homework8-solutions

homework8-solutions - ME 104: Homework 8 Solutions Prof....

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Unformatted text preview: ME 104: Homework 8 Solutions Prof. Karl Hedrick: University of California, Berkeley Chapter 16, Solution 3 For the board to remain in the position shown, we need B $ 0. When the board is just about to start rotating about A, we have B 0. MA ( M A )eff : W (3 ft) cos 78 a g cot 78 W a (3 ft)sin 78 g (32.2 ft/s 2 ) cot 78 a 6.84 ft/s 2 Chapter 16, Solution 5 (a) Four-wheel drive: Fy 0: N A k NA NB W k NB 0 k ( NA NA NB ) NB kW W mg Thus: FA FB Fx 0.80 mg ( Fx )eff : FA FB ma 0.80mg a ma 0.80 g 0.80(32.2 ft/s 2 ) a 25.8 ft/s 2 (b) Rear-wheel drive: MB ( M B )eff : (40 in.)W (100 in.) NA NA (20 in.) ma 0.4W 0.2ma Thus: FA Fx k NB 0.80(0.4W ma 0.2 ma ) 0.32 mg 0.16 ma ( Fx )eff : FA 0.32mg 0.16ma 0.32 g a ma 0.84a 0.32 (32.2 ft/s 2 ) 0.84 a 12.27 ft/s 2 (c) Front-wheel drive: MA ( M A )eff : (100 in.) N B (60 in.)W NB (20 in.) ma 0.6W 0.2ma Thus: FB Fx k NB 0.80(0.6W ma 0.2 ma ) 0.48 mg 0.16 ma ( Fx )eff : FB 0.48mg 0.16ma ma 0.48 (32.2 ft/s 2 ) 1.16 0.48 g 1.16a a a 13.32 ft/s 2 Chapter 16, Solution 13 Kinematics: Assume that the barrel is sliding, but not tipping. a 0 aG a a Since the cord is inextensible, Kinetics: aC Draw the free body diagrams of the barrel and the cylinder. FF kN The barrel is sliding. 0.35 N Assume that tipping is impending, so that the line of action of the reaction on the bottom of the barrel passes through Point B. e 10 in. For the barrel. Fy MG 0: N WB 0 : Ne 4T 0N WB 0 (18)(0.35 N ) T e (18)(0.35) N 4 10 6.3 WB 4 0.925WB Fx WB a: T g 0.35 N a g WB a g T WB 0.35 N WB 0.925 0.35 0.575 For the cylinder: F WC a : WC g T WC T 1 a g a g 0.925WB 1 0.575 (2.1765)(200 lb) WC 435 lb WC Chapter 16, Solution 34 Case 1: (a) M0 ( M 0 )eff : (160 lb) 9 ft 12 (15 lb ft s 2 ) α 8 rad/s 2 (b) 2 10 ft 9 ft 12 2 M0 13.333 rad 2(8 rad/s 2 )(13.33 rad) ( M 0 )eff : (160) 9 12 15 ma 9 12 14.61 rad/s Case 2: (a) 160 9 32.2 12 120 (15 2.795) 120 15 6.7435 rad/s 2 9 12 6.74 rad/s 2 (b) 2 10 ft 9 ft 12 13.333 rad 2 2(6.7435 rad/s 2 )(13.333 rad) 13.41 rad/s (a) Case 3: (460) 9 12 (300) M0 ( M 0 )eff : 9 12 15 460 9 a 32.2 12 460 9 32.2 12 2 300 9 a 32.2 12 300 9 32.2 12 2 120 15 4.2437 rad/s 2 4.24 rad/s 2 (b) 10 ft 9 12 2 13.333 rad 2 2(4.2437)(13.333) 10.64 rad/s 2 Case 4: (a) M0 ( M 0 )eff : (80) 18 12 15 80 18 a 32.2 12 2 80 18 32.2 12 120 (15 5.5901) 120 15 5.828 rad/s 2 5.83 rad/s 2 (b) 2 10 ft 18 ft 12 2 6.6667 rad 2(5.828 rad/s 2 )(6.6667 rad) 8.82 rad/s Chapter 16, Solution 69 Kinetics: Fx ( Fx )eff : k mg ma kg a MG ( M G )eff : Fr k mg ) r I 22 mr 5 5 kg 2r ( α Kinematics: When the ball rolls, the instant center of rotation is at C, and when t t1 v r (1) v0 kg t v v0 0 at t (2) 5 kg t 2r 0 When t t1: r: v0 v0 k gt1 0 Eq. (1) : v 5 kg t1 r 2r 5 2 k gt1 k gt1 0r t1 v0 15 ft/s, 2 (v0 g r 0) kg 4 in. 1 ft 3 (3) 0 9 rad/s, r (a) t1 1 (9) 2 3 7 0.1(32.2) 15 1.5972 s t1 1.597 s PROBLEM 16.69 (Continued) (b) Eq. (2): v1 v0 k gt1 15 0.1(32.2)(1.5972) v1 15 5.1429 9.857 ft/s v1 9.86 ft/s (c) a kg 0.1(32.2 ft/s 2 ) 3.22 ft/s 2 s1 v0t1 12 at1 2 1 (3.22 ft/s 2 )(1.597 s)2 2 23.96 4.11 19.85 ft (15 ft/s)(1.597 s) s1 19.85 ft Chapter 16, Solution 76 a MA ( M A )eff : PL 1 2 (ma ) I L 2 1 mL2 12 I m PL L 2 L 2 1 mL2 12 12 mL 3 3P mL 3(1.5 lb) 4 lb 32.2 ft/s 2 (a) Angular acceleration. (3 ft) 12.08 rad/s 2 12.08 rad/s 2 (b) Components of the reaction at A. Fy ( Fy )eff : Ay W Ay Fx ( Fx )eff : Ax P 0 (N ma 4 lb) Ay 4.00 lb Ax Ax Pm P 2 L L 3P Pm 2 2 mL 1.5 lb 0.75 lb 2 P 2 Ax 0.750 lb Chapter 16, Solution 84 w0a L 2 MA ( M A )eff : W L 2 I ma L 2 mg L 2 L 2 1 mL2 12 12 mL 3 m L 2 3g 2L L 2 mg (b) Reaction at A. Fy ( Fy )eff : A mg ma m L 2 A mg A mg A m L 2 3g 2L 3 mg 4 1 mg 4 A 1 mg 4 (a) Acceleration of B. aB an a B/A 0L aB L 3g 2L 3 g 2 aB 3 g 2 Chapter 16, Solution 87 Masses and lengths: Moments of inertia: m AB 1 kg, LAB 0.6 m, mBC 2 kg, LBC 0.3 m I AB I BC 1 mAB L2 AB 12 1 mBC L2 BC 12 0.62 1 (1)(0.3)2 7.5 10 3 kg m 2 12 1 (2)(0.6)2 60 10 3 kg m 2 12 0.61847 m Geometry: rAB rBC tan 0.152 1 LBC 0.3 m 2 0.15 0.6 14.036 Kinematics: Let α be the angular acceleration of object ABC. (a AB )t rAB ( aB ) n ( aBC )t rAB rBC 2 ( aBC ) n rBC I AB 2 Kinetics: MC ( M C )eff : m AB g LAB 2 rAB m AB ( a AB )t I BC I AB rBC mBC (aBC )t 2 mAB rAB I BC 2 mBC rBC (1)(9.81)(0.15) [(7.5 10 3 ) (1)(0.61847)2 (60 10 3 ) (2)(0.3)2 ] 2.3357 rad/s 2 FX ( FX )eff : Cx m AB ( a AB )t cos m AB ( a AB )n sin 2 mBC ( aBC )t mAB rAB ( cos sin ) mBC rBC 10 2 sin14.035 ) (2)(0.3)(2.3357) (1)(0.61847)(2.3357 cos14.035 17.802 N Fy ( Fy )eff : Cy mAB g mBC g m AB (a AB )t sin m AB (a AB ) n cos mBC (aBC ) n PROBLEM 16.87 (Continued) Cy Cy Cy (mAB mBC ) g 2 mAB rAB ( 2 cos sin ) mBC rBC 2.3357sin14.035 ) (3)(9.81) (1)(0.61847)(102 cos14.035 (2)(0.3)(10) 149.08 N Reaction at C. C 17.8022 149.082 150.1 N tan 17.802 149.08 83.2 C 150.1 N 83.2 Chapter 16, Solution 94 I mk 2 ma mr MC ( M C )eff : (W sin )r (mg sin )r rg sin (ma )r (mr )r (r 2 I mk 2 k 2) rg sin r2 k 2 a r r rg sin r2 k2 a r2 r2 k2 g sin Chapter 16, Solution 141 Data: l I 6 m, m 50 kg, P 1 ml 2 12 1 (50)(6)2 12 aC aC 500 N, 150 kg m 2 k 0.3 Kinematics: aG aC [aC (aG/C )t (aC/G ) n l 2 10 l 2 2 80° (1) Kinetics: Sliding to the right: Ff kN MG ( M G )eff : N l sin10 2 Ff l cos10 2 P l cos10 2 kN h I I 150 Fy ( Fy )eff : N mg N l cos10 2 (3)(sin10 sin10 l sin10 2 0.3cos10 ) N ml 2 2 l cos10 2 500(3cos10 P h 2) (2) ml 2 cos10 ml sin10 2 [(50)(3)sin10 ] N N mg ml 2 2 cos10 (50)(9.81) (50)(3)(1) 2 cos10 (3) Solving Eqs. (2) and (3) simultaneously, 2.5054 rad/s Fx N cos10 277.52 N ml 2 2 ( Fx )eff : P Ff maC ml 2 sin10 PROBLEM 16.141 (Continued) maC 50aC aC ml ml 2 cos10 sin10 2 2 500 (0.3)(277.52) (50)(3)(2.5054) cos10 P kN (50)(3)(1)2 sin10 15.2159 m/s 15.2159 (3)(2.5054) Using Eq. (1), aG 10 (3)(1)2 80 (aG ) x (aG ) y 15.2159 7.5162 cos10 8.3348 m/s 2 3cos80 7.5162sin10 4.2596 m/s 2 3sin 80 aG (8.3348) 2 9.36 m/s 2 (4.2596) 2 tan 4.2586 8.3348 27.1 aG 9.36 m/s 2 27.1° (a) (b) Acceleration at Point G. Normal force. N 278 N ...
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This note was uploaded on 02/06/2010 for the course ME 104 taught by Professor Oreilly during the Fall '08 term at Berkeley.

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