Forces in systems, Atwood machines and friction AP Physics 1 · Lesson 2 of 7

Q: Why is the tension the same on both sides of the pulley?

In AP Physics 1 the pulley is usually ideal: massless and frictionless. A massless pulley needs no net torque to turn, so the string pulls equally on both sides. When the pulley has mass (Unit 5, rotation), the tensions are different.

Q: How do I know which way friction points on a block that is not moving?

Imagine the surface is frictionless and ask which way the block would slide. Static friction points the opposite way. On a table with a hanging mass pulling the block right, static friction on the block points left.

Q: Can μ be bigger than 1?

Yes. μ is a ratio of friction to normal force, and some surfaces (rubber on rough concrete, racing tyres) have μ above 1. Most surfaces in AP problems have μ between 0.1 and 0.8.

Q: Does friction depend on how fast the block is sliding?

In AP Physics 1, no: kinetic friction is μkFN at every speed. Real friction changes a little with speed, but that is outside the course.

Q: Do I have to use the system method?

No, but it is faster. You can always write Newton's second law for each object separately and solve the equations together. The system method gives a in one line; then you need one separate free-body diagram to find the tension.

Q: Why does a heavy box lurch forward when it finally starts to move?

Because μk < μs. The push that just beats maximum static friction is suddenly larger than the kinetic friction that replaces it, so there is a net force and the box accelerates.

Q: Is inertial vs gravitational mass on the AP Physics 1 exam?

Yes. The course framework says inertial and gravitational mass are experimentally equal, and the exam can ask you how each is measured, or why it means all objects fall with the same acceleration.

Q: What is an inertial balance?

A platform held by springs that oscillates side to side. Heavier loads oscillate more slowly. Because it moves horizontally, gravity plays no part, so it measures inertial mass. It works on the International Space Station.

Written by STEM-Tutor's tutors · Updated October 2026 · 5 topics, 17 practice problems, 2 free-response questions

Real AP problems rarely involve one object. This lesson shows you how to treat connected objects as a system, solve Atwood machines and blocks pulled over pulleys, and handle static and kinetic friction correctly, including the friction graph that appears on the exam.

On this page

In this lesson
Forces in systems
Atwood machines
Static vs kinetic friction (and graphing it)
Sliding blocks, with and without friction
Inertial vs gravitational mass
Key equations
Practice problems
Free-response practice
Common mistakes
FAQs
Keep going

In this lesson

This lesson covers forces in systems, Atwood machines, friction with coefficients (static vs kinetic, and the friction graph), sliding blocks with and without friction, and inertial vs gravitational mass:

Forces in systems
Atwood machines
Static vs kinetic friction
Sliding blocks with a pulley
Inertial vs gravitational mass

All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.

Forces in systems

A system is whatever object or group of objects you choose to analyse. Forces between objects inside the system are internal: they come in Newton's third law pairs that cancel when you add up the forces on the whole system. Only external forces change the motion of the system's centre of mass.

a_sys = ΣF_ext / m_total

Two blocks joined by a string, pulled by a 20 N force on a frictionless floor. Treating A and B together as one system, the string's pulls on A and on B are internal and cancel.

The two-step method for connected objects

Treat everything that moves together as one system. Find a from the net external force and the total mass.
Then draw a free-body diagram of one object on its own to find an internal force such as a tension or contact force.

Worked example: tension between two blocks

In the diagram, F = 20 N pulls B (3.0 kg), which is tied to A (2.0 kg). The floor is frictionless. Find the acceleration and the tension in the string.

Whole system: a = 20 / (2.0 + 3.0) = 4.0 m/s².
Block A on its own: the only horizontal force on it is the tension, so T = m_Aa = 2.0 × 4.0 = 8.0 N.
Check with B: 20 − T = 3.0 × 4.0 = 12, so T = 8.0 N. It agrees.

Key idea

The tension (8.0 N) is less than the pulling force (20 N) because the string only has to accelerate A. Pull on A's side instead and the tension would be 12 N, because the string would then accelerate B.

