Equilibrium of forces · 力的平衡
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | 平衡 | píng héng |
| principle of moments/ˈprɪnsɪpl ɒv ˈməʊmənts/ | 力矩原理 | lì jǔ yuán lǐ |
| clockwise/ˈklɒkwaɪz/ | 顺时针 | shùn shí zhēn |
| anticlockwise/ˌæntɪˈklɒkwaɪz/ | 逆时针 | nì shí zhēn |
| tension/ˈtenʃn/ | 张力 | zhāng lì |
| line of action/laɪn ɒv ˈækʃn/ | 作用线 | zuò yòng xiàn |
| vector triangle/ˈvektə ˈtraɪæŋɡl/ | 矢量三角形 | shǐ liàng sān jiǎo xíng |
| resolve/rɪˈzɒlv/ | 分解 | fēn jiě |
A balanced see-saw
- Two children balance a see-saw even when they weigh different amounts.
- The lighter one just sits further out.
- Balance is about both forces and their turning effects.
平衡的跷跷板
- 即使两个孩子体重不同,也能让跷跷板平衡。
- 较轻的那个只要 坐得更靠外。
- 平衡既关乎力,也 关乎它们的转动效果。
Two conditions for equilibrium 平衡
- A body is in equilibrium when both are true:
- the resultant force is zero, and the resultant moment is zero.
A couple: two equal and opposite forces, a distance apart, producing a torque
平衡的两个条件
- 当 两者 都成立时,物体处于 平衡(equilibrium):
- 合 力 为零,而且 合 力矩 为零。
一个力偶:两个大小相等、方向相反、相隔一段距离的力,产生一个力偶矩
Forces in equilibrium · 平衡力
When forces are balanced the resultant is zero — the vectors form a closed loop. Drag the arrows to keep them cancelling. · 当力平衡时，合力为零——矢量形成一个闭合回路。拖动箭头以保持它们相互抵消。
Select both · 两者 conditions a body must meet to be in equilibrium. · 选出物体处于平衡必须满足的 两个 条件。
Equilibrium = zero resultant force and zero resultant moment. A body can be in equilibrium while moving at constant velocity. · 平衡 = 合力为零 而且 合力矩为零。物体在匀速运动时也可以处于平衡。
A body with zero resultant force must be in equilibrium. · 合力为零的物体一定处于平衡。
Not necessarily — a couple gives zero resultant force but still turns the body. You also need zero resultant moment. · 不一定——力偶的合力为零,但仍会使物体转动。你还需要合力矩为零。
Principle of moments 力矩原理
- For a body that is not turning: total clockwise 顺时针 moment = total anticlockwise 逆时针 moment (about any point).
力矩原理
- 对于不转动的物体:总 顺时针 力矩 = 总 逆时针 力矩(对任何一点)。
At balance, the total clockwise moment equals the total ____ moment. · 平衡时,总顺时针力矩等于总 ____ 力矩。
That is the principle of moments — the two turning effects cancel about any chosen point. · 这就是力矩原理——对任何选定的点,两个转动效果互相抵消。
Step one: draw every force
- Before any equation, draw each force as a labelled arrow at the point where it acts.
- The usual cast: weight (down, at the centre of gravity), tension 张力 (along the string, away from the object), normal contact force (at right angles to the surface), friction (along the surface).
- Never draw an arrow for "motion" or "momentum" — they are not forces, and an examiner counts them as errors.
第一步:画出每一个力
- 在列任何方程之前,把每个力画成 带标签的箭头,画在它作用的点上。
- 常见的角色:重力(向下,作用在重心)、张力(tension)(沿着绳子,背离物体)、法向接触力(垂直于表面)、摩擦力(沿着表面)。
- 绝不要为"运动"或"动量"画箭头——它们不是力,阅卷人会把它们算作错误。
A book rests on a table while someone pushes it slowly along. Which of these should not · 不 appear on its force diagram? · 一本书放在桌上,有人慢慢推着它滑动。下面哪一项 不应该 出现在它的受力图上?
Motion is not a force. A force diagram shows only forces: weight, normal contact force, friction and the push. · 运动不是力。受力图只画力:重力、法向接触力、摩擦力和推力。
Solving a balance problem
- Take moments about an unknown force, so its moment is zero and it drops out.
- List each force × its perpendicular distance, then set clockwise = anticlockwise.
- Use "resultant force = 0" if you need a second equation.
求解平衡问题
- 对一个 未知 力取矩,使它的力矩为零而被消去。
- 列出每个力 × 它的垂直距离,然后令顺时针 = 逆时针。
- 如果需要第二个方程,用"合力 = 0"。
To simplify a moments problem, it is smart to take moments about the point where: · 为了简化力矩问题,聪明的做法是对哪个点取矩:
A force acting at the pivot has zero perpendicular distance, so its moment is zero and it drops out of the equation. · 作用在转轴上的力垂直距离为零,所以它的力矩为零,从方程中消去。
Put the steps for solving a moments problem in order. · 把求解力矩问题的步骤按顺序排列。
Forces first, then a pivot that removes an unknown, then the moments equation, and finally the force balance for whatever is left. · 先画力,再选一个能消去未知量的转轴,然后列力矩方程,最后用力的平衡求剩下的量。
Worked example: a beam and a cable
A uniform beam of weight $120\ \text{N}$ and length $2.0\ \text{m}$ is hinged at one end and held horizontal by a vertical cable at the other end. A $300\ \text{N}$ load hangs $0.50\ \text{m}$ from the hinge. Find the tension in the cable and the vertical force at the hinge.
- Moments about the hinge — its unknown force then has zero moment.
- Clockwise: $120 \times 1.0 + 300 \times 0.50 = 270\ \text{N m}$ (the beam's weight acts at its middle).
- Anticlockwise: $T \times 2.0$.
- So $T = \dfrac{270}{2.0} = 135\ \text{N}$.
- Vertical forces balance: $H + 135 = 120 + 300$, so the hinge pushes up with $H = 285\ \text{N}$.
- Check: the cable, further from the load, carries less than the hinge — as expected.
例题:横梁与缆绳
一根重 $120\ \text{N}$、长 $2.0\ \text{m}$ 的均匀横梁一端用铰链固定,另一端由一根竖直缆绳拉住保持水平。一个 $300\ \text{N}$ 的重物挂在离铰链 $0.50\ \text{m}$ 处。求缆绳的张力和铰链处的竖直力。
- 对铰链取矩——它的未知力的力矩就为零。
- 顺时针:$120 \times 1.0 + 300 \times 0.50 = 270\ \text{N m}$(横梁的重力作用在它的 中点)。
- 逆时针:$T \times 2.0$。
- 所以 $T = \dfrac{270}{2.0} = 135\ \text{N}$。
- 竖直方向的力平衡: $H + 135 = 120 + 300$,所以铰链向上推 $H = 285\ \text{N}$。
- 检查: 缆绳离重物更远,承担的力比铰链小——符合预期。
A child of weight $200\ \text{N}$ sits $1.5\ \text{m}$ left of a see-saw pivot. What weight, $1.0\ \text{m}$ to the right, balances it? · 一个体重 $200\ \text{N}$ 的孩子坐在跷跷板转轴左边 $1.5\ \text{m}$ 处。右边 $1.0\ \text{m}$ 处多大的重力能使它平衡?
Clockwise = anticlockwise: $W \times 1.0 = 200 \times 1.5$, so $W = 300\ \text{N}$. · 顺时针 = 逆时针:$W \times 1.0 = 200 \times 1.5$,所以 $W = 300\ \text{N}$。
A uniform beam of weight $80\ \text{N}$ and length $4.0\ \text{m}$ is hinged at one end and held horizontal by a vertical cable at the other. A $200\ \text{N}$ load hangs $1.0\ \text{m}$ from the hinge. What is the tension in the cable, in N? · 一根重 $80\ \text{N}$、长 $4.0\ \text{m}$ 的均匀横梁一端铰接,另一端由竖直缆绳拉住保持水平。一个 $200\ \text{N}$ 的重物挂在离铰链 $1.0\ \text{m}$ 处。缆绳的张力是多少(单位 N)?
Moments about the hinge: $T \times 4.0 = 80 \times 2.0 + 200 \times 1.0 = 360$, so $T = 90\ \text{N}$. · 对铰链取矩:$T \times 4.0 = 80 \times 2.0 + 200 \times 1.0 = 360$,所以 $T = 90\ \text{N}$。
The distance in a moment is the perpendicular distance from the pivot to the force's line of action 作用线. For a force at angle $\theta$ to the beam, that is $d\sin\theta$, not $d$. Using the full length of the beam for a slanting cable is the most common lost mark in this topic.
力矩中的距离是从转轴到力的 作用线(line of action) 的 垂直 距离。对于与横梁成 $\theta$ 角的力,这个距离是 $d\sin\theta$,而不是 $d$。对一根斜拉的缆绳用横梁的全长,是本主题最常见的丢分点。
The vector triangle 矢量三角形
- Three forces in equilibrium, drawn tip to tail, form a closed triangle.
- Solve it with the sine/cosine rule, or resolve into perpendicular components instead.
矢量三角形
- 三个处于平衡的力,首尾相接 画出,构成一个 闭合三角形。
- 用正弦/余弦定理求解,或改为分解成互相垂直的分量。
Three forces in equilibrium, drawn tip to tail, form a closed triangle. · 三个处于平衡的力,首尾相接画出,构成一个闭合三角形。
Yes — if they balance, the three arrows return to the start, making a closed vector triangle. · 是的——如果它们平衡,三个箭头会回到起点,构成一个闭合的矢量三角形。
Worked example: two strings, one lamp
A lamp of weight $40\ \text{N}$ hangs from two identical strings, each at $30^\circ$ to the horizontal. Find the tension in each string.
- Resolve 分解 vertically: the two upward components carry the weight.
- Each string pulls up by $T\sin 30^\circ$, so $2T\sin 30^\circ = 40$.
- $\sin 30^\circ = 0.5$, so $T = 40\ \text{N}$.
- Horizontally the two $T\cos 30^\circ$ components cancel, which is why the lamp does not swing sideways.
- Check: flatter strings (smaller $\theta$) give a smaller $\sin\theta$ and a larger tension — a washing line pulled nearly straight can snap.
例题:两根绳子,一盏灯
一盏重 $40\ \text{N}$ 的灯由两根相同的绳子挂着,每根绳子与水平方向成 $30^\circ$。求每根绳子的张力。
- 竖直方向 分解(resolve):两个向上的分量承担重力。
- 每根绳子向上拉 $T\sin 30^\circ$,所以 $2T\sin 30^\circ = 40$。
- $\sin 30^\circ = 0.5$,所以 $T = 40\ \text{N}$。
- 水平方向 两个 $T\cos 30^\circ$ 分量互相抵消,这就是灯不会左右摆的原因。
- 检查: 绳子越平(θ 越小),$\sin\theta$ 越小,张力 越大——拉得几乎笔直的晾衣绳会断。
You've got it
- equilibrium needs both: zero resultant force and zero resultant moment
- principle of moments: clockwise = anticlockwise about any point — take moments about an unknown force
- a moment uses the perpendicular distance to the line of action
- three balanced forces close into a vector triangle, or resolve into components
你掌握了
- 平衡需要 两者:合力为零 而且 合力矩为零
- 力矩原理:对任何一点,顺时针 = 逆时针——对未知力取矩
- 力矩用的是到作用线的 垂直 距离
- 三个平衡的力闭合成一个 矢量三角形,或者分解成分量