. A textbook on electric lighting and railways. International correspondence schools, Scranton, Pa. v. 1-3, 5 . ig. 8 correspond to the two weights of 20 pounds inFig. 9, and if we treat the 200 horsepower as weights andfind their center of gravity, that center of gravity will bethe center of load or the correct location for the powerhouse. Since the two loads or weights are equal, the centerof gravity or load must, as in Fig. 8, be at point A, midwaybetween O and O § 20 ELECTRIC RAILWAYS. 19 21. Take another ease. Suppose that there are threeweights (Fig. 10): /r= 10 pounds; IC = 50 pounds; a
. A textbook on electric lighting and railways. International correspondence schools, Scranton, Pa. v. 1-3, 5 . ig. 8 correspond to the two weights of 20 pounds inFig. 9, and if we treat the 200 horsepower as weights andfind their center of gravity, that center of gravity will bethe center of load or the correct location for the powerhouse. Since the two loads or weights are equal, the centerof gravity or load must, as in Fig. 8, be at point A, midwaybetween O and O § 20 ELECTRIC RAILWAYS. 19 21. Take another ease. Suppose that there are threeweights (Fig. 10): /r= 10 pounds; IC = 50 pounds; andIF = 10 pounds; further, suppose that the distance fromIf to Jf is G miles; from W to W, 7 miles; and fromjr to Jl, 1 miles. Where is the center of gravity sit-uated ? The way to ascertain this is to first find the centerof gravity between any two of the weights, and then, sup-posing the sum of the two weights to be situated at thispoint, to find the center of gravity between this and thethird weight. Let us first find the center of gravitybetween weights IV — 10 and IV = 10, where the distance -Qr 40. Fig. centers is 7 miles. This distance of 7 miles mustbe divided into 2 parts, such that JV X I— W X /, where/ and / are the distances of W and \V\ respectively,from the center of gravity for these two bodies. To solvethe problem graphically, lay out the plan to scale on paper;that is, represent the 7 miles by 7 inches, and so on, andlet a difference in the sizes of the circles represent thedifference in weights, as shown in the diagram. Call L thedistance from fF to J/, and let the distance from \V tothe center of gravity, t(j be found, be represented by /; thenthe distance of W from the center of gravity will berepresented by the difference, or L—l\ and since JFx /;= IF X {L-l), we have /F/= \V L - IV/, or IV L 20 ELECTRIC RAILWAYS. § 20 IV L= IV/-^ JJ^ /, and /= ,„■ Substituting for the weights 7 X 10and for L the numerical values given, we have / =
Size: 2018px × 1239px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1900, booksubjectelectri, bookyear1901
