Inclined planes, interacting blocks and centre of mass AP Physics 1 · Lesson 4 of 7
Slopes, blocks pushing on blocks, and the centre of mass. This lesson shows you how to tilt your axes on an incline, find the minimum coefficient of friction that stops something slipping, measure μ with a tilting board, and handle contact forces between blocks.
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This lesson covers forces on inclined planes, friction on slopes (including the minimal μs to prevent slipping and evaluating coefficients of friction), interacting blocks and centre of 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 on inclined planes
On a slope, tilt your axes: take x along the slope and y perpendicular to it. Then the normal force and friction lie along the axes, and only gravity needs to be split up. Gravity has a component mg sinθ down the slope and mg cosθ into the slope, where θ is the angle of the slope.
Perpendicular: FN = mg cosθ Frictionless: a = g sinθ down the slope
Key idea
Sense check the trig with extreme angles. At θ = 0 (flat), mg sinθ = 0 (no push down the slope) and FN = mg. At θ = 90° (vertical wall), mg sinθ = mg (free fall) and FN = 0. If your answer does the opposite, you have sin and cos swapped.
Worked example 1: frictionless ramp
A 3.0 kg cart is released on a frictionless 20° ramp. Find its acceleration and the normal force.
- a = g sinθ = 9.8 sin 20° = 3.35 m/s² down the ramp. The mass does not matter.
- FN = mg cosθ = 3.0 × 9.8 × cos 20° = 27.6 N.
Worked example 2: how far up?
A puck is given a speed of 6.0 m/s up a frictionless 15° ramp. How far along the ramp does it go before stopping?
- Up the ramp positive: a = −g sin 15° = −2.54 m/s².
- 0 = 6.0² + 2(−2.54)d, so d = 36/5.07 = 7.1 m.
Friction on slopes: minimal μs and measuring μ
Friction on a slope acts along the slope, opposite to the motion (or the tendency to move). Because FN = mg cosθ, friction is μmg cosθ.
| Situation | Friction points | Acceleration (down the slope positive) |
|---|---|---|
| Sliding down | Up the slope | a = g(sinθ − μk cosθ) |
| Sliding up (moving up, slowing) | Down the slope | a = −g(sinθ + μk cosθ) |
| At rest | Up the slope, Ff = mg sinθ | 0 |
Minimum μs to prevent slipping. A block stays at rest only if static friction can match mg sinθ: mg sinθ ≤ μsmg cosθ. The mass and g cancel:
μs ≥ tanθ
So the minimum coefficient of static friction to stop a block slipping on a slope of angle θ is tanθ, for any mass. Equally, if you tilt a ramp slowly, the block starts to slip at the critical angle where tanθc = μs.
Evaluating coefficients of friction with a tilting ramp
- Place the block on a board and raise one end slowly. Record the angle at which the block just starts to slide: μs = tanθc.
- Lower the board a little, tap the block to start it, and find the angle at which it slides down at constant speed: then mg sinθ = μkmg cosθ, so μk = tanθ.
- Repeat several times and average. The method needs only a protractor, and the result does not depend on the block's mass.
Worked example 1: sliding down with friction
A 5.0 kg box slides down a 30° ramp with μk = 0.25. Find its acceleration.
- a = g(sin 30° − 0.25 cos 30°) = 9.8(0.500 − 0.217) = 2.78 m/s² down the ramp.
Worked example 2: will it slip?
A phone rests on a book tilted at 25°. μs between phone and book is 0.40. Does it slip? What is the friction on the 0.20 kg phone?
- tan 25° = 0.466, which is greater than 0.40. Friction cannot hold it: the phone slips.
- At 20° instead, tan 20° = 0.364 < 0.40, so it would stay put, with friction = mg sin 20° = 0.20 × 9.8 × 0.342 = 0.67 N (not μsFN).
Worked example 3: pushed up the slope
A 2.0 kg block is launched up a 30° ramp at 5.0 m/s. μk = 0.20. How far up does it go?
- Moving up, friction acts down the slope with gravity's component: a = −g(sin 30° + 0.20 cos 30°) = −9.8(0.500 + 0.173) = −6.60 m/s².
