Inclined Planes and Atwood Machines — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Free-body diagram construction for inclines and Atwood machines, gravity component resolution on inclines, tension analysis, system acceleration derivation, connected object motion, and friction effects on both systems.

You should already know: Newton's first and second laws of motion, how to draw free-body diagrams, how to resolve vectors into perpendicular components.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Inclined Planes and Atwood Machines?

Inclined planes and Atwood machines are two core constrained and connected motion systems tested extensively in AP Physics 1 Unit 2: Dynamics, which contributes 12–18% of the total exam score. Both topics assess your ability to apply Newton’s second law ($\sum F=ma$) to non-standard coordinate systems and connected masses, so they appear regularly in both multiple choice (MCQ) and free response (FRQ) sections, often as standalone problems or embedded in larger multi-concept questions. An inclined plane is a sloped surface that constrains an object to move only along its surface, requiring rotation of the standard xy-coordinate system to simplify force calculations. An Atwood machine is a system of two or more masses connected by a massless, inextensible string running over a massless, frictionless pulley, used to analyze uniform acceleration of connected objects where tension is the same throughout the ideal string. Both systems emphasize consistent sign convention and correct force decomposition, foundational skills that transfer to almost every other unit in AP Physics 1.

2. Inclined Planes Without Friction

When an object moves along an inclined plane, the only force that requires decomposition is weight (gravity), because normal force acts perpendicular to the surface, and acceleration acts parallel to the surface. The standard, exam-preferred convention is to rotate your coordinate system so that the positive $x$-axis runs parallel to the incline (pointing down the incline by default, though you can flip it as long as you are consistent with signs) and the positive $y$-axis is perpendicular to the incline, pointing outward from the surface.

Weight $mg$ always acts straight down toward the center of the Earth. Using similar triangles, the angle between the weight vector and the negative $y$-axis equals the incline angle $\theta$ (measured from the horizontal). This gives the parallel (x) component of weight as $mg\sin \theta$, and the perpendicular (y) component as $mg\cos \theta$. Since the object cannot accelerate through the incline or fly off the surface, $\sum F_y=0$, so normal force $N=mg\cos \theta$. Along the $x$-axis, net force equals $mg\sin \theta =ma$, so acceleration simplifies to $a=g\sin \theta$ for a frictionless incline. This passes intuition checks: when $\theta =0^{\circ }$ (flat ground), $a=0$, and when $\theta =90^{\circ }$ (vertical), $a=g$ (free fall), which matches our expectations.

Worked Example

Problem: 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?

Solution:

1. Draw a free-body diagram with weight $mg$ downward, normal force $N$ perpendicular to the incline. Rotate coordinates so $x$ is parallel down the incline, $y$ is perpendicular outward.
2. Decompose weight into components: parallel (down incline) = $mg\sin \theta$, perpendicular (into incline) = $mg\cos \theta$.
3. Sum forces in the $y$-direction: $\sum F_y=N-mg\cos \theta =0$, so $N=mg\cos \theta$.
4. Plug in values: $N=(5\text{kg})(9.8{\text{m/s}}^2)(\cos 30^{\circ })\approx 42.4\text{N}$.
5. Sum forces in the $x$-direction: $\sum F_x=mg\sin \theta =ma$, so $a=g\sin \theta =9.8(0.5)=4.9{\text{m/s}}^2$ down the incline.

Exam tip: If you mix up $\sin$ and $\cos$ for components, test the edge case $\theta =0^{\circ }$ (flat ground) to catch the mistake in 2 seconds: acceleration should be zero, so parallel component must use $\sin \theta$.

3. Inclined Planes With Friction

Nearly all real inclined planes have friction, so we add a friction force to the free-body diagram that always opposes motion or the tendency for motion. Static friction acts on stationary objects, with maximum magnitude $f_s\le {\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. Kinetic friction acts on sliding objects, with fixed magnitude $f_k={\mu }_{k}N$, where ${\mu }_k<{\mu }_s$ for almost all material pairs.

Normal force does not change when friction is added: for rigid inclines with no acceleration perpendicular to the surface, $N=mg\cos \theta$ still holds. If an object slides down the incline, friction acts up the incline, so net force becomes $\sum F_x=mg\sin \theta -{\mu }_{k}mg\cos \theta =ma$, giving acceleration $a=g(\sin \theta -{\mu }_k\cos \theta )$. If an object slides up the incline, friction acts down, so $a=g(\sin \theta +{\mu }_k\cos \theta )$ (deceleration if up is positive). For a stationary object, static friction equals $mg\sin \theta$ up the incline, until $mg\sin \theta >{\mu }_{s}mg\cos \theta$, which simplifies to $\tan \theta >{\mu }_s$. This is the critical condition for sliding: any incline steeper than ${\theta }_{crit}=\arctan ({\mu }_s)$ will make the object slide.

