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Unformatted text preview: 6—2 Draw a ﬂeebed}F diagram of the diving beard
Shem‘11 in Fig. P62. The surface at B is smooth.
Neglect the weight of the diving beard. SOLUTION 6—5 Draw a freebody diagram of the bracket shown
in Fig. Poﬁ. The eoirraet surfaces between The
cylinders and bracket are smooth. SOLUTION 6—6 Forces P are applied To the handles of the pipe
pliers shown in Fig. PMS6. Draw ﬂeebody
diagrams of each handle. SOLUTION 6—8 Forces P are applied to the handles of the bolt
cutter shown in Fig. PGS. Draw a freebody
diagram of (a) The lower handle. (13) The lower cutter jaw. SOLUTION 6—11 Draw a freebody diagram of the bar shown in
Fig. P611. The supme atA is ajoumal bearing
and the supports at B and C are ball bearings. SOLUTION 6—12 Draw a ﬂeebody diagram of the shaft shown in
Fig. PIS12. The hearing at A is a thrust bearing.
and the hearing at D is a ball bearing. Neglect
the weights of the shaft and the lEYEl‘S. SOLUTION 6715 A 3011) force P is applied to the brake pedal of an automobile as shoer in Fig. P615. Determine the force Q
applied to the brake cylinder and the reaction at support A. SOLUTION From a freebody diagram of the brake pedal. the equilibrium
equations are solved to get the forces OEMA =0: 5.59—(30cos30°)(11)—(305i11300)(4)= 0 Q=62.8711b
aEﬁ20: Ax—Q+30COS3OO:0
AI :36.8901b
T 2F). : 0: A’saosnmoo : 0
A}_:151b
A=39.81b 4 22.130 Q=6291b e 154 6716 A beam is loaded and supported as shovm in Fig. Fiﬁ16. Determine the reactions at supports A and B.
SOLUTION From a free—body diagram of the beam. the equilibriinn equations are solved to get the forces +21%:0: Ax:0
own. :0: 4.53{(300)(1., —[l(4oo)(1.s) 2 [(300)(1.5)]—E A}. : 441.667 N B : 308 N T 6719 Three pipes are supported in a pipe rack as shown in Fig. P619. Each pipe weighs 100 1b. Determine the
reactions at supports A and B. SOLUTION
From a freebody diagram of the pipes and the rack. the equilibrium equations are solved to get the forces OXMA =0:
lag—(looms300)(9)—(100c05300)(15)
7(10000530°)(21)73(100si1130°)(4) O
3:249.840 lb
aZF‘ =0: AI+Bsin30°=0
AX=—124.920 lb T2550: AA—300+Bcos30°=0
4:83.632113 A=150.3 lb It 33 800 B : 250 lb C 60° 622 A pipe strut BC is loaded and supported as shown in Fig. PGll. Determine
(a) The reactions at supports A and C'. (b) The shearing stress on a cross section of the 101mii—dianieter pin at C. which is in double shear.
(c) The elongation of cable .43 if it is made of aluminum alloy (E = 73 GPa) and has a diameter of 6 mm.
SOLUTION (a) From a freebody diagram of the pipe shut. the equilibrium equations are solved to get the forces 021116 : 0: 1000133 7800(750) : 0 afﬁzﬂ (TI—114320
CY : 600 N
157 6725 A lever is loaded and supported as shown in Fig. P625. Determine
(a) The reactions at A and C'. (b) The normal stress in the 1frillsdiameter rod CD.
(c) The shearing stress in the litinrdiaineter pin atA. which is in double shear. (d) The change in length ofi‘od CD. which is made ofa material with a modulus of elasticity of 30(106) psi. S OLIITION 9 :tah’l (6/12): 265650 LCD : J62 +122 : 13.416 in.
(a) From a freebody diagram of the lever. the equilibrium equations are solved to get the forces OEMA =0: (rm cos6)(6)—40(4)—60(s)+80(12) =0 TC :759628 lb
a21220: AI+125+TCDc05320 =—71667 lb
T2550: A’.740760+807TCDsi116:0
A) = —6.666 lb
A:72.01b :7 531° ................................ ..Aus.
To = 59.6 lb (C) (b) Dividing the normal force in the rod CD by its crosssectional area gives the normal stress
0a, = Jug—D = —304 psi = 304 psi (C) tr(0.5)‘,:’4 (c) Dividing the shear force on the pin by twice its crosssectional area (since it is in double shear) gives the shear
stress VJ 72.0
1. : . A 214s : 2[;t(0.5)3f4] (:1) Finally. the change in length of the rod CD is given by 159 PL E4 (i30><10"’")[;r(0.5)3j4] :7l.358><104i11.:1.358><104i11.(shrink).. 626 The wood plane shown in Fig. P646 moves with a constant velocity when subjected to the forces shown. Determine
(a) The shearing force of the wood on the plane. (b) The noimal force. and its location. of the wood 01] the plane. SOLUTION From a freebody diagram of the plane. the equilibrium
equations are solved to get the forces aZR=0: F740c05700770003160=0 T 21F“: 0: N—40sin'i'0"—'."Osi1116O = 0 OZMA=0: (70c08160)(75)7(703i11160)(220)
+(40cos70°)(60)+Nd :0 (1727285 1.11111
31.5 1111.11 (from front ofplane) 6732 Pulleys A and B ofthe chain hoist shown in Fig. P6—32 are connected and rotate as a unit. The chain is
continuous. and each of the pulleys contains slots that prevent the chain from slipping. Determine the force F
required to hold a 450kg block W in equilibrium if the radii of pulleys A and B are 90 111111 and 100 111111. respectively.
SOLUTION A
From a ﬁ'eefbody diagram of the lower pulley. T T
vertical equilibrium equation gives the tensron t25,:0: 2r7450(9.31):0
T: 2207.25 N Then. from a freebody diagram of the lower pulley.
nioinent equilibrium equation gives the force F Hi 01
0 C35.) 162 ND MECHANICS OF MATERIAL TURG S AND MORRIS o = 0 : 100T — 90r—100F = 0 6736 The le\'er shown in Fig. P636 is formed in a quarter circular arc of radius 450 mm. Determine the angle 9 if
neither of the reactions atA or B can exceed 200 N. SOLUTION
From a freebody diagram of the lever. the equilibrium equations are HER=02 AI—Bsit16=0 T2550: A,—125+Bcosa=0 OZMD = 0: —A) (450): 0 The fourth needed equation is either A : . + A? : 200 N or
B : 200 N . In either case. the moment equation gives
A}. : 0 N
Now. guessing that B = 200 N gives
9 = 51 .320 A], : 15.61 N *A‘ :156.1N<200N Therefore. the guess was correct. 637 A man is slowly raising a 20ftlong homogeneous pole weighing 1501b as sliovm in Fig. PGST’. The lower
end of the pole is kept in place by smooth surfaces. Determine the force exerted by the man to hold the pole in the position shown.
SOLUTION From a freebody diagram] of the pole.
the equations of equilibrium give the tension OEM’A :0: (Tsi1130°)(20)7150(10cos600) 0 ...
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 Fall '08
 Brody

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