Atwood machines

An Atwood machine is two masses hanging from a string over a pulley. The pulley turns the string, so the two masses have accelerations of the same size in opposite directions. With an ideal (massless) string and an ideal (massless, frictionless) pulley, the tension is the same on both sides.

An Atwood machine: two masses on one string over a light, frictionless pulley. If m₂ > m₁, m₂ accelerates down and m₁ up, both with the same size of acceleration. The free-body diagram for m₂ has the weight longer than the tension.

Treat the two masses as one system and take “m₂ going down, m₁ going up” as the positive direction. The tensions are internal. The only external forces that drive the motion are the two weights, which pull opposite ways along the string:

a = (m₂ − m₁)g / (m₁ + m₂)

Then one free-body diagram gives the tension. For m₁: T − m₁g = m₁a, so T = m₁(g + a). Putting in a gives T = 2m₁m₂g / (m₁ + m₂).

Worked example: Atwood machine

Masses of 3.0 kg and 5.0 kg hang over an ideal pulley and are released from rest. Find the acceleration, the tension and the speed after 1.5 s.

a = (5.0 − 3.0)(9.8) / (3.0 + 5.0) = 19.6 / 8.0 = 2.45 m/s².
Tension from the 3.0 kg mass: T = 3.0(9.8 + 2.45) = 36.75 N, about 37 N.
Check with the 5.0 kg mass: 5.0(9.8 − 2.45) = 36.75 N. It agrees.
Speed: v = at = 2.45 × 1.5 = 3.7 m/s.

Sense check

Check limiting cases. If m₁ = m₂, a = 0: the masses balance. If m₁ is tiny compared with m₂, a → g: m₂ just falls freely. The tension is always between the two weights: more than m₁g (it accelerates m₁ up) and less than m₂g (m₂ accelerates down).

In the classic demo, the masses are nearly equal so the acceleration is small enough to time with a stopwatch. Measuring a for different mass differences and plotting a against (m₂ − m₁)/(m₁ + m₂) gives a straight line whose gradient is g.

Static vs kinetic friction (and graphing it)

There are two kinds of friction, and a block can only have one of them at a time.

	Static friction	Kinetic friction
When	The surfaces are not sliding over each other	The surfaces are sliding
Size	Whatever is needed to stop sliding, up to a maximum: F_f ≤ μ_sF_N	Fixed: F_f = μ_kF_N
Direction	Opposite to the way the object would slide without friction	Opposite to the velocity relative to the surface

For most surfaces μ_s > μ_k, which is why it is harder to get a heavy box moving than to keep it moving. The AP equation sheet writes both as |F_f| ≤ μ|F_N|: the equals sign applies to kinetic friction and to static friction only at the point of slipping.

Friction on a crate as a horizontal push slowly increases. Static friction matches the push (the 45° line) until it reaches its maximum, μ_sF_N = 6.0 N. The crate then starts to slide and friction drops to the kinetic value μ_kF_N = 4.2 N, which stays the same however hard you push.

Worked example 1: does it move?

A 20 kg crate sits on a floor with μ_s = 0.50 and μ_k = 0.35. A worker pushes horizontally with 80 N. What is the friction force?

Maximum static friction: μ_smg = 0.50 × 20 × 9.8 = 98 N.
The push (80 N) is less than 98 N, so the crate does not move. Static friction is only as big as it needs to be: 80 N, not 98 N.

Worked example 2: once it slides

The worker pushes harder until the crate just starts to move, then keeps pushing with that same force. Find the crate's acceleration.

The crate starts to move when the push just exceeds 98 N.
Once sliding, friction is kinetic: 0.35 × 20 × 9.8 = 68.6 N.
a = (98 − 68.6) / 20 = 1.47 m/s². The crate lurches forward because friction drops when it starts to slide.

Common trap

Never write F_f = μ_sF_N unless the question says the object is on the point of slipping (“just starts to move”, “maximum”, “minimum force”). Otherwise static friction is found from Newton's laws.

Sliding blocks, with and without friction

A favourite AP set-up is a block on a table connected over a pulley to a hanging block. With a light string and pulley, both move with the same size of acceleration and the tension is the same at both ends.

