- Kinematics
- Dynamics
- Work and Energy
- Pressure
- Significant Digits in Numerical Values
- Properties of Matter
- Dimensional Analysis

The **position** of a particle may be given as a function of time: x(t). The standard unit for measuring position is **one meter** (1m). The standard unit of time is **one second** (1s).

The **velocity** of the particle is: v(t) = dx(t)/dt. The standard unit for velocity is one meter per second (1m/s).

The **acceleration** of the particle is: a(t) = dv(t)/dt = d^{2}x(t)/dt^{2}. The standard unit of acceleration is one meter per second per second (1m/s^{2}).

We may plot graphs of x(t), v(t), and a(t) against t. Then v(t) is the slope of the graph of x(t), and a(t) is the slope of the graph of v(t).

Suppose the particle starts at time t = 0 with initial position x(0) = 0, initial velocity v(0) = 0, and constant acceleration a(t) = a. Then at time t_{1} the velocity of the particle is given by

v(tand the position of the particle is given by_{1}) = Integral(0 to t_{1}): a.dt = a.t_{1},

x(tGraphs of x(t), v(t), and a(t) satisfying these conditions are shown in Fig. 1._{1}) = Integral(0 to t_{1}): v(t).dt = Integral(0 to t_{1}): a.t.dt = (½)a.t_{1}^{2}.

Fig. 1. Graphs of the position x(t), velocity v(t), and acceleration a(t) of a particle starting from rest at t = 0, x = 0 with a constant acceleration.

At the surface of the Earth all bodies fall with the same constant **acceleration due to gravity**, g = 9.8m/s^{2}, except when air resistance is important. Air resistance is important for light objects (such as the leaves of trees) and heavy objects falling at high speed.

**Newton's Law**: *If a force F acts on a body of mass m, then the acceleration of the body is given by F = m.a, where a is the acceleration.*

The standard unit of force is **one newton**: 1N = 1kg.m/s^{2}.

The **weight** of a body is the force on the body due to gravity: W = m.g.

**Friction** is a contact force that opposes the relative motion of two surfaces. If a body resting on a surface is pulled along at a constant speed (zero acceleration), then there is a friction force f equal and opposite to the applied force F, as shown in Fig. 2.

Fig. 2. The applied force F on a body in contact with a surface, and the friction force f.

The **work** W done by a constant force F which moves a body through a distance x in the same direction as the force is defined by

W = F.x.

The standard unit of work is **one joule**: 1J = 1N.m = 1kg.m^{2}/s^{2}.

By Newton's law, when there is no friction we have F = m.a. Therefore, the work done is W = m.a.x. After a time t_{1} the work done is given by

W = m.a.(½)a.t_{1}^{2}= (½)m(a.t_{1})^{2}= (½)m.v^{2}.

The quantity (½)m.v^{2} is called the **kinetic energy** of the body. The kinetic energy is a measure of the amount of work needed to increase the speed of the body from zero to v.

The work done to lift a body of weight m.g through a height h is m.g.h. The body gains **potential energy** V = m.g.h. If the body falls back from the height h, then it gains kinetic energy (½)m.v^{2} = m.g.h, and loses its potential energy. The sum of the kinetic energy and the potential energy remains constant. This is called **the conservation of energy**.

The general principle of conservation of energy states that *energy cannot be created or destroyed*. If energy seems to disappear, it is in fact changed into a different form. For example, if a falling body hits the ground, then its kinetic energy is converted into random vibrations of the particles in the body. These random vibrations make the body hotter.

When a force F is applied at right angles to a surface, and is distributed over an area A of the surface, we define the **pressure** P on the surface to be the force per unit area: P = F/A, see Fig. 3. The standard unit of pressure is **one pascal**: 1Pa = 1N/m^{2} = 1kg/m.s^{2}. Another common unit of pressure is **one atmosphere**, which defined as follows: 1atm = 101.325kPa.

Fig. 3. Pressure: P = F/A.

Numerical values obtained from measurements generally have some uncertainty. For example, the result of a measurement may be 15.6m with an uncertainty of 2%. Since 2% of 15.6 is approximately 0.3, the result of the measurement is 15.6 ± 0.3m. The true value is likely to be between 15.3m and 15.9m.

Instead of an explicit statement of uncertainty, the accuracy of a result may be indicated by the number of digits in the value given. We say that 15.6 has three **significant digits**, with the understanding that the last decimal digit 6 may not be completely certain. The addition of more digits (for example, 15.620478) would be meaningless.

