Machinery's Handbook, 31st Edition
Mechanics 159 As mentioned earlier, Newton’s second law of motion states that force is proportional to mass times acceleration. Because an unsupported body on the earth’s surface falls with acceleration g (32 ft/s 2 approximately), the pound force (lbf) is that force which will im- part an acceleration of g ft/s 2 to a 1 pound mass (lbm). Similarly, the kilogram force (kgf) is that force which will impart an acceleration of g (9.8 m/s 2 approximately), to a mass of 1 kg. In the SI, the newton is that force which will impart unit acceleration (1 m/s 2 ) to a mass of one kilogram. The newton is therefore smaller than the kilogram (force) in the ratio 1: g (about 1:9.8). This fact has important consequences in engineering calculations. The factor g now disappears from a wide range of formulas in dynamics, but appears in many formulas in statics where it was formerly absent. It is however not quite the same g , for reasons which will now be explained. In the article on page 186 , the mass of a body is referred to as M , but it is immediately replaced in subsequent formulas by W / g , where W is the weight in pounds (force), which leads to familiar expressions such as WV 2 /2 g for kinetic energy. In this treatment, the M is really expressed in terms of the slug (page 157 ), a unit normally used only in aero- nautical engineering. In everyday engineers’ language, weight and mass are regarded as synonymous and expressions such as WV 2 / 2 g are used without pondering the distinc- tion. Nevertheless, on reflection it seems odd that g should appear in a formula which has nothing to do with gravity at all. In fact the g used here is not the true, local value of the acceleration due to gravity, but an arbitrary standard value chosen as part of the definition of the pound force and is more properly designated g 0 (page 157 ). Its function is not to indicate the strength of the local gravitational field but to convert from one unit to another. In SI the unit of mass is the kilogram , and the unit of force (and weight) is the newton . The following are typical statements in dynamics expressed in SI units: A force of R newtons acting on a mass of M kilograms produces an accelera- tion of R / M m/s 2 . The kinetic energy of a mass of M kg moving with velocity V m/s is 1 ∕ 2 MV 2 kg (m/s) 2 or 1 ∕ 2 MV 2 joules. The work done by a force of R newtons moving a dis- tance L meters is RL N·m, or RL joules. If this work were converted entirely into kinetic energy we could write RL = 1 ∕ 2 MV 2 and it is instructive to consider the units. Remember- ing that the N is the same as the kg·m/s 2 , we have (kg·m/s 2 ) m = kg (m/s) 2 , which is obvi- ously correct. It will be noted that g does not appear anywhere in these statements. In contrast, in many branches of engineering where the weight of a body is important, rather than its mass, using SI units, g does appear where formerly it was absent. Thus, if a rope hangs vertically supporting a mass of M kilograms the tension in the rope is Mg N. Here g is acceleration due to gravity, and its units are m/s 2 . The ordinary numeri- cal value of 9.81 will be sufficiently accurate for most purposes on earth. The expression is still valid elsewhere, for example, on the moon, provided the proper value of g is used. The maximum tension the rope can safely withstand (and other similar properties) will also be specified in terms of the newton, so that direct comparison may be made with the tension predicted. Words like load and weight have to be used with greater care. In everyday language we might say “a lift carries a load of five people of average weight 70 kg,” but in precise tech nical language we say that if the average mass is 70 kg, then the average weight is 70 g N, and the total load (that is force) on the lift is 350 g N. If the lift starts to rise with acceleration a m/s 2 , the load becomes 350 ( g + a ) N; both g and a have units of m/s 2 , the mass is in kg, so the load is in terms of kg·m/s 2 , which is the same as the newton. Pressure and stress: These quantities are expressed in terms of force per unit area. In SI the unit is the pascal (Pa), which expressed in terms of SI derived and base units is the newton per meter squared (N/m 2 ). The pascal is very small—equivalent to 0.15 3 10 − 3 lb/in 2 — hence the kilopascal (kPa = 1000 pascals) and the megapascal (MPa =
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