hw12soln

# hw12soln - Witia,czua:0 Mar\C PROBLEM 6.126 400 mm The...

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Unformatted text preview: Witia ,czua, :0 Mar /\C PROBLEM 6.126 400 mm The press shown is used to emboss a small seal at E. Knowing that P = 250 N, determine (a) the vertical component of the force exerted on the seal, (19) the reaction atA. ir_ SOLUTION ' FBD Stamp D: (S a Q “Fa'u‘ \‘X’C’ka 50 20 u = 0: E —FBD c0520° = 0, E =FBD cos20° um FBD ABC: '; éT , 1‘7 ‘ LEM}, = 0: (0.2 m)(sin30°)(FBD cos 20°) + (0.2 m)(cos 30°)(FBD sin 20°) 1 — [(0.2 m) sin 30° + (0.4 m)coslS°](250 N) = 0 F BD = 793.64 N C and, from above, E = (793.64 N) cos 20° (a) E = 746 N 4 “~25, = 0: A, —(793.64 N)sin20° = 0 A, = 271.44 N—» , T213. = 0: A), + (793.64 N)c0520° - 250 N = 0 A), =495.73Nl so (b) A=565N=q613°4 .2. _ 3 m. .} AHE/‘ﬁmltvﬁvla COR-"ti UR: ‘(t‘kC/i' Cur “30 ‘5 ‘3‘ I E q rd“) 0 A . * v' Q). ‘l’O dh’tuu ‘01—) F N Frs‘r s 30 m Ovﬁa and \ WUOJH - ﬂ— 1&1 Sﬁﬁmc'vﬂ ‘Px{=‘Cxx “Nd 3% = gee, + a? Cab-\ste/w Ag CMQ‘T a membw) '3 A, 3y 3% SMMMz—ﬁj PM }= |ssx\ “Wad A“: 6‘1 Aggx EFH 1: S’OOT ;AY :0 Arm: an in = Bx as 4-80 \to (‘5’) “Die MC” VP.” B ' I“ ‘ , ‘1 *eacﬂﬁuw ﬁn" JKA (5 he...) ragarwﬂ 3 T 353*} Z r L g ME PymwrF (9)11)“ (nu-r3 ;) + '3» (a‘r ‘5) =0 z] gut-QC.) 1» \‘J F 1- lbs 1300 I x I h “use; *0 F ‘ -—..q~€«o iv) in“ i X .- - - . ‘b F“ 4‘10 F '5. F = x Fx-rEx —_9X :0 M440téﬁe+3xq+80 :C} Ex:rc\?ol \b(_~—>> Er - _ y TEV 16 ‘m ‘1' Eva—+00 _;O Eyre \v) “ch PROBLEM 6.149 l_____________i ‘P——-]U iII,—————>§‘——]0 in. a i 1 V ' L I _ ‘ __ Knowing that the frame shown has a sag at B of a =1in., determine ‘ the force P required to maintain equilibrium in the position shown. SOLUTION We note that AB and BC are two-force members ﬂ, (an... iow 99 ..._ w *4 Free body: Toggle 9x _._.—-—-‘-""" a. P - By symmetry: Cy = 3 I Q _ i E 10in. a szncvhzzﬁ a ' a 2 a Z Free body: Member CDE +j)ZME = 0: Cx (6 in.) - Cy (20 in.) — (50 1b)(] 0 in.) = 0 55%) — 5(20) = 500 a 2 30 P ——10 =500 (a J i For a=1.0in. P[3—19—10j=500 20P=500 P:°5.01bi< 240 lb 05 = 035 V m. = 0,25 PROBLEM 8.1 ark—— SOLUTION Harem {Yaw—w, - Assume eguilibrium: Maximum friction force: l5 \KEFX = O: F+(2401b)sin 25°—(1501b)cos 25° :0 F =+34.5181b *7! BF}. = 0: N—(2401b)cos 25°—(1501b)sin 250 = 0 N = +280.91 1b Fm = ,uSN = 0.35(280.91 1b) = 98.3191b Since F < Fm, (b) it: Mare, '1'V\d'~.+ ("J I? ‘2 Hera, ! PROBLEM 8.11 SOLUTION (a) Free body: 20-kg block W1 2(20 kg)(9.81 111/53): 196.2 N F] = ySNI = 0.4(196.2 N) = 78.48 N LZFr—O: T—F1 =0 T=F1 278.48N Free body: 30-kg block W2 = (30 kg)(9.81m/sz)= 294.3 N N2 2196.2 N + 294.3 N = 490.5 N 5 “we “PP‘ F:=#SN2:0.4(490.5N)=196.2N JLZF=O: P—Fl—FZ—T=0 P = 78.48 N +1962 N+78.48 N =353.2 N Free body: Both blocks Blocks move together W = (50 kg)(9.81 m/sz) =490.5 N in =0: P—F=0 P = ,uSN = 0.4(490.5 N) = 196.2 N A F = 34.5181bi\ Determine whether the block shown is in equilibrium and ﬁnd the magnitude and P direction of the friction force when 19 = 25° and P = 150 1b. N = 280.911b / Block is in equilibrium 4 F = 34.5 lb\ 4 The coefﬁcients of friction are ,us =O.4O and ,uk =O.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as show, (b) is removed. P=353N‘*—< w=m~$77 Lm '~* fag/v fly:¥2>.r~ % > PROBLEM 8.105 The coefﬁcient of static friction between block B and the horizontal surface and between the rope and support C is 0.40. Knowing that mA 2 12 kg, determine the smallest mass of block B for which equilibrium is maintained. SOLUTION Support at C: FBD block B: 773 (31'$+u~rn : rad 1—6 \$953 fr; Z; = h”! '3 rs WA = mg = (12 kg)g 12F), =0: NB —WB =0 or NB =WB x Impending motion: FB = ,uSNB = 0.4NB = 0.4 WE —>ZFX =0: FE —TB =0 or TB =FB =O.4WB At support, for impending motion of WA down: ’ WA = TBeﬂ-vﬂ so TB = WAe‘ﬂsﬂ = (12 kg)g*‘~°-‘”’“'2 = (6.4019 kg)g NOW W K-Z'B—z g 0.4 0.4 W l . sothat m3: B=M :1“; 2.7“, mB=l6.00kg4 g g m, ...
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hw12soln - Witia,czua:0 Mar\C PROBLEM 6.126 400 mm The...

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