- 0 = 25 + 2(−6.60)d, so d = 1.89 m. Then check whether it slides back: tan 30° = 0.577. It slides back down only if μs < 0.577.
Interacting blocks: contact forces and stacked blocks
When blocks touch or are stacked, they push on each other with contact forces (a normal force and sometimes friction). These come in Newton's third law pairs: equal and opposite, acting on different blocks. Use the same two-step method as in Lesson 2: system first for a, then one block alone for the force between them.
Worked example 1: blocks in contact
A 24 N force pushes block A (4.0 kg), which pushes block B (2.0 kg), on a frictionless floor. Find the force A exerts on B.
- System: a = 24/6.0 = 4.0 m/s².
- B alone: the only horizontal force on B is the contact force from A. FAB = mBa = 2.0 × 4.0 = 8.0 N.
- By Newton's third law, B pushes back on A with 8.0 N. Check A: 24 − 8.0 = 4.0 × 4.0. It agrees.
Key idea
If the same 24 N push is applied to B instead (from the right), the contact force becomes mAa = 16 N, because now it must accelerate the heavier block. The force between blocks depends on which block is being pushed.
For stacked blocks, the top block can only be accelerated by friction from the block below it. The largest acceleration static friction can give it is amax = μsmtopg/mtop = μsg. If the bottom block is pulled harder than that allows, the top block slips.
Worked example 2: stacked blocks
In the diagram, μs = 0.30 between the blocks and the floor is frictionless. What is the largest force F that can pull the bottom block without the top block slipping?
- The top block's largest acceleration: amax = μsg = 0.30 × 9.8 = 2.94 m/s².
- Both blocks move together with that acceleration: F = (6.0 + 2.0) × 2.94 = 23.5 N.
Centre of mass
The centre of mass of a system is the average position of its mass. For objects along a line:
xcm = (m1x1 + m2x2 + …)/(m1 + m2 + …)
For a uniform, symmetric object, the centre of mass is at its geometric centre. The centre of mass does not have to be inside the object: a ring's centre of mass is in the empty middle.
Why it matters for forces: a system moves as if all its mass were at the centre of mass and all the external forces acted there. Internal forces, however large, cannot change how the centre of mass moves. If the net external force on a system is zero, its centre of mass moves at constant velocity (or stays still).
Worked example 1: three masses
Find the centre of mass of the three masses in the diagram.
- xcm = (2.0 × 0 + 1.0 × 2.0 + 3.0 × 4.0)/(2.0 + 1.0 + 3.0) = 14/6.0 = 2.33 m from the left end.
Worked example 2: people on a boat
A 60 kg person stands at one end of a 4.0 m, 140 kg boat floating at rest. Ignore water resistance. The person walks to the other end. How far does the boat move?
- No net horizontal external force, so the system's centre of mass stays put.
- The person moves 4.0 m relative to the boat. If the boat moves d the other way, the person moves (4.0 − d) over the water.
- 60(4.0 − d) = 140d, so 240 = 200d and d = 1.2 m.
Key idea
On the AP exam, “system” questions often ask about the centre of mass: a firework that explodes in mid-air still has its centre of mass follow the original parabola, because the explosion forces are internal.
Key equations
| Weight components on a slope | mg sinθ (along), mg cosθ (into) |
|---|---|
| Normal force on a slope | FN = mg cosθ |
| Sliding down with friction | a = g(sinθ − μk cosθ) |
| Point of slipping | μs = tanθc |
| Centre of mass | xcm = Σmixi/Σmi |
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.
Hint
Only mg sinθ acts along the slope.
Worked solution
- a = g sin 25° = 9.8 × 0.423 = 4.14 m/s².
Hint
Perpendicular to the slope, FN balances mg cosθ.
Worked solution
- FN = mg cos 35° = 39.2 × 0.819 = 32.1 N.
Hint
FN = mg cosθ and, at rest, Ff = mg sinθ.
Worked solution
- As θ increases, cosθ decreases and sinθ increases.
- So the normal force decreases and the static friction increases. The answer is B.
