力学

A force 力 is a push or a pull. It is a vector, so it has size and direction. Because it is a vector, you can split a force into components 分量 (usually horizontal and vertical), and you can add several forces into one resultant 合力.

A particle is in equilibrium 平衡 when the forces are balanced: the vector sum of the forces is zero. In practice this means the components in any direction add to zero.

Friction

When two surfaces touch, the contact force 接触力 between them has two parts: the normal reaction 法向反作用力 $R$, at right angles to the surface, and the friction 摩擦力 $F$, along the surface, which opposes sliding. A "smooth" surface is a model with no friction.

Friction can only grow up to a maximum. At that maximum the body is in limiting equilibrium 极限平衡, about to slip, and the friction is limiting friction 最大静摩擦力. The maximum is set by the coefficient of friction 摩擦系数 $\mu$:

By Newton's third law 牛顿第三定律, the two surfaces push on each other with equal and opposite forces.

Worked example. A block of weight $20\text{ N}$ rests on a rough horizontal table with coefficient of friction $\mu = 0.4$. Find the largest horizontal force that can be applied before the block slides.

The table's normal reaction balances the weight, so $R = 20\text{ N}$. The block is on the point of slipping when friction reaches its maximum $F = \mu R = 0.4 \times 20 = 8\text{ N}$. In equilibrium the applied force equals the friction, so the largest force it can resist is $8\text{ N}$.

Worked example: resolving. A particle of weight $20\text{ N}$ hangs in equilibrium from two strings. One string makes $30^\circ$ with the horizontal and the other makes $60^\circ$, on opposite sides. Find the two tensions $T_1$ and $T_2$.

Resolve horizontally and vertically; each direction must sum to zero:

The first equation gives $T_2 = T_1\dfrac{\cos 30^\circ}{\cos 60^\circ} = \sqrt3\,T_1$. Substitute: $\tfrac12 T_1 + \sqrt3 T_1\cdot\tfrac{\sqrt3}{2} = 2T_1 = 20$, so $T_1 = 10\text{ N}$ and $T_2 = 10\sqrt3 = 17.3\text{ N}$. The steeper string carries more of the weight, as you would expect.

Common mistakes. Mixing up $\sin$ and $\cos$: the component next to the angle uses $\cos$. Check with the angle you actually drew, not the one in the question.

Distance 距离 and speed 速率 are scalars (size only). Displacement 位移, velocity 速度 and acceleration 加速度 are vectors (size and direction).

On a velocity-time graph 速度时间图, the area under the graph is the displacement and the gradient is the acceleration (deceleration 减速度 means acceleration opposite to the velocity, reducing speed). On a displacement-time graph, the gradient is the velocity. More generally, differentiate with respect to time to go from displacement to velocity to acceleration, and integrate to go back.

For motion with constant acceleration 匀加速, use these formulae (the "suvat" equations):

Worked example. A car starts from rest and accelerates at $2.5\ \text{m s}^{-2}$ for $4\ \text{s}$. Find its speed and the distance travelled.

Worked example: acceleration that changes. A particle moves in a straight line with velocity $v = 6t - t^2\ \text{m s}^{-1}$ for $0 \leqslant t \leqslant 6$. Find its greatest speed and the distance it travels.

The acceleration is not constant, so suvat does not apply; use calculus. Differentiate: $a = \dfrac{dv}{dt} = 6 - 2t$, which is zero at $t = 3$, so the greatest speed is $v = 18 - 9 = 9\ \text{m s}^{-1}$. The velocity is never negative on this interval, so the distance is

If $v$ changed sign, you would integrate each part separately and add the sizes.

The linear momentum 动量 of a body is $\text{mass} \times \text{velocity}$. It is a vector. In a direct collision of two bodies, the total momentum is unchanged when the net external impulse is zero or negligible. This is the conservation of linear momentum 动量守恒:

Worked example. A body of mass $2\ \text{kg}$ moving at $3\ \text{m s}^{-1}$ hits a stationary body of mass $1\ \text{kg}$, and they stick together. Find their common speed afterwards.

Worked example: moving towards each other. A $3\ \text{kg}$ body moving at $4\ \text{m s}^{-1}$ meets a $2\ \text{kg}$ body moving at $5\ \text{m s}^{-1}$ in the opposite direction. After they collide, the $3\ \text{kg}$ body moves at $0.5\ \text{m s}^{-1}$ in its original direction. Find the velocity of the $2\ \text{kg}$ body.

Take the $3\ \text{kg}$ body's direction as positive, so the other body's velocity before is $-5$:

It is positive, so the $2\ \text{kg}$ body now moves in the $3\ \text{kg}$ body's original direction. Forgetting the minus sign for the opposite direction is the usual error.

Newton's laws of motion 牛顿运动定律 connect force and acceleration. The key one is: resultant force $=$ mass 质量 $\times$ acceleration,

The weight 重力 of a body is the force of gravity on it: $W = mg$. Forces in a problem may include weight, friction, tension 张力 in a string, the thrust 推力 (push) in a rod, and air resistance 空气阻力.

For motion on an inclined plane 斜面, split each force into a part along the slope and a part at right angles to it, then apply $F = ma$ along the slope. For connected particles 连接质点 (joined by a string), apply $F = ma$ to each body, or to the whole system.

Worked example. A block of mass $12\ \text{kg}$ is pulled up a rough plane by a rope parallel to the slope. The plane is at $20^\circ$ to the horizontal, the coefficient of friction is $0.4$, and the acceleration is $2\ \text{m s}^{-2}$. Find the tension in the rope.

The normal reaction is $R = mg\cos 20^\circ = 120\cos 20^\circ = 112.8\ \text{N}$, so the friction is $F = \mu R = 0.4 \times 112.8 = 45.1\ \text{N}$. Along the slope, $T - mg\sin 20^\circ - F = ma$:

Worked example: connected particles. Block $A$ of mass $5\ \text{kg}$ lies on a rough horizontal table ($\mu = 0.2$). A light inextensible string joins it, over a smooth pulley at the table's edge, to block $B$ of mass $3\ \text{kg}$, which hangs freely. The system is released from rest. Find the acceleration and the tension.

For $A$, $R = 5g = 50\text{ N}$, so friction is $0.2 \times 50 = 10\text{ N}$. Write $F = ma$ for each block in its own direction of motion:

Add them: $20 = 8a$, so $a = 2.5\ \text{m s}^{-2}$, and then $T = 30 - 3(2.5) = 22.5\text{ N}$. Check with $A$: $22.5 - 10 = 12.5 = 5(2.5)$. Note that $T$ is less than $B$'s weight; if it were equal, $B$ would not accelerate.

The work done 功 by a constant force is the scalar product of force and displacement — the force times the distance moved in the direction of the force: $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the motion. Work is measured in joules (J).

Energy comes in forms you can calculate:

• Kinetic energy 动能 (energy of movement): $\text{KE} = \tfrac12 mv^2$.
• Gravitational potential energy 重力势能 (energy of height): $\text{PE} = mgh$.

The net work done on a particle equals its change in kinetic energy. When only gravity does work, kinetic plus gravitational potential energy stays constant. More generally, total energy is conserved when all transfers, including heating by friction and work by other forces, are accounted for — the conservation of energy 能量守恒.

Power 功率 is the rate of doing work. For a force pulling in the direction of motion, $P = Fv$ (power $=$ force $\times$ velocity). Power is measured in watts (W).

Worked example. A car engine works at $12\ \text{kW}$ while the car moves at $20\ \text{m s}^{-1}$ on a level road. Find the driving force.

Worked example: energy with friction. A $5\ \text{kg}$ box slides from rest $10\ \text{m}$ down a slope at $30^\circ$ to the horizontal. A constant friction force of $12\ \text{N}$ acts. Find its speed at the bottom.

The box drops a height $h = 10\sin 30^\circ = 5\ \text{m}$, so it loses $mgh = 5(10)(5) = 250\ \text{J}$ of potential energy. The work done against friction is $12 \times 10 = 120\ \text{J}$. The rest becomes kinetic energy:

Use the vertical height for potential energy ($5$ m) but the distance along the slope for the work against friction ($10$ m).

• Draw a clear force diagram and resolve into perpendicular components; for equilibrium, each direction sums to zero.
• Use SUVAT only for constant acceleration and keep a consistent positive direction.
• Apply $F = ma$ along the direction of motion, including friction: $F\leqslant\mu R$ for static contact, with equality at limiting friction; use the stated sliding-friction model during sliding.
• State your assumptions (light inextensible string, smooth pulley, particle) — they are often worth a mark.