A block on a table pulled by a string over a pulley to a hanging block. The string makes them move together with the same size of acceleration, and the tension is the same all along a light string over a light, frictionless pulley.

Take the direction of motion (A to the right, B down) as positive. Treating A and B as one system, the external forces along the string are B's weight forward and friction on A backward:

a = (m_Bg − μ_km_Ag) / (m_A + m_B)

Without friction, set μ_k = 0: a = m_Bg / (m_A + m_B). Then find the tension from B alone: m_Bg − T = m_Ba.

Worked example 1: without friction

Block A (4.0 kg) is on a frictionless table and block B (2.0 kg) hangs. Find a and T.

a = (2.0 × 9.8) / (4.0 + 2.0) = 19.6/6.0 = 3.27 m/s².
From A: T = m_Aa = 4.0 × 3.27 = 13.1 N, less than B's weight of 19.6 N, as it must be for B to accelerate down.

Worked example 2: with friction

Now the table is rough, with μ_k = 0.20. Find the new acceleration and tension.

Friction on A: 0.20 × 4.0 × 9.8 = 7.84 N.
a = (19.6 − 7.84) / 6.0 = 1.96 m/s².
From B: T = m_B(g − a) = 2.0(9.8 − 1.96) = 15.7 N. The tension is larger than before, because B accelerates down less.

Worked example 3: will it start?

If A starts at rest and μ_s = 0.60, does the system start to move?

Maximum static friction on A: 0.60 × 4.0 × 9.8 = 23.5 N.
B's weight pulls with only 19.6 N, which is less than 23.5 N. The system stays at rest, with static friction of 19.6 N.

Inertial vs gravitational mass

Mass shows up in two different laws, and they measure two different things:

	Inertial mass	Gravitational mass
Defined by	How hard it is to accelerate: m = ΣF/a	How strongly gravity pulls on it: F_g = mg
How to measure it	Apply a known force, measure a (cart and force probe, or an inertial balance that oscillates horizontally)	Weigh it, or balance it against known masses on a pan balance
Works without gravity?	Yes, e.g. on the International Space Station	No

Experiments show these two masses are equal for every object to very high precision. That is why all objects fall with the same acceleration (ignoring air resistance): the gravitational force is proportional to the gravitational mass, the resistance to acceleration is proportional to the inertial mass, and they cancel, a = (m_gg)/m_i = g.

Worked example: measuring inertial mass

Astronauts on a space station push an object with a steady 6.0 N force and measure an acceleration of 1.5 m/s². A bathroom scale would read zero. What is the object's mass?

The scale reads zero because everything in orbit is in free fall, so gravitational methods do not work.
Inertial mass: m = F/a = 6.0/1.5 = 4.0 kg. On Earth it would weigh 4.0 × 9.8 = 39 N.

Key idea

On the Moon, an object has the same inertial and gravitational mass, but its weight is smaller because g is smaller there (about 1.6 m/s²). Kicking a heavy box hurts just as much on the Moon as on Earth.

Key equations

System acceleration	a = ΣF_ext/m_total
Atwood machine	a = (m₂ − m₁)g/(m₁ + m₂)
Static friction	F_f ≤ μ_sF_N
Kinetic friction	F_f = μ_kF_N
Table block and hanging block	a = (m_B − μ_km_A)g/(m_A + m_B)

Practice problems

17 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.

Q1Warm-up

Blocks of 4.0 kg and 6.0 kg are tied together by a light string on a frictionless floor. A 30 N force pulls the 6.0 kg block. What is the tension in the string?

Hint

Find a for the whole system first, then look at the block that only the string pulls.

Worked solution

a = 30/(4.0 + 6.0) = 3.00 m/s².
The string alone pulls the 4.0 kg block: T = 4.0 × 3.00 = 12.0 N.

Q2Warm-up

Two blocks are pushed together across a frictionless floor by a force on the left block. When analysing the two blocks as a single system, which force is internal?

Hint

Internal forces are exerted by one part of the system on another part.

