140 An Investigation of the Pressure sustained ly 



will be as ay— xy — —a^ + ax — ~ a;% the curve being supposed 

 to differ but little from a straight line : hence the fluxion of the 

 inclination will be as ayx—yxi— ~ a^ i + axi— — x'i, the flu- 

 ent ayx rV-"'^ — T ("-^^-^ ~ "^^ — ?" ^'j which requires no 



correction : and in the same manner the fluent of the ordinate 



will be found — ayx^ -y^^— t a^x^+ - ax^ — — x^, which 



must vanish when x—a, since the ends arc supposed to be abso- 

 lutely fixed, or 0= - ahj— -^ a'y — -^ ^'^ + -^ «■*— '^^ "■* — 

 ~ y o, and y = -^a, which is --■ of 2a, the whole pressure; 



3 



so that the two ends support — of the whole pressure, and leave 



-^ for the middle. 



B. In order that a flexible bar, equably loaded, may rest 

 equally on each of three fixed supports, their distance must be 

 •3472 of the whole length. 



If the half length be a, and the distance m, the strain, be- 



tween the supports, will be y«("^— ^) r(«— ■^)^j the incli- 



2 1 1 



nation, by taking the fluent, is found — amx— — ox* — T'^^'^'^ 

 — ax^—^x\ and the ordinate — amx^— --ax^— -r-a^x'' + 

 — «j;' x^i which must vanish when x = m, and -am-^ — 



6 24 ^ J 9 



ani — -ra^+—am — --m^, or r^am -a^—-- m^ must 



28 ^ 14 142 



vanish also; whence m^— ■^am= — 6a% m — ^a= + V-^o, 



and »n = '——^^~a = •6944a; that is, about — - of the whole 



length more than if the bar were composed of three pieces, and 

 each point supported an equal share. 



C. If a flexible bar, equably loaded, rest on five fixed points, 



at the distance of — of the lenglh from each other, and— from the 



ends, the pressures will be as 59, 52, and 58, or as 2107, 1857, 

 2071, 1857, and 2107; and if the middle support be removed, 

 the pressure on the remaining points will be as 11 and 21, or a? 

 1719,3281,3281,1719. 

 Calling the whole length 10a, and the pressure on the lateral 



fulcrums. 



