# 摩擦力与张力

> AP 物理 1 · 第2单元：动力学
> 来源: https://www.owlsprep.com/zh/study/ap-physics-1-u2-friction-and-tension/

涵盖AP物理1动力学中的静摩擦力、动摩擦力、理想绳索和定滑轮的张力，以及连接物体系统的分析，包含分步讲解的例题和考试专属技巧。

**先修:** [牛顿三大运动定律](https://www.owlsprep.com/zh/study/ap-physics-1-u2-newtons-laws/); 绘制和解读受力图; 力分量的矢量分解

## 学习目标

- 区分静摩擦力和动摩擦力
- 计算静止和滑动物体的摩擦力
- 分析理想绳索和定滑轮中的张力
- 解决涉及摩擦力和张力的连接系统问题

## 摩擦力与张力的核心定义

摩擦力是阻碍接触的两个固体表面之间相对运动的接触力，而张力是通过拉伸的柔性介质（例如绳索、细线或钢缆）传递的拉力。本知识点约占AP物理1第2单元：动力学内容的三分之一，该单元占AP考试总分的12–18%，在选择题和自由作答题中都经常出现。

**摩擦力** — 阻碍接触的两个固体表面之间相对（或即将发生的相对）运动的接触力

*记法:* $f$, $f_s$ (static), $f_k$ (kinetic)

*例:* 木箱在地面上滑动时，动摩擦力阻碍其运动，使木箱减速

**张力** — 沿拉伸柔性介质传递的拉力，在介质两端大小相等

*记法:* $T$

*例:* 悬挂静止重物的绳索对重物的向上拉力等于重物的重力，该拉力就是张力

在AP物理1中，除非题目明确说明，否则我们几乎总是假设绳索是理想绳索（无质量、不可伸长），滑轮是理想滑轮（无质量、无摩擦）。这简化了分析，因为理想绳索的张力处处相等。

## 静摩擦力与动摩擦力

摩擦力根据接触面是否存在相对运动分为两类。静摩擦力作用在无相对运动的情况下，其大小会调整为恰好抵消外力的平行分量，直到达到最大阈值。动摩擦力作用在相对滑动的接触面之间，对于给定的接触面对和正压力，其大小恒定。

最大静摩擦力的公式为：

$$f_{s,max} = \mu_s N$$

where $\mu_s$ is the dimensionless coefficient of static friction (dependent on the two surface materials), and $N$ is the magnitude of the normal force perpendicular to the contact surface. Kinetic friction follows the formula:

$$f_k = \mu_k N$$

For any pair of surfaces, $\mu_k < \mu_s$, which means it takes more force to start moving an object than to keep it moving at constant speed. A common misconception is that normal force always equals an object’s weight; this is only true for horizontal surfaces with no additional vertical forces. $N$ must always be calculated from Newton’s second law in the direction perpendicular to the contact surface.

**例题:** A 12 kg wooden crate rests on a horizontal concrete floor, with $\mu_s = 0.6$ and $\mu_k = 0.4$. What is the magnitude of friction when a horizontal 50 N force pushes on the stationary crate?

1. Calculate the normal force: no vertical acceleration, so

   $$N = mg = 12 \times 9.8 = 117.6 \text{ N}$$
2. Calculate maximum static friction

   $$f_{s,max} = \mu_s N = 0.6 \times 117.6 = 70.56 \text{ N}$$
3. Compare the applied force to the maximum threshold: $50 \text{ N} < 70.56 \text{ N}$, so the crate remains stationary
4. For stationary objects not at the sliding threshold, static friction matches the applied parallel force

   $$f_s = 50 \text{ N}$$

> **tip**
>
> Always compare the applied force to $f_{s,max}$ before assuming friction is kinetic. AP exam questions regularly trick students into automatically using $f_k$ when the object is not moving.

