text{N}$
3
Calculate maximum static friction:
4
$f_{s,max} = \mu_s N = 0.5(17.76) = 8.88\ \text{N}$
5
Calculate the down-incline component of weight:
6
$mg \sin\theta = (2)(9.8)(\sin 25^\circ) \approx 8.28\ \text{N}$
7
Compare the forces: $8.28 N < 8.88 N$ , so static friction balances the down-incline force.
8
Conclusion: The box does not slide. If it did slide, acceleration would be approximately $0.6 m/s^{2}$ down the incline.

3. Ideal Atwood Machines★★★☆☆⏱ 3 min

📘 Definition

Ideal Atwood Machine

A system of two masses connected by a massless, inextensible string over a massless, frictionless pulley. Key assumptions: same magnitude of acceleration for both masses, same tension throughout the string.

Example:

Used to test application of Newton's second law to connected accelerating systems, a common AP Physics 1 problem type

To solve, write Newton's second law for each mass separately, then eliminate tension. If $m_{1} > m_{2}$ , $m_{1}$ accelerates down, $m_{2}$ accelerates up. Align positive direction with acceleration, add the equations to solve for acceleration:

a = g \frac{m_1 - m_2}{m_1 + m_2}, \quad T = 2g \frac{m_1 m_2}{m_1 + m_2}

📐 Worked Example

An ideal Atwood machine has a 7 kg mass and a 3 kg mass connected over a massless pulley. What is the magnitude of acceleration of the system, and what is the tension in the string?

1
The 7 kg mass accelerates downward, the 3 kg mass accelerates upward, both with the same acceleration magnitude $a$ .
2
Write Newton's second law for each mass, aligning positive direction with acceleration:
3
$7g - T = 7a \quad (\text{7 kg mass, down positive})$
4
$T - 3g = 3a \quad (\text{3 kg mass, up positive})$
5
Add the equations to eliminate tension:
6
$4g = 10a$
7
Solve for acceleration:
8
$a = \frac{4(9.8)}{10} = 3.92\ \text{m/s}^2$
9
Substitute back to find tension:
10
$T = 3g + 3a = 3(9.8 + 3.92) \approx 41\ \text{N}$

4. Modified Atwood Machines (Incline + Atwood Combinations)★★★★☆⏱ 4 min

The most common AP exam problem combines both topics: one mass on an incline connected to a hanging mass over a pulley, called a modified Atwood machine. The same ideal assumptions apply: uniform tension, equal acceleration magnitude for both masses.

📐 Worked Example

A modified Atwood machine has a 4 kg block on a 20° frictionless incline connected by an ideal string to a 3 kg hanging block over a pulley at the top of the incline. Find the acceleration magnitude and direction of the system.

1
Compare net pulling forces: Hanging weight = $3 g = 29.4 N$ , down-incline weight component of the 4 kg block = $4 g sin 2 0^{\circ} \approx 13.4 N$ . The hanging mass will accelerate downward, pulling the block up the incline.
2
Write Newton's second law for each mass:
3
$m_2 g - T = m_2 a \quad (\text{hanging mass, down positive})$
4
$T - m_1 g \sin\theta = m_1 a \quad (\text{block, up incline positive})$
5
Add equations to eliminate tension:
6
$m_2 g - m_1 g \sin\theta = a(m_1 + m_2)$
7
Solve for acceleration:
8
$a = 9.8 \frac{3 - 4(0.342)}{4+3} \approx 2.3\ \text{m/s}^2$
9
Positive acceleration confirms direction: 3 kg mass accelerates downward, 4 kg block accelerates up the incline.