Hint
Friction acts up the slope: a = g(sinθ − μk cosθ).
Worked solution
- a = 9.8(0.500 − 0.25 × 0.866) = 2.78 m/s².
Hint
At the point of slipping, mg sinθ = μsmg cosθ.
Worked solution
- μs = tan 22° = 0.404.
Hint
At the critical angle, μs = tanθc.
Worked solution
- μs = tan 31° = 0.60.
- The answer is A.
Hint
Does the critical angle depend on mass?
Worked solution
- tanθc = μs: the mass cancels.
- Same material means same μs, so the same angle. The answer is C.
Hint
Constant speed means the forces along the slope balance.
Worked solution
- mg sinθ = μkmg cosθ.
- μk = tan 18° = 0.325.
Hint
Moving up, both gravity's component and friction act down the slope.
Worked solution
- Size of deceleration: g(sin 30° + 0.20 cos 30°) = 6.60 m/s².
- d = v0²/(2a) = 25/13.2 = 1.89 m.
Hint
Kinetic friction opposes the velocity relative to the surface.
Worked solution
- The block moves up the ramp, so kinetic friction acts down the ramp. The answer is B.
Hint
System for a, then B alone.
Worked solution
- a = 30/5.0 = 6.0 m/s².
- FAB = mBa = 2.0 × 6.0 = 12 N.
Hint
The contact force accelerates whichever block is not pushed directly.
Worked solution
- Pushing A: the contact force accelerates B (light), so it is mBa.
- Pushing B: the contact force accelerates A (heavy), so it is mAa, with the same a. Larger. The answer is A.
Hint
Static friction is the only force that accelerates the top block, so its largest acceleration is μsg.
Worked solution
- amax = 0.30 × 9.8 = 2.94 m/s².
- F = (6.0 + 2.0) × 2.94 = 23.5 N.
Hint
Weighted average of the positions.
Worked solution
- xcm = (1.0 × 0 + 3.0 × 0.80)/4.0 = 0.600 m.
Hint
Are the explosion forces internal or external?
Worked solution
- The explosion forces are internal: they cancel in pairs for the system.
- Only gravity acts externally, as before, so the centre of mass keeps following the same parabola. The answer is B.
Hint
The centre of mass of student plus raft cannot move.
Worked solution
- 50(3.0 − d) = 100d.
- d = 150/150 = 1.00 m in the opposite direction.
Hint
a = g sinθ on a frictionless slope.
Worked solution
- The mass cancels: a = 9.8 sin 20° = 3.35 m/s² for both. The answer is C.
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.
Mathematical Routines (MR) style · 8 points · about 14 minutes
A child on a sled (combined mass M) starts from rest at the top of a straight, snowy hill inclined at angle θ to the horizontal. The coefficient of kinetic friction between the sled and the snow is μ. The hill is d long, measured along the slope.
- (a) Draw a free-body diagram for the sled and child while they slide down the hill.
- (b) Derive an expression for the speed of the sled at the bottom of the hill in terms of θ, μ, d and physical constants.
- (c) Calculate the speed at the bottom when θ = 25°, μ = 0.12 and d = 40 m.
- (d) A heavier adult rides the same sled down the same hill. A student says the adult will reach the bottom faster because a heavier object has a larger force pulling it down the slope. Is the student correct? Explain using your answer to (b).
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)
Dot with three forces: weight Mg straight down, normal force perpendicular to the slope, away from it, and kinetic friction up the slope.
Part (b)
Along the slope: Mg sinθ − μMg cosθ = Ma, so a = g(sinθ − μ cosθ). From rest, v² = 2ad, so v = √[2gd(sinθ − μ cosθ)].
Part (c)
a = 9.8(sin 25° − 0.12 cos 25°) = 3.08 m/s². v = √(2 × 3.08 × 40) = 15.7 m/s.
Part (d)
No. The force down the slope (Mg sinθ) is larger, but so is the friction (μMg cosθ) and so is the mass to be accelerated. In the expression from (b), M does not appear, so the speed at the bottom is the same (ignoring air resistance).