Worked Example

Problem: 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?

Solution:

1. Draw the free-body diagram, calculate normal force: $N=mg\cos \theta =(2)(9.8)(\cos 25^{\circ })\approx 17.76\text{N}$.
2. Calculate maximum static friction: $f_{s,max}={\mu }_{s}N=0.5(17.76)=8.88\text{N}$.
3. Calculate the down-incline component of weight: $mg\sin \theta =(2)(9.8)(\sin 25^{\circ })\approx 8.28\text{N}$.
4. Compare forces: $8.28\text{N}<8.88\text{N}$, so static friction balances the down-incline force, and the box does not slide.
5. If the box did slide, acceleration would be $a=g(\sin \theta -{\mu }_k\cos \theta )\approx 0.6{\text{m/s}}^2$ down the incline.

Exam tip: Always check if the object is moving first: compare the down-incline weight component to maximum static friction before using kinetic friction to calculate acceleration.

4. Ideal Atwood Machines

An ideal Atwood machine is a connected system of two masses that relies on three standard AP Physics 1 assumptions: the string is massless and inextensible, and the pulley is massless and frictionless. An inextensible string means both masses have the same magnitude of acceleration: if one moves up 1 meter, the other moves down 1 meter, so acceleration magnitude $a$ is identical for both. A massless string and massless frictionless pulley mean tension $T$ is the same on both sides of the pulley, with no losses to pulley rotation or friction.

To solve for acceleration, we write Newton’s second law for each mass separately, then eliminate tension. Let $m_1>m_2$, so $m_1$ accelerates down and $m_2$ accelerates up. Align positive direction with acceleration: for $m_1$, down is positive: $m_{1}g-T=m_{1}a$. For $m_2$, up is positive: $T-m_{2}g=m_{2}a$. Add the two equations to eliminate $T$: $g(m_1-m_2)=a(m_1+m_2)$, so $a=g\frac{m_1-m_2}{m_1+m_2}$. Solving for tension gives $T=2g\frac{m_1{m}_2}{m_1+m_2}$. This passes intuition checks: equal masses give $a=0$ (equilibrium), and a zero-mass $m_2$ gives $a=g$ (free fall).

Worked Example

Problem: 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?

Solution:

1. The 7 kg mass is heavier, so it accelerates downward, and the 3 kg mass accelerates upward, with the same acceleration magnitude $a$.
2. Write Newton's second law for each mass: $7g-T=7a$ (7 kg, down positive) and $T-3g=3a$ (3 kg, up positive).
3. Add the two equations to eliminate $T$: $4g=10a$.
4. Solve for $a$: $a=\frac{4(9.8)}{10}=3.92{\text{m/s}}^2$.
5. Solve for tension: substitute $a$ back to get $T=3g+3a=3(9.8+3.92)\approx 41\text{N}$.

Exam tip: Always align your sign convention with the direction of acceleration: if you take down as positive for the heavier mass, up must be positive for the lighter mass to get the correct sign for acceleration.

5. Modified Atwood Machines (Incline + Atwood Combinations

One of the most common AP Physics 1 problems combines both topics: a modified Atwood machine with one mass on an incline and a second hanging mass connected by an ideal string over a pulley at the top of the incline. The same ideal assumptions apply: tension is uniform, and acceleration magnitude is the same for both masses.

The solution process follows the same steps as separate incline and Atwood problems: draw free-body diagrams for each mass, decompose weight for the mass on the incline, solve for normal force, add friction if applicable, write Newton's second law for each mass, then eliminate tension to solve for acceleration. A common mistake is guessing the wrong direction of acceleration, but this is easy to fix: if you get a negative acceleration after solving, it just means your initial direction guess was wrong, and you can flip the direction while keeping the magnitude.

Worked Example

Problem: 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.

Solution:

1. Compare the net pulling force: the weight of the hanging mass is $3g=29.4\text{N}$, and the down-incline weight component of the 4 kg block is $4g\sin 20^{\circ }\approx 13.4\text{N}$, so the hanging mass will accelerate downward, pulling the block up the incline.
2. Write Newton's second law for each mass: $m_{2}g-T=m_{2}a$ (hanging mass, down positive) and $T-m_{1}g\sin \theta =m_{1}a$ (block, up incline positive).
3. Add the equations to eliminate tension: $m_{2}g-m_{1}g\sin \theta =a(m_1+m_2)$.
4. Solve for $a$: $a=9.8\frac{3-4(0.342)}{4+3}\approx 2.3{\text{m/s}}^2$.
5. The positive acceleration confirms the direction: the 3 kg mass accelerates downward, and the 4 kg block accelerates up the incline.

