- State Newton's three laws and tell action–reaction pairs from balanced forces
- Apply F = ma along and across the motion, including inclines with friction
- Solve connected-body problems (strings, pulleys, the Atwood machine) with the whole-system and isolation methods
- Explain overweight and weightlessness in lifts and find the instantaneous acceleration when a string or spring is cut
Stand on a bathroom scale in a lift: as the lift starts upward the reading jumps, and as it stops at the top the reading drops, although your mass has not changed. Why? The answer is Newton's second law, the most used formula of the CSCA physics test. In the lesson «Forces and equilibrium» the net force was zero; now it is not, and the body accelerates.
Newton's three laws
惯性定律)A body stays at rest or keeps moving in a straight line at constant speed unless a net force acts on it. Inertia is the tendency of a body to keep its velocity, and its measure is the mass. So a force is needed not to keep a body moving, but to change its velocity.
- ΣFthe net (resultant) force of all forces on the body (
合外力), N - mmass, kg
- aacceleration, m/s², always in the direction of ΣF
Newton's second law (牛顿第二定律). It is a vector law and holds at every instant: when the net force changes, the acceleration changes at the same moment. 1 N = 1 kg · m/s².
1) A 1200 kg car brakes from 20 m/s to rest in 4 s. Find the braking force.
2) Forces of 6 N and 8 N at right angles act on a 2 kg body on a smooth horizontal table. Find a.
3) A 5 kg body is lifted by a rope with a tension of 60 N (g = 10 m/s²). Find a.
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2) ΣF = √(36 + 64) = 10 N, a = 10/2 = 5 m/s², along the resultant.
3) ΣF = T − mg = 60 − 50 = 10 N, a = 10/5 = 2 m/s², upward. The answer 60/5 = 12 m/s² forgets gravity.
作用力与反作用力)When body A exerts a force on body B, B exerts on A a force equal in size and opposite in direction, along the same line. The two forces act on different bodies, appear and disappear together and are of the same kind (both elastic, both friction or both gravitational).
| Action–reaction pair | Balanced forces | |
|---|---|---|
| Act on | two different bodies | one body |
| Kind | always the same | may differ (gravity and normal force) |
| Do they cancel? | never (different bodies) | yes, the net force is zero |
| Example | book presses on table ↔ table pushes book | Earth pulls book (mg) and table pushes book (N) |
A horse pulls a cart forward, and the cart speeds up. Which statement is correct?
A) The horse pulls the cart harder than the cart pulls the horse.
B) At every moment the cart's pull on the horse is equal and opposite to the horse's pull on the cart.
C) The two pulls cancel, so the cart cannot accelerate.
D) The cart pulls the horse only after it starts moving.
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Using F = ma along and across the motion
- 1Choose the body and draw its forces
As in «Forces and equilibrium»: gravity, normal forces, friction, tensions, springs.
- 2Axes: along a and across a
Put one axis along the acceleration (along the motion or along the incline) and the other perpendicular to it.
- 3Across the motion: ΣF = 0
There is no acceleration across the motion, so these forces balance. This gives N, and then the friction μN.
- 4Along the motion: ΣF = ma
Forces in the direction of a are positive, those against it negative.
- 5Link to kinematics
If a distance, speed or time is asked, finish with the formulas of «Kinematics: displacement, velocity, acceleration, free fall».
A 10 kg sledge is pulled along level snow by a rope at 37° above the horizontal with a tension of 50 N; μ = 0.2 (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8). Find the acceleration.
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Along: ma = 50 · 0.8 − 14 = 26 N ⇒ a = 2.6 m/s².
Trap: taking N = mg gives f = 20 N and a = 2 m/s².
- θthe angle of the incline
- μthe coefficient of kinetic friction
- athe size of the acceleration; in both cases it points down the slope
A block on a rough incline with no other forces. Sliding down, friction points up the slope and is subtracted; sliding up, both the gravity component and friction point down the slope and add up. If μ ≥ tan θ, a block at rest does not start to slide.
A block is launched up a 37° incline at 8 m/s; μ = 0.25 (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8).
1) Find the acceleration on the way up and the distance it travels up the slope.
2) Does it slide back? If so, with what acceleration?
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2) tan 37° = 0.75 > μ = 0.25, so friction cannot hold it: it slides back with a = 10 · (0.6 − 0.2) = 4 m/s². The trip down is slower than the trip up.
Connected bodies: the whole-system and isolation methods
Bodies joined by strings or pressed together move with the same acceleration (as long as the strings stay taut). First treat them as one system (整体法): internal forces such as the tension between them cancel out, and a = (external force) / (total mass). Then isolate one body (隔离法) to find the internal force.
- Fthe external pull on the front body m₁, N
- m₂the mass of the body pulled by the string (behind), kg
- Tthe tension in the string between the bodies, N
Two blocks on a smooth horizontal floor, F pulling the front block m₁, a string to m₂. The internal force is shared in proportion to the mass that is pulled. T stays the same if both blocks have the same μ, and also along an incline.
1) Blocks of 3 kg (in front) and 1 kg are joined by a string on a smooth floor; a horizontal force of 12 N pulls the front block. Find a and T.
2) A 4 kg block on a smooth table is joined by a string over a smooth pulley at the edge to a hanging 1 kg block (g = 10 m/s²). Find a and T.
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2) The weight of the hanging block drives the system: a = m₂g / (m₁ + m₂) = 10/5 = 2 m/s². Isolate the table block: T = 4 · 2 = 8 N, less than the hanging weight of 10 N; otherwise the hanging block could not speed up downward.
