HW_solu_09-27-04

HW_solu_09-27-04 - pm“: TN- WML 0‘? .W’B'D pMiL-ELJV...

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Unformatted text preview: pm“: TN- WML 0‘? .W’B'D pMiL-ELJV Satin i Far 3-) HH. (-41.0. .0043 chasm [mats 1:0,: '-" m3 : at N 1T 1:35 m l w m Jaw)? 231nm =9 mix («J/3+ raw $0=(— .3m($ui\e).’l‘ + .3m(m9)j))x4?81331fi} 90= .3Nfi3M(;§.‘c~t')L—._3m F (ABC :3 r : 475' N “A. F" 9790’?- mouflMo pull Hi railw- f TE ‘35! .L Efifieo=> ficx (-003) + r: x (-529 L BummxsamSk L-SMIF' SSMOK K :3 r’ = 91W; m. i3 I” 415N491”? 9.1:.24' Uh 4.1 12 Given: Cellar door prepped open as shown; Weight of door = 40 lb Find: Force in the supporting stick Solution: Free body diagram: There is a force directed along QP exerted on the door at P; the weight of the door downward at W, and then couples and forces at each of the hinges as shown. We have six possible equilibrium conditions to use for finding F8. However, note if we choose to sum the moments with respect to O, we can eliminate many of the unknowns: 2M0 =Zrmi xF, +20, =O21.5flix(Fy,j+Fnk)—1.5fiix(Fy1j+Fz1k)+(0.9j+1.2k)?x(— 40m) +(~ 3i+1.8j+2.4k)x|Fs|((— 3i —2.2j+2.4k)/ 32 +2.22 +2.42 )+ Cy1j+Cz1k+ Cy2j+szk = o This is particularly true if we look just at the x-component for the moments with respect to O, that is, if we not the moment with respect to thex-axis = 0: (2mm: (Zrm xF,)-i+(ZC,)oi =0 2 (1.5m X(Fy1j+ Fz1k))oi — (1 .5fl‘i X(Fy1j+ Fz1k))-i +((0.9j + 1 21¢}: x (— 40tbk))oi + k— 3i+1.8j+2.4k)>< th|((— 3i — 2.2j+2.4k)/J32 + 2.22 +2.42 )1. i +(Cy1j+ 621k + 6,2] + szk)oi = o The first two terms can be seen to be zero, either by calculating the cross product or by noting the lines of action of the forces run through the x—axis, so the moment about the x-axis should be zero. The last four terms can be simply seen to be zero when the dot products are calculated. When the other dot products are calculated, we get 0 + o + (ask-40):? o lb +|Fs|k(1.8)(2.4)+(2.4)(2.2))/ 32 + 2.22 +2.42 J+ o = o r0? : o; I "IQ—3 cm 3'53) *"Q‘Ysrkfigb w: 553} ,a‘f-‘éoscfi -S«n55'jj t A . + w 3 ~40 k :rRT... " ~60Ib J: g H: xC-30l63)+ (00“ T _. __ _.__+F K n ' i Fm_l..._..x F51" on 0P. ‘\ llrO I A + C16+C1LL . .. K . A ‘ (dz k _ __:rt..)-(6V1\X Elji'EV-cym ‘3 H IA=O flw.~fbus&,ga.t@‘fi '75 2N0 1 u .A a Z *' 30"»;35 “I? +£9.55 ms wIS' x hwy-0 '* " 38" _ . fi __ {0.5, “1‘30.” W - .55 I 44506 f ) GO [30».35'Jsa t (57293530 _ i... a ‘V‘I‘MM . I : J HM 1.5ft ' 1.5 ft' 31: __ r‘ M Figurn mm Cot-1594. - MM In, (NJ :0 . -’->N+T-I'~{O(b=0 (D new}: z‘Ffiw 'N +T— [0015+0V=0 zfifia-b ‘Z FMKKE; + «9220.:0 _ => - 3m in we)? +(-v.§¢)x(-N3~)=o => -5T+ zoom +4§N=~0 C39 3x0) *CQ 9.7.S‘N -' (2.0/13 =0 6 M‘_’1°"’zs um U¢T=HO-fu “Mb mN-MWG'PG-LJ Map—Lu MW NOIBJSB thfisvut ...
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HW_solu_09-27-04 - pm“: TN- WML 0‘? .W’B'D pMiL-ELJV...

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