Worked solution

The two blocks push on each other: a third-law pair, both inside the system.
All the others are exerted by things outside the system (the hand, Earth, the floor). The answer is B.

Q3Warm-up

An Atwood machine has masses of 3.0 kg and 5.0 kg on an ideal pulley. What is the size of their acceleration?

Hint

Net driving force is the difference in weights; total mass is the sum.

Worked solution

a = (m₂ − m₁)g/(m₁ + m₂) = (2.0)(9.8)/8.0 = 2.45 m/s².

Q4Exam-style

Masses of 2.0 kg and 2.5 kg hang on either side of an ideal pulley. What is the tension in the string?

Hint

Find a, then use Newton's second law for the 2.0 kg mass, which accelerates upward.

Worked solution

a = (0.5)(9.8)/4.5 = 1.09 m/s².
T − m₁g = m₁a, so T = 2.0(9.8 + 1.09) = 21.8 N.

Q5Warm-up

In an Atwood machine with masses m and 3m, how does the tension T in the string compare with the weights?

Hint

Which mass accelerates up, and which down?

Worked solution

The lighter mass m accelerates up, so T > mg.
The heavier mass 3m accelerates down, so T < 3mg. The answer is C (in fact T = 1.5mg).

Q6Warm-up

A 10 kg box rests on a floor with μ_s = 0.40. A horizontal push of 25 N does not move it. What is the friction force on the box?

Hint

Static friction is only as large as it needs to be.

Worked solution

Maximum static friction is 0.40 × 10 × 9.8 = 39.2 N, but the push is only 25 N.
The box stays at rest, so friction balances the push exactly: 25 N. The answer is A.

Q7Warm-up

What is the smallest horizontal force that will start a 30 kg crate moving on a floor where μ_s = 0.45?

Hint

At the point of slipping, static friction is at its maximum.

Worked solution

F = μ_smg = 0.45 × 30 × 9.8 = 132 N (just over this value starts it moving).

Q8Exam-style

A force sensor pulls a block across a table with a slowly increasing force. The block starts to slide when the sensor reads 6.0 N, and then slides at constant velocity with the sensor reading 4.5 N. The block's weight is 15 N. What are μ_s and μ_k?

Hint

μ = friction / normal force, and F_N = 15 N.

Worked solution

Starting to slide: 6.0 = μ_s(15), so μ_s = 0.40.
Constant velocity: friction = 4.5 N, so μ_k = 4.5/15 = 0.30. The answer is A.

Q9Warm-up

Block A (4.0 kg) sits on a frictionless table, connected over an ideal pulley to block B (2.0 kg), which hangs. What is the acceleration of the blocks?

Hint

Only B's weight drives the system; the total mass is accelerated.

Worked solution

a = m_Bg/(m_A + m_B) = 19.6/6.0 = 3.27 m/s².

Q10Exam-style

A 5.0 kg block on a table (μ_k = 0.25) is pulled by a string over a pulley to a hanging 3.0 kg block. The blocks are moving. What is their acceleration?

Hint

Driving force: hanging weight. Opposing force: kinetic friction on the table block.

Worked solution

Friction = 0.25 × 5.0 × 9.8 = 12.2 N; hanging weight = 29.4 N.
a = (29.4 − 12.2) / 8.0 = 2.14 m/s².

Q11Exam-style

For the same system (5.0 kg on a table with μ_k = 0.25, 3.0 kg hanging), what is the tension in the string?

Hint

Use the hanging block's free-body diagram: weight down, tension up, accelerating down.

Worked solution

m_Bg − T = m_Ba, so T = m_B(g − a).
T = 3.0(9.8 − 2.14) = 23.0 N.

Q12Exam-style

A block on a rough table is tied over a pulley to a hanging mass and the system accelerates. If the hanging mass is increased, what happens to the friction force on the table block?

Hint

What does kinetic friction depend on?

Worked solution

Kinetic friction is μ_kF_N, and F_N for the table block is its own weight.
Neither changes, so friction stays the same, even though the acceleration increases. The answer is C.

Q13Exam-style

A 6.0 kg block on a table with μ_s = 0.50 is tied over a pulley to a hanging mass. What is the largest hanging mass for which the system stays at rest?

