PROBLEM 4.1. M B 0: kn F B kn kn F C kn : kn 0?
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1 Chapter 4 PROBLEM 4. The boom on a 4300-kg truck is used to unload a pallet of shingles of mass 600 kg. Determine the reaction at each of the two (a) rear wheels B, (b) front wheels C. W m g 600 kg9.8 m/s 5696 N W kn W G m g G 4300 kg 9.8 m/s 4 83 N W G 4.83 kn (a) rom f.b.d. of truck with boom MC 0: kn cos5 m B m (b) rom f.b.d. of truck with boom 4.83 kn 0.5 m B 4.66 kn 5. M B 0: kn 6cos5 4.3 m 4.83 kn m C m C kn 5. 0: kn 0? y Check: B C.3 kn 6.8 kn kn 0 ok
2 675 N 675 N 75 mm PROBLEM 4.4 the beam and loading shown, determine the range of values of the distance a f which the reaction at B does not eceed 5 N downward 450 N upward..5 N 00 mm 50 mm 50 mm 675 N 675 N To determine a ma the two 675 N fces need to be as close to B without having the vertical upward fce at B eceed 450 N rom f.b.d. of beam with B = 450 N.5 N 450 N ÂM D = 0: (675 N) (a ma 00 mm) (675 N) (a ma 5 mm) (.5 N) (50 mm) + (450 N) (00 mm) = 0 a ma = 5 mm 675 N 675 N 5 N.5 N To determine a min the two 675 N fces need to be as close to without having the vertical downward fce at B eceed 5 N. rom f.b.d. of beam with B = 5 N ÂM D = 0: (675 N) (00 mm a min ) (675 N) (a min 5 mm) (.5 N) (50 mm) (5 N) (00 mm) = 0 Therefe, a min = 5 mm 5 mm a 5 mm t
3 50 mm 0 mm 40 mm D C B E 0 mm PROBLEM 4.5 follower BCD is held against a circular cam by a stretched spring, which eerts a fce of N f the position shown. Knowing that the tension in rod BE is 4 N, determine (a) the fce eerted on the roller at, (b) the reaction at bearing C. 60 Note: rom f.b.d. of BCD = cos60 (a) rom f.b.d. of BCD sin 60 y 3 MC 0: 40 mm N 40 mm 4 N 0 mm 0 8 N 8.0 N (b) rom f.b.d. of BCD 0: C 4 N 8 N cos N C 8.0 N C 0: C N 8 Nsin 60 0 y y N C 3.5 N C y y Then y C C C N Cy and tan tan 6.68 C 8 C 8. N 6.6
4 B 90 N 90 N 300 mm 300 mm B 90 Nm 50 mm 9 Nm50 mm 45 N N 50 mm 50 mm 5 mm 5 mm (a) (b) PROBLEM 4.4 steel rod is bent to fm a mounting bracket. each of the mounting brackets and loadings shown, determine the reactions at and B. (a) 90 N (a) rom f.b.d. of mounting bracket  M = 0: B (00 mm) (90 N) (300 mm) + (45 N) (50 mm) 9000 N.mm = 0 \ B = N 9000 N.mm 45 N  = 0: N 45 N + = 0 \ = 3.75 N B = 68.8 N t = 3.8 N  y = 0: 90 N + y = 0 \ y = 90 N y = 90 N y Then = + = = 3.9 N and q = tan y b g b g KJ = tan KJ =.834 = t (b) (b) rom f.b.d. of mounting bracket 90 N  M = 0: (B cos 45 ) (00 mm) (90 N) (300 mm) 9000 N.mm + (45 N) (50 mm) = N.mm 45 N \ B = N B = t  = 0: + ( 38.65) cos N = 0 \ = 3.75 N = 3.8 N
5 PROBLEM 4.4 CONTNUED Â y = 0: y (38.65 N) sin N = 0 \ y = N y = 78.8 N y b g b g Then = + = = N and q = tan y KJ = tan KJ = 0.5 = 7.8 N 0. t
