A complete and practical solution book for the common school teacher . two men at the hand-stick may carry f of the log, or twice as much as the one at the end of thelog, the hand-stick must be placed \ as far from thecenter of the stick, or 4 of 6 = 3 ft. .•. The hand-stick must be placed 3the log. ft. from the end of NOTE.—This seems to be more of a mechanical than an arithmet-ical problem, for the above is solved by a well known principle ofmechanics. PROBLEM 471. A string- is wound spirally 100 times around a cone 100 ft. in diam-eter at the base: through what distance will a duck swim in
A complete and practical solution book for the common school teacher . two men at the hand-stick may carry f of the log, or twice as much as the one at the end of thelog, the hand-stick must be placed \ as far from thecenter of the stick, or 4 of 6 = 3 ft. .•. The hand-stick must be placed 3the log. ft. from the end of NOTE.—This seems to be more of a mechanical than an arithmet-ical problem, for the above is solved by a well known principle ofmechanics. PROBLEM 471. A string- is wound spirally 100 times around a cone 100 ft. in diam-eter at the base: through what distance will a duck swim in unwindingthe string, keeping it taut at all times, the cone standing on its baseand at right angles to the surface of the water? Solution. (1) Let/; = height, r=radius of base, /=V (//2+r2)= slantheight, n = num-ber of spirals. (2) Let D, E be thetwo consecutivepoin t s in theducks path. (3) K,L = correspond-i n g consecutivepoints of tangencyof string to cone. (4) F, G = c o r r e s-ponding consecu-tive points in baseof cone. (5) Let CD=^, FK=x. (6) DE=^, ML=^.. FIG. 122. 238 FAIRCHIL&S SOLUTION BOOK. (7) ^2 = DN2+NE2 . . (1). (8) From the similar triangles GOF and DFE, r : GF=FD : DN. (9) .-. DN=()-^ . . (2). (10) From the triangles LMK and KFD, dx : MK=^ : DF. (11) FD=^.MK^^ . . (3). (12) MK : (/—*) = GF :/. (13) MK=[(/-j).GF]^/ ... (4). (14) But CF : x=2irrn : I. (15) .. GF : dx :: ^rn : l. (16) GY—^)^ ... (5). (17) (5) in (4) gives, MK=[2tt^(/—x)dx]-±l* ... (6). (18) (6) in (3) gives, ¥T>=[2irrnx{l—^)]-=-/2 . . (7). (19) (7) and (5) in (2) give, PN=[4tt W(/—x)xdx]+rl* (8). (20) The increment of FD is, (GH + NE) =FG+NE. (21) . (7), <tf(FD)=[2w/w(/— 2*)^]^/2 = (FG+NE). . (9). (22) (5) in (9) gives, NE=[—47rmxdx]-+l* . . (10). (23) (8) and (10) in (1) gives, ds-jf-^\+-±-(l-xYdx. XirVH CI I 7r2^2 (24) :, s = yPJ Jl+^L(/_^)2^=(4r/37r;0 + (2r/3)[7r;2—(2/7r«)]Vl+^2^2+2Hog(7r^+Vl+7r2?^). (25) In the problem, r=l, ;*=100. (26) .. J=(l/7
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