Hint

At the limit, the hanging weight equals the maximum static friction.

Worked solution

m_hangg = μ_sm_Ag; g cancels.
m_hang = 0.50 × 6.0 = 3.00 kg.

Q14Warm-up

Which experiment measures an object's inertial mass rather than its gravitational mass?

Hint

Inertial mass is about resistance to acceleration, not the pull of gravity.

Worked solution

The spring scale, pan balance and vertical spring all depend on gravity's pull, so they measure gravitational mass.
m = F/a from a horizontal track does not involve gravity. The answer is C.

Q15Warm-up

Why do a bowling ball and a tennis ball dropped together (with negligible air resistance) hit the ground at the same time?

Hint

Compare F_g = mg with a = F/m.

Worked solution

The bowling ball has a larger weight but is equally harder to accelerate.
a = mg/m = g for both, because inertial and gravitational mass are equal. The answer is B.

Q16Warm-up

A 25 kg box is pulled across a floor (μ_k = 0.30) by a horizontal rope with tension 120 N. Find its acceleration.

Hint

Kinetic friction = μ_kmg.

Worked solution

Friction = 0.30 × 25 × 9.8 = 73.5 N.
a = (120 − 73.5)/25 = 1.86 m/s².

Q17Challenge

Blocks of 2.0 kg (front) and 3.0 kg (back) are tied together and pulled forward across a rough floor (μ_k = 0.20 for both) by a 25 N horizontal force on the 2.0 kg block. What is the tension in the string between them?

Hint

System first for a. Then the back block alone: the string pulls it forward, friction pulls it back.

Worked solution

Total friction = 0.20 × 5.0 × 9.8 = 9.8 N; a = (25 − 9.8)/5.0 = 3.04 m/s².
Back block: T − 0.20(3.0)(9.8) = 3.0 × 3.04, so T = 9.12 + 5.88 = 15.0 N.

Free-response practice

Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.

Qualitative/Quantitative Translation (QQT) style · 8 points · about 15 minutes

Block A of mass M sits on a horizontal table. The coefficient of kinetic friction between A and the table is μ. A light string attached to A passes over an ideal pulley at the edge of the table to block B of mass m, which hangs. Block B is released and the blocks move, with B falling.

(a) Draw free-body diagrams for block A and for block B. Draw each force as an arrow starting on a dot.
(b) Derive an expression for the acceleration of the blocks in terms of M, m, μ and physical constants, as appropriate.
(c) Block A is replaced with a block of mass 2M made of the same material. Without using equations, explain whether the acceleration increases, decreases or stays the same.
(d) Show that your expression in part (b) agrees with your reasoning in part (c).

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

A: weight Mg down, normal force up (equal lengths), tension T to the right, kinetic friction to the left (shorter than T, since A speeds up). B: weight mg down, tension T up, with mg longer than T.

1 point for: A has four correct forces: weight, normal, tension toward the pulley, friction away from it
1 point for: B has tension up and weight down, with weight longer than tension

Part (b)

System of A and B: the external forces along the string are mg forward and μMg backward. (M + m)a = mg − μMg, so a = (m − μM)g / (M + m).

1 point for: a correct Newton's second law equation for A, B or the system
1 point for: friction written as μMg
1 point for: correct final expression

Part (c)

Decreases. The heavier block is pressed harder into the table, so friction (which opposes the motion) is larger, while the hanging weight that drives the motion is unchanged. There is also more total mass to accelerate. Smaller net force and larger mass both reduce the acceleration.

1 point for: states the acceleration decreases
1 point for: reasoning: friction increases (larger normal force) while the driving force is unchanged, and/or more mass to accelerate

Part (d)

With 2M in place of M, a = (m − 2μM)g/(2M + m). The numerator is smaller and the denominator is larger, so a is smaller. This agrees with part (c).

1 point for: substitutes 2M and explains that the numerator decreases and the denominator increases, consistent with (c)

Your mark: 0 of 8

Translation Between Representations (TBR) style · 5 points · about 10 minutes

A student pulls a 2.0 kg wooden block across a level table with a force sensor. The force increases slowly from zero. The graph shows the sensor reading against time.

