Newton's laws of motion
| English | Chinese | Pinyin |
|---|---|---|
| resultant force | 合力 | hé lì |
| weight | 重力 | zhòng lì |
| tension | 张力 | zhāng lì |
| inclined plane | 斜面 | xié miàn |
| connected particles | 连接质点 | lián jiē zhì diǎn |
| inextensible | 不可伸长 | bù kě shēn cháng |
The law that launched rockets
- Every rocket, every car, every falling apple obeys the same rule: $F = ma$.
- Newton's second law is the foundation of all mechanics. Master it and you can predict the motion of anything.
Newton's second law
- The key law: resultant force 合力 = mass × acceleration:
- Weight 重力 is the force of gravity: $W = mg$ (take $g = 10\text{ m s}^{-2}$).
- Forces include weight, friction, tension 张力 in a string, and the push in a rod.
Worked example. A 4 kg box is pushed with a horizontal force of 20 N on a smooth surface. $F = ma \Rightarrow 20 = 4a \Rightarrow a = 5\text{ m s}^{-2}$.
Resultant force, not just any force. $F = ma$ uses the net (resultant) force. If there's friction, you must subtract it: $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$.
Resultant force
F = ma
The resultant force sets the acceleration. Balanced forces ⇒ no acceleration.
A resultant force gives a 5 kg mass an acceleration of 2 m/s². What is the force (F = ma), in N?
F = ma = 5 × 2 = 10 N.
Taking g = 10 m/s², what is the weight of a 2 kg object (W = mg), in N?
W = mg = 2 × 10 = 20 N.
F = ma uses the resultant (net) force, not just any single force.
F = ma requires the net force. If friction is present, subtract it from the applied force first.
Slopes and inclined planes 斜面
- On an inclined plane: split forces along and perpendicular to the slope, then apply $F = ma$ along the slope.
- Component of weight along slope: $mg\sin\theta$. Component perpendicular: $mg\cos\theta$.
To analyse motion on an inclined plane, you should:
Resolve along the slope (apply F = ma there) and perpendicular to it (for the normal reaction).
Connected particles 连接质点
- For connected particles (joined by a string): apply $F = ma$ to each body, or to the whole system.
- The string has the same tension throughout (if light and inextensible 不可伸长).
- Both bodies have the same acceleration.
Connected particles share one tension T and one acceleration; apply F = ma to each mass.
A 3 kg and 2 kg mass over a smooth pulley. Acceleration = g/5. With g = 10, find a (m/s²).
a = g/5 = 10/5 = 2 m/s².
For the 3 kg and 2 kg masses over a pulley (a = 2 m/s²), find the tension T from T − 2g = 2a (g = 10).
T − 2(10) = 2(2) → T = 20 + 4 = 24 N.
Worked example — connected particles
- A 3 kg and 2 kg mass are connected by a light string over a smooth pulley. Find the acceleration.
- For the 3 kg mass (moving down): $3g - T = 3a$.
- For the 2 kg mass (moving up): $T - 2g = 2a$.
- Adding: $3g - 2g = 5a \Rightarrow a = \dfrac{g}{5} = 2\text{ m s}^{-2}$.
- Newton's laws of motion give $F=ma$; forces may include thrust and air resistance.
You've got it
- F = ma; weight $W = mg$
- on a slope, resolve forces along and perpendicular to the slope
- for connected particles, apply $F = ma$ to each body or the whole system