## Tension in Ideal Ropes and Pulleys

Tension is a pulling force that acts along the length of a rope, pulling equally on both objects connected to the rope. For AP Physics 1, all ropes and pulleys are assumed ideal unless stated otherwise, with the following properties:

- Ideal rope: massless and inextensible. Inextensible means all connected objects have the same magnitude of acceleration, even if acceleration directions differ. Massless means net force on the rope is zero, so tension is uniform along the rope.
- Ideal fixed pulley: massless and frictionless. It only changes the direction of tension, not its magnitude, so tension is equal on both sides of the pulley.

**例题:** A 5 kg mass hangs vertically from an ideal rope that runs over a fixed ideal pulley, connected to an 8 kg block resting on a frictionless horizontal table. What is the magnitude of tension in the rope?

1. Assign acceleration: the hanging mass accelerates downward, the block accelerates to the right, with equal magnitude $a$
2. Write Newton's second law for the 8 kg block (horizontal direction)

   $$\sum F = T = 8a$$
3. Write Newton's second law for the 5 kg hanging mass (downward as positive)

   $$\sum F = mg - T = 5a = 49 - T$$
4. Substitute $T=8a$ into the second equation and solve for $a$

   $$49 - 8a = 5a \rightarrow 13a = 49 \rightarrow a \approx 3.77 \text{ m/s}^2$$
5. Solve for tension

   $$T = 8 \times 3.77 \approx 30.2 \text{ N}$$

> **tip**
>
> If a pulley is accelerating (e.g., a movable pulley in a system), you must include forces on the pulley itself in your analysis; only fixed ideal pulleys have equal tension on both sides.

## Combined Tension-Friction Connected Systems

Most AP Physics 1 problems involving both friction and tension are connected object systems, where one or more objects rest on a frictional surface, connected by a rope and pulley to a hanging object. Follow this systematic approach to solve these problems:

1. Draw a separate free-body diagram for every object in the system
2. Resolve forces into components aligned with the direction of possible motion
3. Write Newton's second law for each object, using equal tension and equal acceleration magnitude for ideal systems
4. Check if the system is stationary or accelerating by comparing the applied pulling force to maximum static friction, then solve the system of equations

**例题:** Block A (mass 4 kg) rests on a horizontal table, connected by an ideal rope over a fixed ideal pulley to hanging Block B (mass 3 kg). $\mu_s = 0.35$ and $\mu_k = 0.25$ between Block A and the table. Is the system stationary, or does it accelerate? If it accelerates, what is the tension?

1. Calculate maximum static friction on Block A

   $$f_{s,max} = \mu_s m_A g = 0.35 \times 4 \times 9.8 = 13.72 \text{ N}$$
2. Compare to the pulling force from Block B: the required tension for equilibrium would equal $m_B g = 29.4 \text{ N}$. Since $29.4 \text{ N} > 13.72 \text{ N}$, static friction cannot hold the system, so it accelerates
3. Write Newton's second law for Block A (right positive)

   $$T - f_k = m_A a, \quad f_k = \mu_k N = 9.8 \text{ N} \rightarrow T - 9.8 = 4a$$
4. Write Newton's second law for Block B (down positive)

   $$29.4 - T = 3a$$
5. Add equations to eliminate tension, then solve for $a$ and $T$

   $$19.6 = 7a \rightarrow a = 2.8 \text{ m/s}^2, \quad T = 21 \text{ N}$$

> **tip**
>
> Always confirm the direction of friction: friction opposes impending or actual motion, so if the system is pulling a block up an incline, friction acts down the incline, and vice versa.

**概念自测**

Test your understanding of friction with an angled applied force:

1. A 10 kg box rests on a horizontal surface with $\mu_s = 0.5$ and $\mu_k = 0.3$. A person pulls the box with a 30 N force at an angle of 30° above the horizontal. What is the magnitude of friction acting on the box?

   - 0 N
   - ~26 N
   - ~36 N
   - ~41 N

   *解析:* First calculate the reduced normal force from the upward pull component, then check if the applied horizontal force is less than maximum static friction. Since it is, static friction equals the applied horizontal component, giving ~26 N.

## 常见错误

- **错误做法:** Using $f_s = \mu_s N$ for static friction when the object is not at the point of sliding
  - 原因: Students memorize the maximum static friction formula and apply it to all static friction cases, forgetting static friction adjusts to match the applied force
  - 正确做法: Only use $f_s = \mu_s N$ if the problem states the object is just about to slide; for all other stationary cases, use $f_s = F_{applied,parallel}$
- **错误做法:** Assuming normal force $N$ equals the object's weight $mg$ in all cases
  - 原因: Students generalize from simple horizontal surface problems to all cases, including angled forces and inclines
  - 正确做法: Always calculate $N$ from Newton's second law in the direction perpendicular to the surface, accounting for angled applied forces or inclines before calculating friction
- **错误做法:** Assigning different acceleration magnitudes to connected objects on an ideal inextensible rope
  - 原因: Students confuse different acceleration directions with different magnitudes of acceleration
  - 正确做法: For any two objects connected by an ideal rope, set the magnitude of acceleration equal when writing your system of equations
- **错误做法:** Changing the magnitude of tension when it goes around an ideal fixed pulley
  - 原因: Students assume pulleys change tension magnitude, when they only change direction for ideal fixed pulleys
  - 正确做法: For any ideal massless, frictionless fixed pulley, tension has the same magnitude on both sides of the pulley
- **错误做法:** Using kinetic friction when the applied force is less than maximum static friction
  - 原因: Students rush to use the kinetic friction formula without checking if motion actually occurs
  - 正确做法: Always compare the net applied force trying to move the object to $f_{s,max}$ first; only use $f_k$ if the applied force exceeds $f_{s,max}$

## 速查表

| Category | Formula/Rule | Key Notes |
| --- | --- | --- |
| Maximum Static Friction | $f_{s,max} = \mu_s N$ | Only applies when object is just about to slide; $f_s \leq f_{s,max}$ for all stationary objects |
| Kinetic Friction | $f_k = \mu_k N$ | Applies when surfaces slide relative to each other; $\mu_k < \mu_s$ for all surface pairs |
| Static Friction (non-maximum) | $f_s = F_{applied,parallel}$ | Matches the parallel applied force for stationary objects not at the sliding threshold |
| Tension in ideal rope | $T_1 = T_2$ | Equal tension magnitude at both ends of a massless inextensible rope |
| Connected object acceleration | $\|a_1\| = \|a_2\|$ | Equal magnitude acceleration for all objects connected by an ideal inextensible rope |
| Tension over ideal fixed pulley | $T_{left} = T_{right}$ | Ideal fixed pulleys only change tension direction, not magnitude |
| Static friction direction | Opposes impending relative motion | Points opposite to the direction the object would slide if friction were removed |
| Kinetic friction direction | Opposes actual relative motion | Points opposite to the direction the object is sliding relative to the surface |

## 下一步

Mastering friction and tension is the foundation for all subsequent dynamics problems in AP Physics 1, and these concepts are immediately applied to nearly all future units. In Unit 3: Circular Motion and Gravitation, friction provides the centripetal force for objects like cars turning on flat roads, and tension acts as the centripetal force for objects moving in vertical circles. Friction also appears later in energy problems, where it does non-conservative work that changes the total mechanical energy of a system. In rotational dynamics, analyzing rolling motion without slipping relies entirely on static friction to provide the torque needed for rotation. Solid skills here will make all more complex force problems much easier to solve.

- [斜面与阿特伍德机](https://www.owlsprep.com/zh/study/ap-physics-1-u2-inclined-planes-and-atwood-machines/)
- [圆周运动与万有引力概述](https://www.owlsprep.com/zh/study/ap-physics-1-u3-overview/)
- [匀速圆周运动中的加速度](https://www.owlsprep.com/zh/study/ap-physics-1-u3-acceleration-in-uniform-circular-motion/)

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