HW 16 Solution

# HW 16 Solution - PROBLEM 5.25 A T50 = g unifunn ﬂ mm as...

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Unformatted text preview: PROBLEM 5.25 A T50 = g unifunn smelredishcminmacimularmofradiuasoﬂ mm as shown-Therodissupporbedbyapin utA andthe curdBC.Determinn (a}1he tension in the cord, (1:) the reaction at A. SOLUTION ‘ D Firstnote, from Figm'e 5.83: J? = 0’5 “Ema 1.5 =—-—m v: = mg = (0.?5 kg){9.31m!az) = 7.353 N Also note that A ABD is an equilateral uimla Equilibrium then (squirm (0}ZMA = D: [0.5 m - [1?5 m]cos30“](7.358 N) — [(0.5 mm soﬂrx = n or rm 21.4693N or Tm, 21.470»: 4 (b) ESP: 2 0: A, + [1.4698 Njcosﬁﬁ" = 0 or A; = 43.734911 2.3 = m 4y — ”LESS-N +{l.4698 N]sin60° = 0 or Ay=6.085N thus A=6.13N 5.33.1“ d PROBLEM 5.23' ' - : ' an: lhehpmngenemgswimiﬁCDishmtasshmmdisaﬂmhedwahmge at C. Determine the lengthl. forwhichtheportionBCD ofthe wire is horizontal. SOLUTION Fixst mm: that for equilibrimn the center ofgravify ofthe wire must lie on a vertical line through C. ther. because them is hambgeneous, its center of gravity will coincide. with the centroid ofd'le curl-Impending line. Thus; EME- = 0, which implies that f = I} ﬁsh-=0 gu) (—3 m.}{s in.) + [4; in.)[lﬁ in.) = o L2 #144 in2 or;L=12.00in. 4 PROBLEM 5.58 ' \ E f -. Dam-name mec'apacity, in gallona,'ofﬂ1epmh bowl shown if R =12 in. - L—u—A‘ SOLUTION The volume can be gmted by rotating the triangle and circular sector shown about the y axis Applying the second theorem ofPappus~GuIdinus mdusmgFig 5 am, we have Since 1331 = 231m3 . 3526.03 in’ V = —-—-— =1526 23lin3fgal 3‘1 V=15.26gaH PROBLEM 5.61 For the beam and loading showi'l. detemiine (a) the magnitude and mu lel‘r location of the resultant of the distributed load, {b}tl1e reactions at'the beam supports. - SOLUTION .- Resultam R = R1 + R:- (:1) Have R] =‘(40lh’ﬁ)(l8ﬁ'} = 7201b I 122 = %{1201wn}(13 ft) = 1080113 or R = 18001]: The resultant is located at the centmid C of the distributed load 3 Have +3313: [1800 ME = [401m)[1m)(9ﬁ)+%(120 Ibm](18 fl](12 ft] 01' E = 10.8011 ' R = 1300 11:4 3 = 10.3% (b) ”amaze: A,=u +125 :0; Ay—tsoulbzo, Ay=13001b A=18001b§4 +321 A = o: M,{ —(1-suolb}{10.s ﬂ] = 0 MT =19.4441b-ﬁ or M{ =19.44kip-ft ) 1 ...
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