6 PROBLEM 4.3 Neglecting friction, determine the tension in cable BD and the reaction at suppt C. 65 mm 35 N 50 mm 50 mm rom f.b.d. of inverted T-member  M C = 0: T (65 mm) T (50 mm) (35 N) (50 mm) = 0 \ T = 90 N T = 90 N t  = 0: C 90 N = 0 35 N \ C = 90 N C = 90 N  y = 0: C y + 90 N 35 N = 0 \ C y = 45 N C y = 45 N y bg bg Then C = C + C = = 00.6 N and q = tan C C y KJ = tan K J = C = 00.6 N 6.6 t
7 PROBLEM 4.39 B q O C 5 N 60 Rod BCD is bent in the shape of a circular arc of radius 80 mm and rests against frictionless surfaces at and D. Knowing that the collar at B can move freely on the rod, determine (a) the value of f which the tension in cd OB is as small as possible, (b) the cresponding value of the tension, (c) the reactions at and D. 45 D 45 (a) rom f.b.d. of rod BCD M 0: 5 N cos60 d Tcos d 0 (b) rom Equation () E OE OE.5 N T cos () T is minimum when cos is maimum, 0.5 N T.5 N cos0 T.50 N min (c) 0: N cos 45 N cos 45.5 N D 5 N cos60 0 N N 0 ND N D () 0: N sin 45N sin 45 5 N sin 60 0 y D N N N (3) Substituting Equation () into Equation (3), D N N N and N 5.3 N 45.0 N 5.3 N 45.0 D
8 0.45 m 0.45 m 0.45 m B B B PROBLEM N weight can be suppted in the three different ways shown. Knowing that the pulleys have a 00 mm radius, determine the reaction at in each case. (a) 90 N 90 N 90 N (b) (c) (a) rom f.b.d. of B 90 N 0: 0  y = 0: y 90 N = 0 y = 90 N and = 90 N t  M = 0: M (90 N) (0.45 m) = 0 \ M = N.m M = 33.8 N.m t 90 N rom f.b.d. of B  = 0: 90 N = 0 = 90 N 90 N  y = 0: y 90 N = 0 y = 90 N y bg bg Then = + = = 7.79 N \ = 7.3 N 45 t  M = 0: M + (90 N) (0. m) (90 N) (0.45 m + 0. m) = 0 \ M = N.m M = 33.8 N.m t
9 PROBLEM 4.45 CONTNUED (c) rom f.b.d. of B 0: 0  y = 0: y 90 N 90 N = 0 y = 80 N 90 N 90 N and = 80 N t  M = 0: M (90 N) (0.45 m 0. m) (90 N) (0.45 m + 0. m) = 0 \ M = 67.5 N.m M = 67.5 N.m t
10 PROBLEM 4.54 slender rod B, of weight W, is attached to blocks and B, which move freely in the guides shown. The blocks are connected by an elastic cd which passes over a pulley at C. (a) Epress the tension in the cd in terms of W and. (b) Determine the value of f which the tension in the cd is equal to 3W. (a) rom f.b.d. of rod B l MC 0: Tlsin W cos Tlcos 0 T W cos cos sin Dividing both numerat and denominat by cos, T W tan T W tan (b) T 3 W, 3W W tan tan 6 tan
11 00 mm 00 mm PROBLEM 4.63 Hizontal and vertical links are hinged to a wheel, and fces are applied to the links as shown. Knowing that a 75 mm, determine the value of P and the reaction at. 00 mm 95 N s shown on the f.b.d., the wheel is a three-fce body. Let point D be the intersection of the three fces. 75 mm rom fce triangle 5 = P 00 = \ P = (95 N) = 6.67 N 95 N P = 6.7 N t 95 N and = 5 75 q = tan (95 N) = N K J = \ = 58.3 N 36.9 t
