Forces, density and pressure

Centre of gravity

The weight 重力 of a large object can be treated as acting at one single point, called the centre of gravity 重心 (the same as the centre of mass 质心 in a uniform gravitational field). For a uniform, regular shape — a rectangle, a sphere, a uniform rod — the centre of gravity is at the middle.

When you draw a free-body diagram 受力图, always put the weight arrow at the centre of gravity.

Moment of a force

The moment 力矩 of a force about a point is

where $F$ is the size of the force 力 and $d$ is the perpendicular distance 垂直距离 from the point to the line of action 作用线 of the force. Unit: $\text{N m}$.

If the force acts at angle $\theta$ to a lever arm 力臂 of length $r$ from the pivot 支点, then $d = r\sin\theta$, so $M = F r\sin\theta$. Only the part of the force perpendicular 垂直 to the lever arm makes it turn.

Worked example. A force of $50\ \text{N}$ is applied to the end of a spanner $0.40\ \text{m}$ long, at $30°$ to the spanner. Find the moment of the force about the nut.

The moment depends on the perpendicular distance $d$ from the pivot to the line of action of the force

A moment is either clockwise 顺时针 or anticlockwise 逆时针 about the chosen point.

Couple and torque

A couple 力偶 is a pair of forces that are:

• equal in size,
• opposite in direction,
• with their lines of action a perpendicular distance apart.

A couple makes the body turn only — its resultant force is zero, so it gives no straight-line acceleration.

The torque 力偶矩 of a couple is the turning effect it makes:

where $F$ is the size of one force and $d$ is the perpendicular distance between the two lines of action. Unit: $\text{N m}$. The torque is the same about any point — a special property of couples.

A couple: two equal and opposite forces, a distance apart, producing a torque

A common multiple-choice trap: two equal forces in the same direction are not a couple (they have a resultant force and cause translation 平动). A couple needs equal size and opposite direction.

Conditions for equilibrium

A body is in equilibrium 平衡 when:

1. the resultant force 合力 is zero (no straight-line acceleration), AND
2. the resultant moment about any point is zero (no angular acceleration 角加速度).

Both must hold. A body with no resultant force can still be turning; a body with no resultant moment can still be moving in a straight line.

Principle of moments

For a body that is not turning, the total clockwise moment about any point equals the total anticlockwise moment about the same point. This is the principle of moments 力矩原理.

To solve a balance problem:

1. Choose a pivot — usually where an unknown force acts, so that force drops out (its distance is zero).
2. List every force and its perpendicular distance from the pivot.
3. Set $\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}$.
4. Use $\sum F = 0$ if you need a second equation.

A ruler balanced on a pivot with masses on each side is solved this way. For a heavy uniform rod, remember to include its weight acting at its centre of gravity.

Worked example. A uniform beam of length $4.0\ \text{m}$ is pivoted at its centre. A $40\ \text{N}$ weight hangs $1.5\ \text{m}$ from the pivot on one side. How far from the pivot, on the other side, must a $30\ \text{N}$ weight hang to balance the beam?

The beam's own weight acts at the centre (the pivot), so it has no moment. Setting clockwise $=$ anticlockwise moments:

Weights on a rule balanced at a pivot — used to test the principle of moments

Vector triangle

Three forces in the same plane that are in equilibrium can be drawn as a closed vector triangle 矢量三角形 — drawn tip-to-tail, the three arrows come back to the start. This is a drawing method instead of splitting into components 分量.

Three forces in equilibrium form a closed vector triangle

Use the sine rule 正弦定理 or the cosine rule 余弦定理 on the triangle to find unknown sizes or directions, or draw the triangle to scale on graph paper.

You can also split each force into horizontal 水平 and vertical 竖直 components and set $\sum F_{x} = 0$ and $\sum F_{y} = 0$.

An iceberg floats with most of its volume hidden: ice is slightly less dense than water.

Density 密度 is the mass per unit volume:

Unit: $\text{kg m}^{-3}$ (or $\text{g cm}^{-3}$; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$). Density is a scalar 标量.

Some useful densities to know:

• water: $1000\ \text{kg m}^{-3}$
• air at room conditions: $\sim 1.2\ \text{kg m}^{-3}$
• iron / steel: $\sim 7800\ \text{kg m}^{-3}$

Pressure 压强 is the force per unit area, where the force acts at right angles to the surface:

Unit: $\text{Pa} = \text{N m}^{-2}$. Pressure is a scalar.

A precision aneroid barometer measures atmospheric pressure

Hydrostatic pressure

Take a column of fluid 流体 with density $\rho$, cross-sectional area 横截面积 $A$ and height $\Delta h$. Its weight is

This weight presses down on the area $A$ at the bottom, so the extra pressure at the bottom compared with the top is

This is the hydrostatic pressure 流体静压强 equation. It depends only on the density of the fluid and the depth 深度 — the shape of the container does not matter.

Worked example. Find the extra pressure due to the water at the bottom of a swimming pool $2.5\ \text{m}$ deep. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

A column of liquid of area $A$: its weight sets the extra pressure at the depth below

For a submarine at depth $h$ below the surface, the pressure from the water is $\rho_{\text{seawater}}\, g\, h$. For the total pressure, add the atmospheric pressure 大气压强 at the surface (about $1.0 \times 10^{5}\ \text{Pa}$).

Upthrust and Archimedes' principle

When an object is submerged 浸没 in a fluid, the pressure at the bottom of the object is greater than the pressure at the top (by $\rho g \Delta h$, where $\Delta h$ is the object's height). This difference gives a net upward force called the upthrust 浮力.

Upthrust arises because the pressure on the bottom of the object is greater than on the top

For an object of volume $V$ (the volume of fluid displaced 排开), the upthrust is

This is Archimedes' principle 阿基米德原理: the upthrust on a body in a fluid equals the weight of the fluid it pushes aside.

Worked example. A metal block of volume $2.0 \times 10^{-3}\ \text{m}^{3}$ is fully submerged in water. Find the upthrust on it. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

For a fully submerged object, $V$ is its full volume. For a floating 漂浮 object, $V$ is only the volume below the surface — the object floats when the upthrust on the part below the surface equals its weight.

A floating object sinks until the upthrust on the submerged part equals its weight

Force balance with upthrust

A block held under water by a string tied to the bottom of the container is in equilibrium under three vertical forces: weight (down), tension 张力 (down), upthrust (up). Set $F_{\text{upthrust}} = W + T$ to find the tension.

A submerged block hanging from a newton meter 弹簧测力计 reads less than its weight in air, because of the upthrust: reading $= W - F_{\text{upthrust}}$.

The upthrust depends on the fluid density and the displaced volume, not on the object's material or depth (for an incompressible 不可压缩 fluid). On a planet with smaller $g$, the upthrust is smaller in the same ratio as the weight, so a floating object still floats with the same fraction below the surface.

• Take moments about a chosen pivot; for equilibrium, total clockwise $=$ total anticlockwise moments and the resultant force is zero.
• A body in equilibrium under three forces gives a closed triangle of forces.
• Fluid pressure $= \rho g h$; upthrust $=$ weight of fluid displaced (Archimedes).
• Distinguish mass, weight and density, and give the base unit each time.