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InClassLecture21

# InClassLecture21 - T-BQ A pin-eenneeted system of bars...

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Unformatted text preview: T-BQ A pin-eenneeted system of bars supports a EDD- lb lead as shown in Fig. PT-EQ. Determine the reactions at supports s and E and the force exerted by the pin at C on member ACE. EGLUTIDN From a free-body diagram for the complete system: + C'EMA = -B!{2ﬂ} -3Dﬂ{30} = D B = -450 1b = 45H 1h *- I + -+ EF = s + EJlr : A - 45D = ﬂ 1'. H K s = 45D it = 450 lb -+ II Frem a free-body diagram for pin F: + —s EF . = T - 30s sin 45° = ﬂ 1r EF TEF : 212.1 1b 3 212 1b [T] From a free-hsdy diagram fer bar ACE: + c EH: = —Ayilﬂi + dﬁﬂilﬂl - 212.1[1Dfees 45”] = n s3 = lﬁﬂ.ﬂ lb = 15D.ﬂ ]h T + T SE = n + e - 212.1 sin 45“ ‘11" ‘1'" 1" = 150 u + CH - 212.1 sin is“ = D c = ﬂ ?*BQ {Continued} Fran the free-hudy diagram far the canplete system: +TEF =13. +A-3ﬂﬂ 3 3 3 = E? + 150.ﬂ ~ Bﬂﬂ = D E = iﬁﬂ.ﬂ lb = 150.0 lb T ".Ir r—-’——‘ 1 = {13312 + {Aria = {1133.313 + 1133.1113 = 131.3 13 a 171 lb _ -1 1513.0 _ I:- _ a 3A - tam 451M _ 13.131 I as 131 13 .3 13.13 Ana. 2 2 E 2 3 = 1’13“: +131“: = 111-1333} + 1153.1}: = 131.313 3: 13-1111 _ ~1 133.3 _ a a BB - tan 4513.0 .- 131.33 E a: 131 11:. 3 13.13 .1113. C = ﬁﬂﬂ 1b +—- Ans. E a 212 13 3 15.3" 1n1. T-11ﬂ* A cylinder with a mass of 150 ks; is supported by a two—bar frame as shown in Fig. PT—llﬂ. Determine all forces acting 0n member ACE. rig. P7410 SGLUTI 0N '5 From a free-body diagram I x for the cylinder: 54 = mg = 150(93071 = 14:1.1m +—-EF =nsin45"-Esin45°=u + T BF? =21: cos 45° -1471.1 = o D = E = 10¢0.2 1b E a 1040 1b *5 45.0“ (On menher ACE] From a free-body diagram for the campiete frame: FIWJH 1040.2’1L + 4: ENE = M2! -14?1.1{1}= B A = 735.6 N = \$35.6 N T D E C: I. i T36 [i T Ans. Q c A From a free—body diagram 1* for bar ACE: T. + (,7 EMC = Tm — 735.5(1: - 1040.2{031 = o 7353;,” 1‘ = 156T.E N z 1553 N T as 1563 w —» Ans. + -> 22F?r = c“ + T + 1040.2 cos 45° = :3x + 155?.3 + 1040.5 cos 45" = a c“ = —2303 N 2 2300 N «— + ’1‘ BF? = c? + T35.6 a 1040.2 sin 45° = 0 c :0 Eatzaoumt— Ana. 3" ?-BT* A pin-connected system of levers and hers is used as a teggle fer a press as shown in Fig. P?—3T. Determine the force 3 exerted on the can at A when a force e : 100 lb is applied to the lever at G. Fm.Plﬂ? SOLUTION From a Free—body diagram for the lever: + G EMF = lUﬂiﬂﬂ} - FD£:31 = a FDE = 3T5 1h From a free-body diagram for the pin at D: + *4 EFx = -F :03 6T0 - F Gee TED - 3?5 = U ED CD _ o . H c _ + 1‘ EFT — -FBD am ST + FCD em PB - G ______E FED = -639.5 lb FED = -6{}1.3 lb U -‘E'Z- From a. free-hunch.r diagram nu for the piston at B: T : - .r ' D : + EF? A? 639 J Bin 6? u H: A? = 533.? 1b = 533.? 1b T Force on the can: F 3 533 1b ¢ Ans. T-95* Forces of 25 1b are applied to the handles of the pipe pliers shown in Fig. p7-95. Determine the force exerted on the pipe at D and the farce exerted on handle DAB by the pin at A. SDLUTIDN Fran a free—body diagram for nember DAB: + c Ema = 5:1.25} - 25(51 = D D = 130.ﬂ lb Ans. t .4 22 = D sin 35“ + A = U x x A = —D sin 330 X I! -150.0 sin 22° -llﬂ.32 lb = 110.32 1b *— LI + T EF = 5 — D 555 32“ - 25 = u if .‘I’ 5 cos 32° + 25 .1:- II 150.5 cos 33° + 25 ll 166.34 lb = 156.34 lb T 2 = {53:2 + {5?}: = 21415.22}2 + {155.54}2 = 255.25 15 a 255 15 _ -2 155.34 _ a an _ tan :TTETEE _ 122.55 K = 200 lb 5 55.4“ Ans. T-BE* Fereee pf 5 N are applied he the handlee ef the paper punch ahown in Fig. PT-Bﬁ. Determine the farce exerted on the paper at D and the feree exerted on the pin at E by handle ABC. SDLUTIUN Fran a free-hedy diagram for handle EBB: +GEMB=DidﬂI 'EITU} =U D = 3.?5 N T +——}EF :B :12] x x E = U I + T EF = B + 5 + E.T5 = U r x El:Ir = “IE-T5 1h = 13.?5 1h i Farce exerted on the paper by handle EED: ﬁ = E.T5 N i Ana. Force exerted en the pin at a by handle nae: E = 1am lb l ens. ...
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