12 PROBLEM 4.77 D. m 30 C 75 B 0.4 m small hoist is mounted on the back of a pickup truck and is used to lift a 0-kg crate. Determine (a) the fce eerted on the hoist by the hydraulic cylinder BC, (b) the reaction at. irst note W mg 0 kg 9.8 m/s 77. N rom the geometry of the three fces acting on the small hoist. m cos m D y. m sin m D and y tan m tan m BE D Then ybe 0.4 m tan tan D pplying the law of sines to the fce triangle, W B sin sin sin5 (a) (b) 77. N B sin.634 sin sin5 B 86.9 N B.8 kn N 0.68 kn
13 PROBLEM 4.79 modified peavey is used to lift a 0.-m-diameter log of mass 36 kg. Knowing that 45 and that the fce eerted at C by the wker is perpendicular to the handle of the peavey, determine (a) the fce eerted at C, (b) the reaction at. irst note W mg 36 kg 9.8 m/s N rom the geometry of the three fces acting on the modified peavey. m tan m 0. m pplying the law of sines to the fce triangle, W C sin sin sin N C sin sin sin35 (a) C N C 45.4 N 45.0 (b) N 387 N 85.
14 PROBLEM kg slender rod of length L is attached to collars which can slide freely along the guides shown. Knowing that the rod is in equilibrium and that 5, determine (a) the angle that the rod fms with the vertical, (b) the reactions at and B. (a) s shown in the f.b.d. of the slender rod B, the three fces intersect at C. rom the geometry of the fces where tan y CB BC and CB L sin y Lcos tan tan tan tan BC 5 (b) W mg tan tan kg9.8 m/s 98. N 43.0 rom fce triangle W tan 98. N tan N 45.7 N and B W 98. N 08.4 N cos cos 5 B 08. N 65.0
15 y z 50 N T 50 mm C 00 mm 50 mm E D B 5 mm 90 mm PROBLEM 4.99 the ption of a machine shown, the 00 mm-diameter pulley and wheel B are fied to a shaft suppted by bearings at C and D. The spring of constant 360 N/m is unstretched when 0, and the bearing at C does not eert any aial fce. Knowing that 80 and that the machine is at rest and in equilibrium, determine (a) the tension T, (b) the reactions at C and D. Neglect the weights of the shaft, pulley, and wheel. 300 mm 5 mm 5 mm 50 N 50 mm 50 mm 90 mm irst, determine the spring fce, E, at q = 80 E = k s where k s = 360 N/m = (y E ) final (y E ) initial = (300 mm + 90 mm) (300 mm 90 mm) = 80 mm \ E = (360 N/m) (80/000 m) = 64.8 N (a) rom f.b.d. of machine part  M = 0: (50 N) (50 mm) T (50 mm) = 0 \ T = 50 N T = 50 N t (b)  M D (z-ais) = 0: C y (50 mm) E (50 mm + 5 mm) = 0 C y (50 mm) 64.8 N (75 mm) = 0 \ C y = 9.44 N C y = (9.44 N) j  M D (y-ais) = 0: C z (50 mm) + (50 N) + (00 mm) (50 mm) (00 mm) = 0 \ C z = 0 N C z = (0 N) k and C = (9.4 N) j (0 N)k t
16 PROBLEM 4.99 CONTNUED Â = 0: D = 0 Â M C (z-ais) = 0: D y (50 mm) E (300 mm + 5 mm) = 0 D y (50 mm) 64.8 (35 mm) = 0 \ D y = 84.4 N D y = (84. N) j ÂM C (y-ais) = 0: (50 N) (50 mm) D z (50 mm) = 0 \ D z = 80 N D z = (80 N) k and D = (84. N) j (80 N) k t
17 PROBLEM 4.03 The mm square plate shown has a mass of 5 kg and is suppted by three vertical wires. Determine the mass and location of the lightest block which should be placed on the plate if the tensions in the three cables are to be equal. irst note W m g G p 5 kg 9.8 m/s 45.5 N rom f.b.d. of plate W mg m 9.8 m/s 9.8m N 0: 3T W W 0 () y G M 0: W 00 mm W z T 00 mm T 00 mm 0 G 300T 00W Wz 0 () G M 0: T 60 mm W 00 mm W 0 z Eliminate T by fming 00 Eq. Eq. G 30T 00W W 0 (3) G 00W W z 0 z 00 mm 0 z 00 mm, okay Now, 3 Eq Eq. yields 3 30T 3 00 W 3W 30 3T 30W 30W 0 G G
18 PROBLEM 4.03 CONTNUED 0W 30 3 W 0 G W 0 W 3 30 W The smallest value of W will result in the smallest value of W since W G is given. G G and then Use 00 mm ma W 0 W G W 45.5 N W G N minimum 4 4 W N and m.7857 kg g 9.8 m/s m.786 kg at 00 mm, z 00 mm
19 PROBLEM 4.4. m 0.9 m 0.6 m.8 m n.4 m-long boom is held by a ball-and-socket joint at C and by two cables D and BE. Determine the tension in each cable and the reaction at C. 0.3 m.4 m 880 N rom f.b.d. of boom  M CE = 0: l CE (r /C T D ) + l CE (r /C ) = 0 b g b g where l CE = 0. 6 m j m k = b0. 6g + b0. 9g m \ r /C = (.4 m) i 7. (0.6 j 0.9 k) b g b g b g b g b g b g KJ T D T D = l D T D = - 4. m i m j +. m k TD m = = (880 N) j 7. KJ (.4 i j +. k) T D KJ = 0
20 PROBLEM 4.4 CONTNUED ( ) T D (.6) = 0 \ T D = 60 N T D = 60 N t where l CD = Â M CD = 0: l CD (r B/C T BE ) + l CD (r /C ) b g b g 03. m j+. m k = 53. m (0.3 j +. k) 53. r B/C = (.8 m) i b g b g b g b g b g b g T BE = l BE T BE = - 8. m i m j k TBE = m T. KJ BE (.8 i j 0.9 k) \ T BE = 0 ( ) T BE. + (.88) 880 = 0 Â = 0: C (T D ) (T BE ) = 0 C \ T BE = N T BE = N t KJ KJ = 0 \ C = 4480 N Â y = 0: C y + (T D ) y + (T BE ) y 880 N = 0 C y KJ KJ \ C y = 3.33 N Â z = 0: C z + (T D ) z (T BE ) z = 0 C z KJ N = KJ = 0 \ C z = 30 N C = (4480 N) i (3.3 N) j + (30 N) k t
21 y PROBLEM 4. The rectangular plate shown has a mass of 5 kg and is held in the position shown by hinges and B and cable E. ssuming that the hinge at B does not eert any aial thrust, determine (a) the tension in the cable, (b) the reactions at and B. z irst note W mg 5 kg 9.8 m/s 47.5 N 0.08 mi 0.5 m j 0. mk TE TE ETE T E i j k m 0.33 rom f.b.d. of rectangular plate T M 0: 47.5 N 0. m 0. m 0 E y Nm T E 0. m TE 97.9 N 0: T 0 E T 97. N E N N
22 PROBLEM 4. CONTNUED M -ais 0: 0.3 m T 0.04 m W 0.5 m 0 y E Bz y 0.5 y 0.3 m 97.9 N 0.04 m 47.5 N 0.5 m y N MBy-ais 0: z 0.3 m TE 0. m TE 0.04 m 0 z z0.3 m TE 0. m TE 0.04 m z 0: W T B 0 y y E y y N N 47.5 N 97.9 N B y and 3.5 N i 63.8 N j 7.85 N k B y 9.8 N 0: T B 0 z z E z z N 97.9 N B z B z N and B 9.8 N j 66.7 N k
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