✓ Quick check

Test your understanding of static friction:

A box of mass $m$ is held at rest on a rough incline tilted at angle $θ$ from horizontal. The coefficient of static friction between the box and incline is $μ_{s}$ . Which of the following expressions correctly gives the magnitude of the static friction force acting on the box?
- A) $μ_{s} m g$
- B) $μ_{s} m g sin θ$
- C) $μ_{s} m g cos θ$
- D) $m g sin θ$
Reveal answer
D —
Correct: For a stationary object, static friction balances the net down-incline force, so it equals $m g sin θ$ . Option C gives maximum static friction, not the actual friction acting on the stationary box.

5. Common Pitfalls

Wrong move:

Swapping $sin θ$ and $cos θ$ for weight components, getting $N = m g sin θ$ and $a = g cos θ$ .

Why:

Students forget the similar triangles relationship between the incline angle and force component angle.

Correct move:

After calculating components, test $θ = 0^{\circ}$ (flat ground): acceleration should be 0, so the parallel component must be $sin θ$ .

Wrong move:

Leaving friction pointing up the incline when an object slides up the incline, leading to too small a deceleration.

Why:

Students default to friction opposing the weight component, not the direction of motion.

Correct move:

Always set friction direction opposite to the object's velocity, not opposite to the weight component.

Wrong move:

Assuming tension equals the weight of one mass in an accelerating Atwood machine, so $T = m_{1} g$ .

Why:

Students confuse zero-acceleration equilibrium with accelerating motion.

Correct move:

Always write a separate Newton's second law equation for each mass, never assume tension equals weight unless acceleration is zero.

Wrong move:

Calculating normal force as $m g$ for a mass on an incline, instead of $m g cos θ$ .

Why:

Students default to the flat ground rule, forgetting only the perpendicular weight component presses against the incline.

Correct move:

Always solve for normal force from perpendicular force balance, never assume $N = m g$ for inclines.

Wrong move:

Using different acceleration magnitudes for the two masses in an ideal Atwood system.

Why:

Students forget an inextensible string requires equal displacement for both masses, so equal acceleration magnitude.

Correct move:

Always use the same variable $a$ for acceleration magnitude for both masses in connected ideal string systems.

Wrong move:

Using maximum static friction as the actual friction force for a stationary object on an incline.

Why:

Students memorize $f_{s} = μ_{s} N$ and forget this is only the maximum value.

Correct move:

For stationary objects, actual static friction equals the net force it is opposing, not the maximum value.

6. Quick Reference Cheatsheet

Category	Formula/Expression	Key Note
Incline weight components	$F_{∥} = m g sin θ$ , $F_{⊥} = m g cos θ$	$F_{∥}$ points down incline, $θ$ from horizontal
Incline normal force (no perpendicular acceleration)	$N = m g cos θ$	Never use $N = m g$ for inclines
Frictionless incline acceleration	$a = g sin θ$	Acceleration points down the incline
Kinetic friction (object sliding down incline)	$a = g (sin θ - μ_{k} cos θ)$	Friction acts opposite to motion
Critical sliding angle	$θ_{cr i t} = arctan (μ_{s})$	Object slides if $θ > θ_{cr i t}$
Ideal Atwood acceleration	$a = g \frac{m _{1} - m _{2}}{m _{1} + m _{2}}$	$m_{1} > m_{2}$ , $m_{1}$ accelerates down
Ideal Atwood tension	$T = 2 g \frac{m _{1} m _{2}}{m _{1} + m _{2}}$	Same tension on both sides of the pulley

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Modified Atwood acceleration problem
2022 · FRQ
Friction on incline force analysis
2021 · MCQ
Atwood tension comparison question

What's Next

Inclined planes and Atwood machines are foundational applications of Newton's second law, and the problem-solving skills you learned here transfer to every other unit in AP Physics 1. Coordinate rotation for force resolution is the same technique you will use for circular motion, where you resolve forces into radial and tangential components to find centripetal acceleration. Connected system force analysis also prepares you for work and energy problems, where you can use energy conservation to cross-check acceleration calculations for these systems. Mastering consistent sign convention and force decomposition here will help you avoid common mistakes on all free-response questions across the exam.