Mat 5, 1911] 



SCIENCE 



705 



generally, which make M = W/g. If stu- 

 dents are taught in elementary mechanics that 

 m = 32.174W/5' and in engineering that 

 M = W/g, the resulting confusion will be as 

 great as that caused by the poundal and the 

 gee-pound. 



(p) " The equation F = mA will give incor- 

 rect results if the forces and masses are given 

 in any but absolute units. In particular it 

 should never be used in the ordinary problems 

 of engineering." 



The reasoning that leads up to this con- 

 clusion would also condemn the use of the 

 time-honored formulae FT^MV and FS = 

 1/2MV\ 



The equation F = 3IA is given in all engi- 

 neering text-books that deal with problems in 

 dynamics, and it has been used from time 

 immemorial with English units and with per- 

 fect accuracy. 



(g) " On account of the special character of 

 this equation, F = MA, it is unfortunate that 

 it should be so often taken as the funda- 

 mental equation of dynamics, instead of the 

 general equation F/W ^A/g from which it 

 is derived." 



As commonly used by engineers the equa- 

 tion F = MA has no more a special character 

 than the equations FT = MV and F8=. 

 1/2MV'. It is just as fundamental as 

 F/W^A/g, which as well as F^MA may 

 be derived from FT^MV. If in the last 

 equation we take Y/T^A, then F^MA; 

 and if M=W/g, then F/W = A/g. 



Whether a certain equation should or should 

 not be used in engineering depends, (1) on 

 its logical correctness, (2) upon its usefulness. 

 The three equations F = MA, FT = MV and 

 FS^1/2MV\ are all logically correct, 

 equally in the so-called " absolute " (C.G.S.) 

 system, the poimd-foot system, and the kilo- 

 gram-centimeter system, provided that in the 

 C.G.S. system the unit of quantity of matter 

 is the gram and the unit of force is the dyne, 

 or 1/981 of the force with which gravity 

 attracts a gram of matter at latitude 45° ; 

 that in the pound-foot system M = W/32.2 

 and that in the kilogram-centimeter system 

 M =■ W/981, W being the weight of the body 



in pounds (or kilograms) and the definition of 

 weight being the quantity of matter in a body, 

 or the force with which gravity attracts it at 

 latitude 45°. As to the usefulness of the 

 equations, this has never heretofore been 

 doubted. They have filled the engineer's need 

 for a set of handy formulae for accelerated 

 motion, and they are easily understood by the 

 student. 



More than ten years ago a high-school stu- 

 dent in despair over his problems in dynamics, 

 on account of the obscurity of his text-book 

 and of the teacher's explanations, appealed to 

 the writer for assistance. He was told to for- 

 get the formulae of the text-book, with its 

 poundals and units of mass and to memorize 

 the following : V = -yWh S = 1/2VT, FT = 

 MV, F = MA, FS = 1/2MV and M=W/g, 

 F being force in pounds, M nothing else 

 than W/g = W/32.2, or W/32.174 (if great 

 precision is needed) (no " concept of mass " 

 needed) and W weight in pounds (quantity 

 of matter as weighed on a platform scale). 

 With these equations the student soon solved 

 every problem in the book that referred to 

 bodies uniformly accelerated. Many times 

 since the writer has had occasion to give the 

 same advice, and always with the same result. 



Here is a simple problem in acceleration, 

 with its solution by the engineer's method and 

 by the method of the committee's report. 



What is the draw-bar pull required to accel- 

 erate a railroad car whose weight is 100,000 

 pounds, on a level track, in latitude 30°, at 

 the rate of 1 foot per second per second, fric- 

 tion neglected? The engineer's solution: 

 F=.MA, M = W/g, A = (V^—V,) -i-T 

 F = 100,000 -f- 32.2 X 1/1 = 3,105.6 pounds. 

 The mathematician's solution: First look up 

 the value of g for latitude 30°, =32.131. 

 In latitude 45° 5r = 32.174. Weight is the 

 force with which gravity attracts a body at a 

 given place. The unit of force (pound) is the 

 force with which gravity attracts the standard 

 pound at latitude 45°, where jf^ 32.174. A 

 weight of 100,000 pounds at latitude 30° is 

 therefore a force of 100,000X32.131/32.174 

 pounds, = W. " The formula F = MA should 

 never be used in the ordinary problems of en- 