Sometimes the number of significant digits in a written value is not clear. For example, suppose the measurement of a distance is given as 1200m. If the accuracy is 100m, then there are two significant digits, namely 12; the two zeros 00 are added only to make up the correct magnitude of the measurement. However, if the accuracy is 10m, then the first zero is significant but the second zero is not. If the accuracy is 1m, then both zeros are significant. One way of showing how many zeros are significant in this example is to write 1.2km. or 1.20km, or 1.200km.

In multiplications and divisions, the number of significant digits in the final result should be the same as the number of significant digits in the least accurate factor.

(36.479×2.6)/14.85 = 6.3868956 = approximately 6.4. Although extra digits are kept in the intermediate steps of the calculation, the final result has only two significant digits, because the original factor 2.6 has only two significant digits.

In additions and subtractions, the number of digits after the decimal point in the final result should be the same as the smallest number of digits after the decimal point in the terms of the sums or differences.

17.524 + 2.4 - 3.56 = 16.364 = approximately 16.4. The final result has only one digit after the decimal point because the original term 2.4 has only one digit after the decimal point.

- Find the volume V of a circular cylinder with radius r = 1.26cm and height h = 7.3cm. (V =
*pi*.r^{2}.h) - Find the sum: 0.056×10
^{2}+ 11.8×10^{-1}

Matter usually has three phases: **solid**, **liquid**, and **gas**. Liquids and gases are called **fluids** because they flow. A solid has a fixed shape; fluids have no fixed shape. A liquid has a definite volume, and rests at the bottom of its container. A gas expands to fill the whole volume of its container.

The atoms in a solid vibrate about fixed equilibrium positions. The atoms or molecules in a liquid move about freely and collide frequently with each other. The atoms in a solid or liquid are closely packed, which makes it difficult to reduce their volume by pressure. Therefore, solids and liquids are almost incompressible. The atoms or molecules in a gas are far apart, and they collide less frequently than the particles in a liquid. Gases can be compressed easily.

The average **density** *rho* of an object of mass m and volume V is defined as: *rho* = m/V. The standard unit of density is one kilogram per cubic meter (1kg/m^{3}). Densities of some common substances at 0°C and one atmosphere (except where shown otherwise) are given in Table 1.

Substance | Density |
---|---|

Air | 1.29kg/m^{3} |

Hydrogen | 0.09kg/m^{3} |

Helium | 0.18kg/m^{3} |

Steam(100°C) | 0.6kg/m^{3} |

Water(4°C) | 1.000×10^{3}kg/m^{3} |

Seawater | 1.025×10^{3}kg/m^{3} |

Ethanol | 0.8×10^{3}kg/m^{3} |

Mercury | 13.6×10^{3}kg/m^{3} |

Gold | 19.3×10^{3}kg/m^{3} |

Aluminum | 2.70×10^{3}kg/m^{3} |

Iron | 7.86×10^{3}kg/m^{3} |

Wood | 0.43×10^{3}kg/m^{3} |

A small change dP in pressure causes a fractional change in volume dV/V. The **compressibility** *kappa* of the material is defined to be:

There is a negative sign because an increase in P causes a decrease in V. Compressibilities of some common substances are given in Table 2.kappa= -(1/V).(dV/dP).

Substance | Compressibility |
---|---|

Steel | 7.14×10^{-12}m^{2}/N |

Aluminum | 14.3×10^{-12}m^{2}/N |

Mercury | 0.385×10^{-9}m^{2}/N |

Water | 0.476×10^{-9}m^{2}/N |

Air(1atm) | 9.87×10^{-6}m^{2}/N |

The compressibilities of liquids are almost as low as the compressibilities of solids, which shows that the atoms in a liquid are almost as close to each other as the atoms in a solid. The compressibilities of gases are larger and depend on pressure.

- What is the volume of 1kg of water at 4°C and 1atm pressure?
- What is the change in volume of 1kg of water in a pressure change from 1atm to 2atm at 4°C?

Consider a column of liquid with density *rho*, height h, and cross-section area A, see Fig. 4(a). The weight of the column is W = m.g = A.h.*rho*.g, where g is the acceleration of gravity; g = 9.8m/s^{2}. Let P_{a} be the pressure of the atmosphere at the top surface, and let P be the pressure in the liquid at depth h. The weight W of the column of liquid is supported by the net pressure force (P - P_{a})A. Equating W and (P - P_{a})A we get

P = P_{a}+rho.g.h.