- m₁ > m₂the masses at the two ends of a light string over a smooth fixed pulley (
定滑轮) - Tthe tension in the string, N
The Atwood machine. Check: equal masses give a = 0 and T = mg; as m₂ → 0, a → g and T → 0.
Masses of 3 kg and 1 kg hang over a smooth pulley and are released from rest (g = 10 m/s²). Find a, T, the force on the pulley's axle and the distance each mass moves in 1 s.
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The axle holds both ends of the string: 2T = 30 N, less than the total weight of 40 N, because the heavier side accelerates downward.
In 1 s: s = at²/2 = 2.5 m.
Apparent weight in lifts and sudden cuts
A scale shows the normal force it exerts on you, your apparent weight (视重). With upward positive, N − mg = ma. If the acceleration points up, N = m(g + a) > mg: overweight (超重). If it points down, N = m(g − a) < mg: underweight (失重). In free fall a = g and N = 0: complete weightlessness (完全失重), as for astronauts in orbit (see «Universal gravitation and satellites»). Only the direction of the acceleration matters, not the direction of motion.
- Napparent weight: the reading of the scale (or the tension of a spring balance or rope), N
- athe size of the lift's acceleration, m/s²
Gravity mg does not change in a lift; only the support force does.
| Motion of the lift | Direction of a | Scale reading |
|---|---|---|
| up, speeding up | up | > mg (overweight) |
| up, slowing down | down | < mg (underweight) |
| down, speeding up | down | < mg (underweight) |
| down, slowing down | up | > mg (overweight) |
| constant velocity (either way) | a = 0 | = mg |
A 50 kg student stands on a scale in a lift. The scale reads 400 N (g = 10 m/s²). Which motion of the lift is possible?
A) moving up and speeding up
B) moving down at constant speed
C) moving up and slowing down
D) moving down and slowing down
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Sudden cuts. Because F = ma holds at every instant, the acceleration right after a cut is found from the forces that exist at that instant. A light, inextensible string (or a rod, or a contact) can change its force instantly, even to zero. A spring cannot: its force depends on its extension, and the extension needs time to change. So at the moment of cutting, spring forces keep their old values, while the force of a cut string disappears (瞬时加速度).
Balls A and B, each of mass m, hang one below the other: A hangs from the ceiling by a string, and B hangs from A by a light spring. The string is cut. Find the accelerations of the balls at that instant.
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Just after the cut the spring still pulls A down with mg and holds B up with mg.
A: mg + mg = ma ⇒ a = 2g, downward. B: mg − mg = 0 ⇒ a = 0.
If A and B were joined by a string instead of a spring, both would fall freely with a = g.
A ball of mass m is held at rest by a horizontal string and by a light spring that makes angle θ with the vertical. The horizontal string is cut. What is the ball's acceleration at that instant?
A) g B) g sin θ C) g tan θ D) g / cos θ
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B) g sin θ is the answer if the spring is replaced by a string (its force changes at once and the ball starts to swing); A forgets the spring; D) g / cos θ is the spring force divided by m.
How CSCA asks about this
- Net force → acceleration: two forces at an angle (add them as vectors), a force at an angle with friction (N ≠ mg).
- Inclines: a = g(sin θ ∓ μ cos θ), the mass cancels; "at rest on an incline" belongs to «Forces and equilibrium».
- Connected bodies: a from the whole system, the internal force from one isolated body; force sharing by mass; the Atwood and table–pulley formulas.
- Lifts: only the direction of a matters; reading = m(g ± a); free fall gives 0.
- Cuts: springs keep their force, strings lose it; typical answers are 0, g, 2g, g tan θ.
- Third law: concept items — equal and opposite, on different bodies, of the same kind.
- Time-savers (75 s per item): the mass usually cancels, so if the options contain no m, do not carry it; check limiting cases (μ = 0, θ = 0, equal masses).
| Term | 中文 | Pinyin |
|---|---|---|
| inertia | 惯性 | guànxìng |
| Newton's second law | 牛顿第二定律 | Niúdùn dì-èr dìnglǜ |
| Newton's third law | 牛顿第三定律 | Niúdùn dì-sān dìnglǜ |
| action and reaction | 作用力与反作用力 | zuòyònglì yǔ fǎnzuòyònglì |
| acceleration | 加速度 | jiāsùdù |
| net external force | 合外力 | héwàilì |
| connected bodies | 连接体 | liánjiētǐ |
| whole-system method | 整体法 | zhěngtǐfǎ |
| isolation method | 隔离法 | gélífǎ |
| fixed pulley | 定滑轮 | dìng huálún |
| apparent weight | 视重 | shìzhòng |
| overweight (apparent weight > mg) | 超重 | chāozhòng |
| underweight (apparent weight < mg) | 失重 | shīzhòng |
| complete weightlessness | 完全失重 | wánquán shīzhòng |
| instantaneous acceleration | 瞬时加速度 | shùnshí jiāsùdù |
Key points
- First law: with zero net force the velocity does not change; mass is the measure of inertia.
- ΣF = ma: F is the net force, a points the same way and changes at the same instant.
- Third law: equal, opposite forces of the same kind act on different bodies and never cancel.
- Incline: a = g(sin θ − μ cos θ) sliding down, a = g(sin θ + μ cos θ) sliding up.
- Connected bodies: a from the whole system, tension from one body; Atwood: a = (m₁ − m₂)g / (m₁ + m₂), T = 2m₁m₂g / (m₁ + m₂).
- Lift: a up → N = m(g + a), a down → N = m(g − a), free fall → 0; at a cut a spring keeps its force, a string loses it.
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