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Unformatted text preview: PROBLEM 6.121 The press shown is used to emboss a small seal at E. Knowing that
the vertical component of the force exerted on the seal must be 240 lb,
P determine (a) the required vertical force P, (b) the corresponding reaction at A. SOLUTION FBD part D:
(a) 1213 = 0; E — Dcoszo° = 0 D = 2401b
cos 20° = 255.40 1b E: 240% A, Q EMA = 0: [(sm.)cosso°]Dcos20° + [(5 in.) cos 60°]Dsinzo° FBD machine:
—[(5 in.) cos60° +10in.)cosl$°]P = 0 P = 30.45311: P = 80.5 lb 1 4
_. 2F): = D:Ax — Dsin20° =
A,c = 87.351]:
12F), = 0:Ay +2401b—80.51b = 0 A), =159.51b
A =181.91b T 613° 4 PROBLEM 6.129 The double toggle mechanism shown is used in a punching machine.
Knowing that links AB and BC are each of length 150 mm, determine the
couple M required to hold the system in equilibrium when ¢ = 20°. SOLUTION FBD piston: $sz = 0:800 —FCEcos20° = 0 F CE F CD FBC _
sin 40° 5111 40° Sin100° (EMA = 0: (0.15 m)(851.34 N)sin60° — M = 0 M=110.6Nm)< 5+~ PROBLEM 6.135 The slab tongs shown are used to Iiﬁ a 960113 block of granite.
Determine the forces exerted at D and F on tong BDF. SOLUTION FBD link A: By symmetry: T1 = T2 = T TZFy=D:9601b—2w8—T=O T1 23405115 Jﬁ By symmetry: EI = F1; E), = Fy TZFy=0:2Fy—9601b=0 Fy=4801b FBD tong BDF: Q EMD = o; (9.4 mgr; — (5 in.)(4801b) — (15.75 in.)[—§—60Jﬁ 1b]
IEM/Fﬂb J17 _1_ —(5.9 in.)[\/15—7 60511:] = 0 F; :162451b F = 1.694 kips "i 16.46° 4 _3_ 60Jﬁ1b—4301b+D =0
J17 y 125:0: =0 15
—~ ZFI = 0: ——60\/17 1b + D +71624.51b = 0
J17 ‘ D: = —2524.5 1b
D = 2.52 kips 4— 4 PROBLEM 6.141 Determine the magnitude of the gripping forces produced when two 501b
forces are applied as shown. 1 in 1.4m. SOLUTION FBI) handle CD: Q EMD = 0: —(4.2 in.)(501b) — (0.2 in.)——:':4 A +(1m.)[—J§T—4A]= 0 A = 477.27xl8.84 113 Q EMD = 0: (4.4 in.)(501b)—(4 m.)J8—1ﬁ—(477.27JEEZ 1b) +(1.2 in.)F = 0
F =1.408kips 4 SOLUTION FBDs
Top block: 260% 17::
£11,. Bottom block:
(W = 15/0 ’l ’9 PROBLEM 8.11 The coefﬁcients of ﬁiction are [.15 = 0.40 and ,uk = 0.30 between all .51: surfaces of contact. Determine the force P for which motion of the 601b
block is impending if cable AB (a) is attached as shown, (5) is removed. (a) Note: With the cable, motion must impend at both contact surfaces.
12F), =0: Nl—401b20 N1 =401b
Impending slip: Pi = xix/V, = 0.4(40 lb) = 161b
—* ZFx =0: T—Fl :0 T—16lb=0 T=161b
IZFy=0:Nl—401b—601b=0 N2=1001b
Impending slip: F2 = [JSNZ = 0.4(1001b) = 4011;
—— ZFX =0: —P+161b+161b+401b=0 P = 72.01b4—i (b) Without the cable, both blocks will stay together and motion will
impend only at the ﬂoor. 12F), =O:N¥401b—601b=0 N=1001b
Impending slip: F = AN = 0.4(100 lb) = 4011: —~EFx=0: 401b~P=0
P=40.0Ib~—4 PROBLEM 8.19 The hydraulic cylinder shown exerts a force of 680 1b directed to the right
on point B and to the left on point E. Determine the magnitude of the ‘ D
23' .7 couple M required to rotate the drum clockwise at a constant speed. 5 m. ET 12. in. SOLUTION FBDs Rotating drum :> slip at both sides; constant speed ::> equilibrium I. E = #kNl = 0'3Nl; F: = lukNZ = 0.3N2 (EMA = 0: (6 in.)(680 1b) + (6 in.)(1~1)—(1s in.)N = 0 F] [13—1311 — 6 in.) = (6 in.)(680 lb) or P} = 75.555115 (2MB = 0: (6 my?2 + (18 1n.)1v2 — (6 in.)(6801b) = o 18 in.
0.3 F2[6 in. + J: (6 in.)(680 1b) or F2 = 61.8181b (EMC:0: r[F1+F2)—M=0 M = (10 in.)(75.555 + 61.818)1b M = 1374115164 PROBLEM 8.49 Two 8° wedges of negligible mass are used to move and position a
240kg block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, determine the magnitude of the force P for which
motion of the block is impending. SOLUTION
at =tan‘1ys = tan10.4 = 21.301° W = 240 kg(9.81 m/sl) = 2354.4N FBD block + wedge: ‘ \
20%?”“4 7b;
32 23 54.4 N sin4l.801° sin38.398° R2 = 2526.6 N P 2526.6 N si1151.602° sin 68199" P = 2132.7 N PROBLEM 8.53 Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefﬁcient of static friction at all surfaces of contact is 0.25 and that (9 = 45°, determine the smallest force P for which
equilibrium is maintained. SOLUTION ¢3 : tan" #5 = tan“‘0.25 214.036" FBD block A: 4507‘2555 =73072.’
R: ’ edgbsf 75‘. 951“ R2 7501b sin(75.964°) sin(73.072°) R2 = 760.561b FBD wedge B: P 760.56 sin16.928° sin104.03 6° P = 228.31b P =2281b4—4 SOLUTION FBD pipe: PROBLEM 8.60 A 15° wedge is forced under a 100—lb pipe as shown. The coefﬁcient of
static friction at all surfaces is 0.20. Determine (a) at which surface slipping of the pipe Will ﬁrst occur, (b) the force P for which motion of
the wedge is impending. (a) QEMC = 0: rFA —rF3 = 0
or FA=FB But it is apparent that NB > NA,so since (115)}: = (ﬂS)B,
motion must ﬁrst impend at A 4 and FB=FA=#SNA=0.2NA (b) QEMB = o: (rsin15°)W+ r(1+sin15°)FA —(rc0515°)NA : o _0.2588(1001b)+1.2588(0.2NA)— 0.9659NA = 0
or N A = 36.24 lb
and FA = 7.2511: \ 213. = 0: NB — NA sin15° — FACOSIS‘” — Wc0515° = 0 N3 = (36.241b)sin15° + (7.2511; +1001b)coslS° = 112.971b
(noteNB > NA as stated, andFB < (“51%)
121+; = 0: NW + (7.251b)sin15° —(112.971b)cos15° = 0
NW =107.241b
Impending slip: FW = ysNW = 0.2(10724) = 21.4511:
—~ 2:}; = 0: 21.45 lb + (7.251b)cos15° + (112.971b)sin15° 2 P = 0 P =57.7lb——4 ...
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 Spring '08
 VALLE

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