Your mark: 0 of 8
Experimental Design and Analysis (LAB) style · 7 points · about 15 minutes
Students want to find the coefficient of kinetic friction between a wooden block and a board. They release the block from rest on the board at different angles and use a motion detector to find its acceleration down the board each time.
| Angle θ | 15° | 20° | 25° | 30° | 35° |
|---|---|---|---|---|---|
| Acceleration (m/s²) | 1.12 | 1.97 | 2.81 | 3.63 | 4.42 |
- (a) Show that, for a block sliding down, a/cosθ = g tanθ − μg.
- (b) State which quantities to graph to give a straight line, and how to find μ from the line.
- (c) Use the data to find μ.
- (d) Explain why the gradient of the line is a useful check on the experiment.
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)
Along the slope: ma = mg sinθ − μmg cosθ, so a = g sinθ − μg cosθ. Dividing by cosθ: a/cosθ = g tanθ − μg.
Part (b)
Plot a/cosθ on the vertical axis against tanθ on the horizontal axis. The gradient is g and the vertical intercept is −μg, so μ = −(intercept)/g, or μ = −intercept/gradient.
Part (c)
tanθ: 0.268, 0.364, 0.466, 0.577, 0.700. a/cosθ: 1.16, 2.10, 3.10, 4.19, 5.40. The best-fit line has gradient ≈ 9.8 m/s² and intercept ≈ −1.5 m/s², so μ ≈ 1.5/9.8 ≈ 0.15.
Part (d)
The theory says the gradient should equal g = 9.8 m/s². If the gradient from the data is close to 9.8 m/s², it supports the model (constant μ, straight board, accurate angles). A gradient far from 9.8 would point to a systematic error, such as the angle being measured wrongly.
Your mark: 0 of 7
Common mistakes
- Swapping sin and cos on the slope. Check with θ = 0: the component along a flat surface must be zero.
- Writing FN = mg on a slope. It is mg cosθ (if no other force pushes into or away from the slope).
- Drawing ‘mg sinθ’ and ‘mg cosθ’ as extra forces on the free-body diagram. They are parts of the weight, not new forces.
- Using Ff = μsmg cosθ for a block at rest that is not about to slip. At rest, friction is mg sinθ.
- Forgetting that friction reverses direction when a block moving up the slope stops and slides back down.
- Thinking the contact force between two pushed blocks equals the push. It is only the part needed to accelerate the other block.
Frequently asked questions
Why tilt the axes on an incline?
So that the acceleration lies along one axis and the normal force and friction lie along the axes too. Only gravity then needs resolving. With horizontal and vertical axes you would have to resolve three forces and the acceleration.
Is the angle between mg and the perpendicular to the slope really θ?
Yes. The slope is turned through θ from the horizontal, and the perpendicular to the slope is turned through the same θ from the vertical. So the angle between the weight (vertical) and the perpendicular to the slope is θ.
Why does μs = tanθ not depend on mass?
Both the force pulling the block down the slope (mg sinθ) and the largest friction (μsmg cosθ) are proportional to the mass, so it cancels. A heavy and a light block of the same material slip at the same angle.
What happens if a block on a slope is pushed sideways or pulled by a rope?
Resolve the extra force along and perpendicular to the slope too. A rope pulling at an angle to the slope changes the normal force, just as an angled rope on level ground does (Lesson 3).
Does the centre of mass have to be inside an object?
No. For a ring, a horseshoe or a boomerang, the centre of mass is in empty space. A high jumper doing the Fosbury flop can clear the bar while their centre of mass passes under it.
How is centre of mass tested in AP Physics 1?
Mostly through systems: finding xcm for a few masses, explaining that internal forces cannot change the motion of the centre of mass, and choosing whether to treat a group of objects as one system.
When should I use the system method for interacting blocks?
Whenever the blocks move together with the same acceleration. Find a from the external forces and total mass, then isolate one block to find the force between them. If they might slip relative to each other, check the friction limit first.
Keep going
Related guides and practice:
- How to draw free-body diagrams
- AP Physics 1 forces summary
- Torque and centre of mass in equilibrium
- AP Physics 1 forces practice problems
All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers