Rotational Equilibrium · 转动平衡
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| rotational equilibrium/rəʊˈteɪʃənl ˌiːkwɪˈlɪbrɪəm/ | 转动平衡 | zhuǎn dòng píng héng |
| net torque/net tɔːk/ | 净力矩 | jìng lì jǔ |
| static equilibrium/ˈstætɪk ˌiːkwɪˈlɪbrɪəm/ | 静平衡 | jìng píng héng |
| extended free-body diagram/ekˈstendɪd friː ˈbɒdi ˈdaɪəɡræm/ | 扩展受力图 | kuò zhǎn shòu lì tú |
How a seesaw finds its balance
- A small child far out can balance a big adult sitting close to the pivot.
- It is not about weight alone -- it is about weight times distance.
- When the turning effects match, the seesaw holds steady.
- That balance of torques is the key to bridges, beams, and cranes.
跷跷板如何找到平衡
- 一个远坐的小孩可以平衡一个近坐在枢轴旁的大人。
- 这不只关乎体重——而是体重乘以距离。
- 当转动效果相互匹配时,跷跷板就保持稳定。
- 这种力矩的平衡是桥梁、梁和起重机的关键。
Rotational equilibrium
- A body is in rotational equilibrium 转动平衡 when the net torque 净力矩 on it is zero.
- It does not start or stop spinning -- its angular velocity stays constant.
- Clockwise torques exactly balance counterclockwise ones.
转动平衡
- 当物体所受净力矩为零时,它处于转动平衡。
- 它不开始也不停止旋转——它的角速度保持不变。
- 顺时针的力矩与逆时针的力矩恰好平衡。
Select all · 所有 conditions true in rotational equilibrium. · 选择转动平衡中成立的所有条件。
Rotational equilibrium means zero net torque and constant angular velocity -- it could be spinning steadily, not necessarily still. · 转动平衡意味着合力矩为零且角速度恒定——它可以稳定旋转，不一定非要是静止的。
Full static equilibrium
- For static equilibrium 静平衡, both conditions must hold: zero net force and zero net torque.
- Zero net force keeps it from accelerating; zero net torque keeps it from spinning.
- Both together mean it stays completely at rest.
完全静平衡
- 对于静平衡,两个条件都必须成立:净力为零且净力矩为零。
- 净力为零使它不平动加速;净力矩为零使它不旋转。
- 两者一起意味着它完全保持静止。
For an object to be in full static equilibrium, which must be true? · 为了使物体处于完全静力平衡，必须满足什么条件？
Static equilibrium requires zero net force and zero net torque. · 静力平衡要求合力为零 且 合力矩为零。
An extended diagram
- Torque depends on where a force acts, so draw an extended free-body diagram 扩展受力图.
- Show each force at its actual point of application, not all at one dot.
- The weight of a uniform beam acts at its center.
扩展的受力图
- 力矩取决于力作用在哪里,所以要画一张扩展受力图。
- 把每个力画在它实际的作用点上,而不是全画在一个点上。
- 均匀梁的重力作用在它的中心。
In rotational equilibrium? · 处于转动平衡状态的条件是？
An object is in rotational equilibrium when the torques cancel. Sort each case. · 当力矩相互抵消时，物体处于转动平衡。对每种情况进行排序。
If the net force on an object is zero, it cannot possibly be rotating faster and faster. · 如果作用在物体上的合力为零，它不可能加速旋转得越来越快。
Two offset equal-opposite forces give zero net force but a net torque -- the object angularly accelerates. · 两个大小相等、方向相反但错开的力给出零合力但存在合力矩——物体会发生角加速度。
On an extended free-body diagram of a uniform beam, its weight is drawn acting at... · 在均匀梁的自由体受力图中，其重力绘制于...
A uniform beam's weight acts at its center of mass -- its geometric center. · 均匀梁的重力作用于其质心——即几何中心。
The solving trick
- Choose your axis cleverly -- put it at an unknown force so that force's torque becomes zero.
- Then $\sum \tau = 0$ has fewer unknowns and is easy to solve.
- Combine with $\sum F = 0$ to find every remaining force.
求解技巧
- 巧妙地选择你的轴——把它放在某个未知力处,使那个力的力矩变为零。
- 这样 $\sum \tau = 0$ 的未知量就更少,容易求解。
- 再结合 $\sum F = 0$,求出其余每一个力。
A $20\ \text{kg}$ child sits $3\ \text{m}$ left of a pivot. A $30\ \text{kg}$ child balances it on the right. How far from the pivot must they sit (in m)? · 一个$20\ \text{kg}$的孩子坐在支点左侧$3\ \text{m}$处。另一个$30\ \text{kg}$的孩子在右侧使其平衡。他们必须离支点多远（单位 m）？
Balance torques, $20 \times 3 = 30 \times x$, so $x = 60/30 = 2\ \text{m}$. · 平衡扭矩，$20 \times 3 = 30 \times x$，因此$x = 60/30 = 2\ \text{m}$。
A smart choice of ____ puts an unknown force's torque to zero, simplifying the equation. · 明智地选择____可以使未知力的力矩变为零，从而简化方程。
Placing the axis at an unknown force makes its lever arm zero, so it drops out of $\sum\tau = 0$. · 将轴置于未知力处会使力臂为零，从而使该项从$\sum\tau = 0$中消失。
A seesaw balances with a $30\ \text{kg}$ child $2\ \text{m}$ left of the pivot and a $40\ \text{kg}$ child at $x$ right.
- Balance torques: $30 \times 2 = 40 \times x$, so $x = \dfrac{60}{40} = 1.5\ \text{m}$.
- The heavier child must sit closer to the pivot -- less distance, same torque.
跷跷板平衡时,一个 $30\ \text{kg}$ 的孩子在枢轴左侧 $2\ \text{m}$,一个 $40\ \text{kg}$ 的孩子在右侧 $x$ 处。
- 平衡力矩:$30 \times 2 = 40 \times x$,所以 $x = \dfrac{60}{40} = 1.5\ \text{m}$。
- 更重的孩子必须坐得更近枢轴——距离更小,力矩相同。
Zero net force alone is not enough for full equilibrium. A pair of equal, opposite forces offset from each other has zero net force but a nonzero net torque -- it would spin. You must check both conditions.
仅仅净力为零不足以达到完全平衡。一对大小相等、方向相反且互相错开的力,净力为零但净力矩不为零——它会旋转。你必须检查两个条件。
Rotational equilibrium means the net torque is zero (no change in spin). Full static equilibrium needs zero net force and zero net torque. Draw an extended free-body diagram (forces at their real locations), and put your axis at an unknown force to simplify $\sum \tau = 0$.
转动平衡意味着净力矩为零(旋转不变)。完全静平衡需要净力为零且净力矩为零。画一张扩展受力图(力画在真实位置),并把轴放在某个未知力处以简化 $\sum \tau = 0$。