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Sol01

# Sol01 - PROBLEM 2.35 Knowing tl'nt a 35° determine t...

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Unformatted text preview: PROBLEM 2.35 Knowing tl'nt a: 35°, determine t]: resultant tithe the: forces shown. SOLUTION IDO-N Force: F; =+(l00 N)ms§5°=+3|.9l5N F? = {100 N)sin35° =—5T.353 N ISD-N Fume: F} =+(ISO b00665" = #33393 N F, =—{150 N)sin65“=—135.946 N 200-01 Force: F1 =—{20ﬂ N}0W35° =-163.830 N F? =-{200 N}sin35°=-l 14.115 N \$1.915 +£53.393 463.330 Rx =-l8.522 R, =-308.02 R =in + Ry] = (-18.52? Nﬁ+(—308.02 N” R tana'=—’ RI _ 300.02 18.522 11:86.59" 0 _ 300.02 N _ sin 86.559 ﬁrz—IB.522J_ R=309N 7 3645“ ‘ PROBLEM 2.49 Two forces Pandeappliedasshowntomajmmﬁ amneclion Knowing that Ihe connecﬁon is in equﬂihrhlmand Ihnt P=500N and 9:650N, determine the magnitudes of the forces exertedouthe rodsA andB. SOLUTION Free-Body Diagram Resolving the forces into x— and y—dimclions: R=P+Q+FJ +F3 :0 Substituting mmpomms: R = {500 M] + [(650 N)ous 50°]i — [(650 N)sin 50°]j +F,I — (F1 MSMI +{FJ sin 50"”: 0 In the y—direcﬁon {one unknown fmee) —500N — (650N)sin50° +FA sinSO“ = (I = 500 N +(650 N)sin 50° 513150” =13011ﬂN F4 {1550l~«I]n::t'.-.'.s,50"+Fﬂ —Fxou550°=0 F, = 1-1, 00550“ — (550 N1m55ﬂ° = (1302.?0N)00550”— {650 mam-50° =419.55N SOLUTION PROBLEM 2.1 11 weight of the plate. Dimemium :in mm A rectangular plate is suppa'ted by three cables as shown. Knowing that the tensim in cable AC is 60 N, detamine the We note that the weigtt of the plate is ecpal in wimde to the fame P exerted by the mppmt on Point A. We have: In Erma mm)I—(430mm)j+(360mm)k AB=6ﬂJmm A—C ={450mmﬁ-{4m mm)]+(36{l mm)k AC =1somm ={250 runny—{4H} mm” —(360 mm)k AD=650 nnn A3 8 . 12 . 9 Ta = Tulsa =73 — =['—' '—J+—k)Tn .43 I? I? I? IE T“ : Ta: 1“ : THE =(0.6I —0.64] +0.48k]TAC To 5 915 1.2 T =T i. =1" —= —‘——‘__k'f “3 “3"” ”AD (13' 13’ IS J‘” Substituting into the Eq. D" = I] and factoring i, j, k: s 5 [—ﬁrﬂmﬁnc +5149} I? 9.6 . +[—ﬁTﬂ _0'64TAC —FTJD +P]j 9 7'2 + —'I' +0.48?" ——T k=0 [IT a at 13 .10] Free Body A: PRIDE-LEE 3.25 Thtmnp ﬁEﬂimnﬂby-mblunmmﬂaﬂﬂ. Thctcmiminmdlnfﬂmm‘ﬂisﬁlﬂnﬂdmnﬁmﬂr Muhamdnfﬂmhmﬁmwmjﬂrmblcatﬂ, {bJﬂJccdﬂan M1 :13»: 1"511135 TEL-I ={2-3' mjj 1'DE =1DETDE _{D.ﬁm]i +{33mjj—{3 m1 [Ell] H] Ina]! +133]2 +{El-fm ={1m mi +[5'94 H11 — {5‘0 ”it i j 1: M1: 0 2.3 a 11-111 Luz 394 —5¢J =43: N -m]i—|f2d-E.4N -m]]: u" 1MJ =—ﬂ242N-m]i—[243N-m]l: «1 M1 = rm 3" 1:313 rm. =[2.‘Tm‘j +[2.3- m1] Tm =3;me = {a m]i+[3-.3- m1] —{3 m1 1E3]! +{3-3? +{3fm :4"): mi Hi“ HJj—{jd-D but i j 1: 11.1,. = 2.1 2.2. u M -m 4m: 59:; 4w =—(12¢2 H -m‘j “[1451! H -m]j+[1352 H my: [310”] cr 1M,I = 4m: M manna“ -m]j H1352 H -m‘ﬂ: I PROBLEM 3.1 13 PullevsA and B are mounted on bracket CDEF. The tmsion on each side of the two belts is as shown. Replace the four forces with a single equivalent force. and determine where its line of action intersects the bottom edge of the bracket. ISBN SD LUTIDN Equivalent force—coup le at .4 due to belts on pulley A Wehave 2F: —lZﬂN—lﬁﬂN=RA R A =23] N 1 We have EM .43 —40 Nail em} 2 MA MA = so New} Equivalent force—couple at 3 due to belts on pulley B Wehave 2F: {21ﬂN+lSﬂ mgr: 25°=Iq3 RB =3-50 b15125“ 5 l I We have DAB: —a-:1 N{1.5 em} = MB “v D M3=9oN-cm) FLJJ H ‘ l Equivalent force—couple at F 1Wehave IF: RF ={—28ﬂN}j+{3ﬁﬂN}{eosZS°i+sin25°j} 43252? N}i 4121.357 N}; R 2 RF = JR; + RE), = #92527? + {121.35 7? =3SCI.43N we Rm. 2 ml {421351) 326.2? =—21399° or RF=R=3SUNWQZL4°1 ‘J PROBLEM 3.156 A hlade held in a brace is used to tighten a screw at A. (a) Determine the tomes exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = —[30 N)i +Ryj + Rik and Mi = 412 N - m]i. (.5) Find the corresponding Tvalues of R! and 32. [e] What isthe orientation ofthe slot in the head of the screw for 1which the blade is least likely to slip when the brace is in the position shown? SOL UTIDN {a} Equivalence requires EF: R = B +C or —[30 N)i+Ryj+R(k=—Bk+[—Cxi+Cyj+Czk) Equaﬁngtheieoeﬁleients i: —3I}N=—C or CI=30N .‘I' Also EMJ: Mﬁ=rmxs+rmxc —{12 N - m]i = [{D.2 m)i +3115 m}j]x[—B}k +(o.4 m)i x [—(30 Nji + c}, j + Czk] Equating eoeﬁleients i: — 12 N -m = —{IEI. 15 m)B or B = 8D N k: D = [0.4 rn}Cy or Cy = 0 j: D = (0.2 m)(8l} N) — ((1.4 mll'C'z or C: = 40 N s = 430.0 N)k c = —[3ﬂ.ﬂ N)i +(4on NM: «1 Now we have for the equivalence of forces —{3s N)i + Ryj +321; = —[ss N)k +[[—3[:- N)i + (40 N)k] Equating eoeﬁleients j: J, = [II R}, = 0 i k: sz=—s[:-+4s or sz=—4s.oud ﬁrst note that R = —(30 N]i — [4t] N)k. Thus, the screw is best ahle to resist the lateral three Rz when the slot in the head ofthe screw is Trertieal. ‘1 ...
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Sol01 - PROBLEM 2.35 Knowing tl'nt a 35° determine t...

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