The force sensor reading. After 3.0 s the block slides at constant velocity.

(a) Describe the motion of the block from t = 0 to t = 3.0 s, and from t = 3.0 s to t = 6.0 s. Explain how you know from the graph.
(b) Calculate the coefficients of static and kinetic friction.
(c) The student repeats the experiment with a 1.0 kg mass stacked on top of the block. Sketch or describe how the graph would change.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

0 to 3.0 s: the block is at rest, held by static friction that grows to match the pull. 3.0 to 6.0 s: the block slides at constant velocity, so the pull (5.9 N) equals kinetic friction. The sudden drop at 3.0 s shows the switch from maximum static to kinetic friction.

1 point for: at rest with static friction matching the pull before 3.0 s
1 point for: constant velocity after 3.0 s, so pull equals kinetic friction

Part (b)

F_N = mg = 2.0 × 9.8 = 19.6 N. μ_s = 7.8/19.6 = 0.40. μ_k = 5.9/19.6 = 0.30.

1 point for: uses F_N = mg = 19.6 N
1 point for: μ_s = 0.40 and μ_k = 0.30

Part (c)

The same shape, but the peak and the flat section are both 1.5 times higher (weight goes from 19.6 N to 29.4 N): the peak is about 11.8 N and the flat part about 8.8 N. If the force still increases at the same rate, the peak is reached later, at about 4.5 s.

1 point for: same shape, with both levels raised in proportion to the normal force (about 11.8 N and 8.8 N)

Your mark: 0 of 5

Common mistakes

Setting the tension equal to the hanging weight. If the hanging block accelerates, T is less than its weight.
Writing F_f = μ_sF_N for an object that is not on the point of slipping. Static friction is usually less than its maximum.
Including internal forces (like the string tension) when finding the acceleration of the whole system.
Using the mass of only one block in a = ΣF/m for a connected system. Both blocks are accelerated, so use the total mass.
Thinking kinetic friction gets bigger when the object is pulled harder or moves faster. It is μ_kF_N, whatever the pull or speed.
Forgetting to check whether a system moves at all before using kinetic friction.
Saying heavier objects fall faster because gravity pulls harder. It does pull harder, but they also have more inertia, so a = g for all.

Frequently asked questions

Why is the tension the same on both sides of the pulley?

In AP Physics 1 the pulley is usually ideal: massless and frictionless. A massless pulley needs no net torque to turn, so the string pulls equally on both sides. When the pulley has mass (Unit 5, rotation), the tensions are different.

How do I know which way friction points on a block that is not moving?

Imagine the surface is frictionless and ask which way the block would slide. Static friction points the opposite way. On a table with a hanging mass pulling the block right, static friction on the block points left.

Can μ be bigger than 1?

Yes. μ is a ratio of friction to normal force, and some surfaces (rubber on rough concrete, racing tyres) have μ above 1. Most surfaces in AP problems have μ between 0.1 and 0.8.

Does friction depend on how fast the block is sliding?

In AP Physics 1, no: kinetic friction is μ_kF_N at every speed. Real friction changes a little with speed, but that is outside the course.

Do I have to use the system method?

No, but it is faster. You can always write Newton's second law for each object separately and solve the equations together. The system method gives a in one line; then you need one separate free-body diagram to find the tension.

Why does a heavy box lurch forward when it finally starts to move?

Because μ_k < μ_s. The push that just beats maximum static friction is suddenly larger than the kinetic friction that replaces it, so there is a net force and the box accelerates.

Is inertial vs gravitational mass on the AP Physics 1 exam?

Yes. The course framework says inertial and gravitational mass are experimentally equal, and the exam can ask you how each is measured, or why it means all objects fall with the same acceleration.

What is an inertial balance?

A platform held by springs that oscillates side to side. Heavier loads oscillate more slowly. Because it moves horizontally, gravity plays no part, so it measures inertial mass. It works on the International Space Station.

Keep going

Related guides and practice:

All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers