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Unformatted text preview: ME 104: Homework 5 Solutions Prof. Karl Hedrick: University of California, Berkeley Chapter 13, Solution 19 (a) Kinematics: xB vB 2 xA 2v A A and B. Assume B moves down. v1 T1 T2 0 0 1 2 mAv A 2 1 2 mB vB 2 2 vB v2 1 (2 kg) B 2 4 T2 U1 2 52 vB 4 mA g (cos 30 )( x A ) mB g (cos 30 ) xB 2m 1m (2)(9.81) 16.99 J 3 [ 1 2] 2 xB xA U1 U1 2 2 Since work is positive, block B does move down. T1 U1 2 T2 52 vB 4 13.59 vB 3.69 m/s 0 16.99 2 vB 60° (b) B alone. v1 T1 v2 0 0 3.69 m/s, (from a )) ( T2 U1 U1 U1 2 1 1 2 mB v2 (2)(3.69)2 13.59 J 2 2 (mB g )(cos 30 )( xB ) (T )( xB ) (2 kg)(9.81 m/s 2 ) 33.98 2T 3 2 (T ) (2 m) 2 2 PROBLEM 13.19 (Continued) T1 U1 2 T2 0 33.98 2T 2T 13.59 20.39 33.98 13.59 T 10.19 N Chapter 13, Solution 41 vA 0 TA 0 TB UAB UAB UAB 12 mvB 2 1 560 2 vB 2g 280 2 vB g W (90) (1 cos 40 ) (560 lb) (90 ft) (0.234) 11, 791 ft lb TB 0 11, 791 280 2 vB g TA UA 2 vB B (11, 791 ft lb)(32.2 ft/s 2 ) (280 lb) 2 vB 1356 ft 2/s2 Newton’s law at B N W cos 40 2 mvB 2 ; vB R 1356 ft 2/s 2 (560 lb) (1356 ft 2/s 2 ) (32.2 ft/s 2 ) (90 ft) N 167.0 lb N N (560 lb)( cos 40 ) 429 262 167.0 lb Chapter 13, Solution 44 Block leaves surface at C when the normal force N 0. mg cos g cos 2 vC man 2 vC n gh cos gy (1) Work-energy principle. (a) 12 mv 2 12 TC mvC 2 TB U B C TC 12 32m mg (h y ) mvC 2 TB 1 m(8) 2 2 32m UB C W (h y) mg (h yC ) Using Eq. (1) 32 g (h yC ) 32 gh yC yC yC 1 g yC 2 3 g yC 2 (32 gh) 3 g 2 (2) (32 (32.2)(3)) 3 (32.2) 2 2.6625 ft (3) 0.8875 27.4 yC h cos cos yC h 2.6625 3 PROBLEM 13.44 (Continued) (b) From (1) and (3) vC vC vC gy (32.2)(2.6625) 9.259 ft/s vC cos (9.259)(cos 27.4 ) 8.220 ft/s At C: (vC ) x (vC ) y y vC sin yC (vC ) y t (9.259) (sin 27.4 ) 12 gt 2 4.261 ft/s 2.6625 4.261t 16.1 t 2 A t E: yE t 0 t2 0.2647t 0.1654 0 0.2953 s h(sin ) (vC ) x t (3) (sin 27.4 ) (8.220) (0.2953) A t E: x x 1.381 2.427 3.808 ft x 3.81 ft Chapter 13, Solution 178 Velocity of A and B after impact mA mB WA g WB g 1.3 32.2 2.6 32.2 0.04037 lb s 2 /ft 0.08075 lb sec 2 /ft Sphere A falls. Use conservation of energy to find v A , the speed just before impact. Use the plate surface as the datum. 1 2 T1 0, V1 mA gh0 , T2 mAv A , V2 0 2 1 2 T1 V1 T2 V2 0 m A gh0 mAv A 0 2 With h0 vA 1.8 ft, 2 gh0 (2)(32.2)(1.8) vA 10.767 ft/s Analysis of the impact. