Mechanics

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 force at angle $\theta$ has a horizontal part $F\cos\theta$ and a vertical part $F\sin\theta$.

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.

Every mechanics problem is a picture like this. Three forces act on the climber — weight straight down, tension along the rope, and a push from the rock — and because they balance, the climber hangs still in equilibrium. Resolving each into horizontal and vertical parts turns the picture into equations

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 $\mu$:

On a rough surface the contact force splits into the normal reaction $R$ and friction $F$ (at most $\mu R$).

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 $8\text{ N}$.

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 (a negative acceleration is a deceleration 减速度). 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.

The shaded area gives the distance travelled; the slope of the line gives the acceleration.

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.

A Newton's cradle demonstrates conservation of momentum in collisions.

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. This is the conservation of linear momentum 动量守恒:

The total momentum is the same before and after the collision.

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.

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.

On a slope, resolve the forces along the slope and at right angles to it.

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, $T - mg\sin 20^\circ - F = ma$:

A roller coaster trades potential energy for kinetic energy as it rises and falls.

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 work done by the outside forces equals the change in the total energy. When no friction acts, the total energy stays the same — 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.

• 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 ($F = \mu R$) on a rough surface.
• State your assumptions (light inextensible string, smooth pulley, particle) — they are often worth a mark.