Exam tip: If you get a negative acceleration after solving, don't redo all your algebra: just flip the direction of acceleration and keep the magnitude you calculated.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Swapping $\sin \theta$ and $\cos \theta$ for weight components on an incline, getting $N=mg\sin \theta$ and $a=g\cos \theta$. Why: Students forget the similar triangles relationship between the incline angle and force component angle. Correct move: After calculating components, test $\theta =0^{\circ }$ (flat ground): acceleration should be 0, so parallel component must be $\sin \theta$.
• 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 $mg$ for a mass on an incline, instead of $mg\cos \theta$. 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 $N=mg$ 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 $f_s={\mu }_{s}N$ 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.

7. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A box of mass $m$ is held at rest on a rough incline tilted at angle $\theta$ from horizontal. The coefficient of static friction between the box and incline is ${\mu }_s$. Which of the following expressions correctly gives the magnitude of the static friction force acting on the box? A) ${\mu }_{s}mg$ B) ${\mu }_{s}mg\sin \theta$ C) ${\mu }_{s}mg\cos \theta$ D) $mg\sin \theta$

Worked Solution: This question tests the distinction between maximum static friction and actual static friction for a stationary object. For the box to remain at rest, net force parallel to the incline must be zero. The down-incline component of weight is $mg\sin \theta$, so static friction must equal this magnitude to balance the force. Options A, B, and C are incorrect: A uses full weight instead of normal force, B uses incorrect components, and C gives the maximum static friction, not the actual friction acting on the stationary box. The correct answer is D.

Question 2 (Free Response)

A modified Atwood machine consists of a 5.0 kg block on a 30° incline, connected by an ideal massless string to a 4.0 kg hanging block over a massless, frictionless pulley at the top of the incline. The coefficient of kinetic friction between the block and the incline is 0.20. (a) Draw a free-body diagram for each mass, labeling all forces. (b) Calculate the magnitude of the acceleration of the system. (c) Is the tension in the string greater than, less than, or equal to the weight of the 4.0 kg hanging block? Justify your answer.

Worked Solution: (a) For the 5.0 kg block on the incline: gravitational force $m_{1}g$ downward, normal force $N$ perpendicular outward from the incline, tension $T$ up the incline, kinetic friction $f_k$ down the incline (opposing motion up the incline). For the 4.0 kg hanging block: gravitational force $m_{2}g$ downward, tension $T$ upward. (b) Normal force for the block: $N=m_{1}g\cos \theta$, so $f_k={\mu }_{k}N={\mu }_k{m}_{1}g\cos \theta$. Newton's second law: $m_{2}g-T=m_{2}a$ (hanging, down positive) and $T-m_{1}g\sin \theta -f_k=m_{1}a$ (block, up positive). Adding equations: $$a=g\frac{m_2-m_1\sin \theta -{\mu }_k{m}_1\cos \theta }{m_1+m_2}=9.8\frac{4-2.5-0.866}{9}\approx 0.69{\text{m/s}}^2$$ (c) Tension is less than the weight of the hanging block. The hanging block accelerates downward, so net force on it is downward: $\sum F=m_{2}g-T=m_{2}a>0$, which rearranges to $T<m_{2}g$.

Question 3 (Application / Real-World Style)

A civil engineer is designing a parked car rest area on a hillside, and needs to ensure that parked cars remain at rest even in icy conditions, when the coefficient of static friction between rubber and ice is 0.15. What is the maximum angle (from horizontal) that the hillside lot can be sloped to meet this requirement? Interpret your result in context.

Worked Solution: For a car to remain at rest, the maximum static friction must be at least equal to the down-slope component of the car's weight. At the maximum angle, these forces are equal: $$mg\sin {\theta }_{max}={\mu }_{s}mg\cos {\theta }_{max}$$ Cancel $mg$ from both sides: $\tan {\theta }_{max}={\mu }_s=0.15$. Take the arctangent: ${\theta }_{max}=\arctan (0.15)\approx 8.5^{\circ }$. This means the rest area can be sloped at no more than ~8.5 degrees from horizontal to prevent icy parked cars from sliding down the hill.

8. Quick Reference Cheatsheet

9. What's Next

Mastering inclined planes and Atwood machines builds the core skill of applying Newton's second law to constrained connected systems, which is required for all subsequent dynamics topics in AP Physics 1. Immediately next, you will apply these same force analysis and coordinate rotation skills to circular motion, where you align your coordinate system with centripetal acceleration just as you aligned it with incline motion here. Without getting comfortable with coordinate rotation and connected system force balancing, you will struggle to correctly identify net centripetal force for circular motion problems, a common high-weight FRQ topic. This topic also feeds into the broader study of work and energy, where you will calculate work done by gravity and friction on inclines, and use conservation of energy to solve connected system problems without algebra for acceleration.

• Newton's Laws of Motion
• Circular Motion and Gravitation
• Work, Energy, and Power
• Rotational Dynamics