Exam 4 solutions

Exam 4 solutions - KEY : Exam4 NAME: ‘ cwm: 1. (30...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: KEY : Exam4 NAME: ‘ cwm: 1. (30 points) Sketch the shear and bending moment diagrams for the beam shown below. Determine tHS a'bsglute values and their locations of the shear and bending moments_ FBI): I 6001b 2Fx : fix": 0 4000 lb'ft 62MB -—= ~(eoo)oo)_ 4000 +20 B -=. o 60004-400 D = [0000 " 20 20 l!)I .~... (moxie): [000 (“:4 Fl 2 -(s'oo)l5)-.- .1500 lL-m. Q: = «(wish -).S'00 IL. (1,)- IV‘M‘f 500W») losxsw M = ZS'ODlL-fi-l- x=l$ l Mme /(d+ 2. (30 points) Determine the tension in each segment of the cable shown below and th bl ’ t 1: II th. 9 eca erBc‘; 515:1?“ Came E7 FBI) Sejmenf)‘ AB 4m % I 15 kN GENE: 4(5)+h,(12)—'2(9)= 0 5 ‘TL-ZO 7c. 3m smfism Ax ‘—‘——-1 :T-i ZFxf-“Ex-hm: 0 fix: GZME: -DY(|9)+4KN(IS)+IS'W(IO) Est-4 mama + RKN (1.) = O + 9Y2 G°fi§°+é=¥fi 92Fx=TBCC089Bc—c.33=o - p - n, KN 1‘ Tm. Cosbhcr 6.3) KM (D 7 .. Trig = 974 493+ E1: 0 “it, = 11"4-Tksm9bc: 0 57 7. 11-41 ‘ Tm, sham: 9K» C3) EV... '0 KNT ®+ ® $419}: 9 't n '-'-‘ "" o Cosg‘c .— 4 8°C \I = Y - Yunnan; __ o B : YE“ S’t‘an $1.01!;9 9132: 3"; S" 6.315. :: 5.68%1lm T361: c—vfl-Vos Swan: = Tm: 1‘0'7-035' _|-- - TN: 10.7.0 luv 3 FBI) Pain"? H‘ 12 kN X 633kN A 9A3 '= Tan fine-RB 1: 0 TM TAB Sun 2 © " ‘ ‘7. o Ckeck: VB: 3.1mm“: 3 +a... 62.;79'7"; 51539.1... \/ FR 'P 'n+ C o ,f b m 2F”: theme”... 10.2035‘cos5'I-6315 -- 0 102:6) CL 94:61) cos. ech :- J @ T*‘-" COhS’h ‘- II- ‘ ' o.- a m 2;}, - T6,, smecb+ mom VI 6325' i9 0 *1;b g; “gab :. 6.4a!an @ ©+© 15me mm“ 2.5 9Cb=418613° C u ""' “63"7? 3'3" ' 6.33 ' Tab COS 47.96230 5 I}, FED Po'm')‘ E ' _ 'm EFX = 6,}3—‘rabcosesbz 0 7 _ TE: C°Yeeb= 6‘33 “6) E :3 = 10-11, may 0 TED Tab S"m 955 2' I 0 ® ®T© tune) = =§ eED=§7.¢sze°_ Eb , . _ 6.33 _' ___ t '84 RN Te» “cm - "33"" m 4 Yb ~: ?_ +0.“ 0% =-. atmsmszfi = 3.571 m C *9“: Yb -= x, C - 9 the“: Irena-1.9623": 3.1774... / - 51nng :: m = (mm m 13c :JGc‘YQ-L'FSI =W= 9.05733.“ ch: W=W= “£24m Q»: “JV: +11 = W: 3.737%m DrHemm-Hve'y) 4; Yb .6M» RHB ‘ 2 “SWA- : \/ “"993 fm 61.l75‘9° Q ~.—. Yc-Ys _ \1—5’.69+7. BC - _—-—..__—..‘—. -,- 9.0mm / SM 65¢ S’in 5163250 ‘ R :7. VC “Vb :- — Cb S‘m 9c.“ 51“ 4135230 - \‘qu‘Hm J z \t b ‘ QM: '7'" = 3.737% m / S‘m 9n S"m 97.6525 L = kcb+QbE = s .42») 1— 9.055314.— mww‘lms = 30.1447 3. (40 points) The static coefficient of friction between the crate and the ramp is us: 0.35. Assume that the crate’s center of gravity is located at the centroid. i) Find the value of P for impending sliding motion up the plane. ii) For what range of heights above the plane may this force be applied if tipping is not to occur? iii)lf the force P is removed, will the crate tip over in a counterclockwise sense? Prove. If the crate will not tip, determine the location of the resultant normal force relative to the left side of the crate. iv)lf the force P is removed, will the crate slide down the ramp? Prove. mm) x L P m F=fiyN 900m 30 2F“ :: N +Ps1nZOO-Wc 0530"; O N .. wcoszot Fs‘mlo" ZFt': Peosloo— F- UUPWBOO: 0 Va “loo—)4th ~Ws~in30°= 0 Y’c «202% S (Wcoc‘so°— Psi n20°) ' "~— W r‘m‘SOO = 0 = M3 = (looks)(‘i.%oL-;—"=-,)x Ham 0 o i=(co 920° +fls s". do" 5 == “30 Mum” ) F_ W(S"ivBO° +J4 gcm‘so" )_ (m I.2N)(sin30°+ 352-0930") " o - o " “we‘r- co§20 +fig§M20 c0510 +.35"s‘m20 Fey no +‘P‘s‘enSJN é F ac'l- 4+ +ke 63M"? Commev a. . - - MEDI- - ~Pc0510 do N o 0 3° 5 -_-.w (gooszn‘so Ms'ocosEO ) r — MA 3' i C0510 kw“: 2 (mbiaxeoosdo’ +450cofio )nqég 106M”. 5 \436.'l4~coslo° r a“. no +:‘~?'.nj ‘por For imrenéanj ’6‘.qu F 5% N ac?" at“- \~e¢4— Garner) b g: M L -= w s‘m30°(Coo)—wcos‘so°(4§o) u t w ($0051n300~450c05300 ) -= HQRN (goolswnBDO—dks'o Mao”) 1' "175)‘141 N‘W‘" “ 1350201... I N'M Fe»: u? uBfi"“ 1M -.-, 475342 N-m. +Nx=0 ukgve N=Wcor30°= "£9.49 N ‘15’ “m- =10}.me ><" m‘fi’s‘ .For *‘Mvené;hj 40% +L? Ham; -F = F: fix N k5 F .ZFnt-ZNHWCOS‘SOo: O N: Wc09300= (\‘lb\l)c05300 N = IMBASN FM «,6 :- .‘s§(\c%.+§)-.:. S‘I‘n‘l’f? M / L‘A' “M43?! F 3S YQcKJcJ} :Ft 2 "F —- UUg‘mgo =-_ O .0. o F ‘-‘- WS‘MBO =1 HHlsinBD ...
View Full Document

Page1 / 6

Exam 4 solutions - KEY : Exam4 NAME: ‘ cwm: 1. (30...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online