Fig. 4. (a) Vertical pressure forces on a column of liquid. (b) A liquid barometer.

In a liquid barometer for measuring atmospheric pressure the liquid column is supported by the pressure at its base, see Fig. 4(b). The height of a column of mercury that supports a pressure of one atmosphere is given by:

h = (P_{a})/rho.g = (1.013×10^{5}[N/m^{2}])/(13.6×10^{3}[kg/m^{3}]×9.80[m/s^{2}]) = 0.760m.

What is the height of a column of water which supports a pressure of 1atm?

An object completely covered by a liquid seems to weigh less than it does in air. This is because fluids exert an upward force on the object. This force causes wood to float on water, and helium-filled balloons to rise in air. The upward force exists because the pressure in the fluid increases with depth, see Fig. 5.

Fig. 5. Forces on an object completely covered by a liquid.

If the object is replaced by the surrounding fluid there is equilibrium. This shows that the upward force F caused by the pressures on the surface of the object of volume V must equal the weight of the fluid with the same volume. Therefore F = *rho*.V.g. This is called **the principle of Archimedes**.

A object made of metal, which looks like gold, has mass 3kg and an apparent weight of 26N under water. What is the density of the object? Is it made of solid gold?

Let *m*, *V*, and *rho*_{m} be the mass, volume, and density of the metal object. Let *rho*_{w} be the density of water. The apparent weight W' of the object equals the real weight m.g minus the weight *rho*_{w}.V.g of the water with the same volume as the object:

W' = m.g -Thereforerho_{w}.V.g.

V = (m.g - W')/(But the density of the metal isrho_{w}.g).

Since the density of gold is 19.3×10rho_{m}= (rho_{w}) / (1-W'/m.g) = 10^{3}[kg/m^{3}] / (1 - 26[N]/(3[kg]×9.8[m/s^{2}])) = 8.6×10^{3}kg/m^{3}.

A 5kg ball of density 6g/cm^{3} is hanging from a thin string. The ball is completely under water. What is the tension in the string?

Each derived unit in mechanics can be written in terms of the basic SI units of **mass** one kilogram [kg], **length** one meter [m], and **time** one second [s]. These factors indicate the **dimensions** of a physical quantity.

- Area has dimensions (length)
^{2}and basic units [m^{2}]. - Velocity has dimensions (length)(time)
^{-1}and basic units [m.s^{-1}]. - Force has dimensions (mass)(length)(time)
^{-2}and basic units [kg.m.s^{-2}].

An equation such as A = B + C has meaning only if the dimensions of all three quantities are the same. For example, we cannot add a distance to a speed. All equations must be *dimensionally consistent*.

Let s be the distance moved in a time t by a particle which starts from rest and has acceleration *a*. Then s = (½)a.t^{2}. On the left s is a distance having units [m]; and on the right we have a.t^{2} having units [m.s^{-2}][s^{2}] = [m]. Both sides have the same dimension of length with basic unit [m]. Therefore the equation is dimensionally consistent.

The periodic time *tau* of a pendulum of length *l* and mass *m* is assumed to be given by

where k is a dimensionless constant, and g is the acceleration of gravity. We can use dimensional analysis to find x, y, and z, and the equation fortau= k.m^{x}.l^{y}.g^{z},

The left side has dimension time, and basic unit [s]. The right side has dimensions (mass)^{x}(length)^{y}(acceleration)^{z}, and basic units [kg^{x}][m^{y}][m^{z}s^{-2z}] = [kg^{x}m^{y+z}s^{-2z.}]. Since the basic units must be the same on both sides, we have

s = kgTherefore x = 0, y + z = 0, -2z = 1. It follows that y = ½, and z = -½, and the equation for the periodic time has the form^{x}.m^{y+z}.s^{-2z}.

The value of the constant k is not given by dimensional analysis.tau= k.l^{½}.g^{-½}= k.sqrt(l/g).

- Bernoulli's equation for a fluid, having density
*rho*and pressure P, and flowing with velocity v at height h, is:P +

where g is the acceleration of gravity. Show that the dimensions of each term in the sum on the left are the same.*rho*.g.h + (½)*rho*.v^{2}= constant, - The Van der Waals equation for a gas with pressure P and volume V at a constant temperature is
(P + a/V

where a and b are constants. Find the dimensions of a and b.^{2})(V - b) = constant,

*
By R. H. B. Exell, 1998. King Mongkut's University of Technology Thonburi.
*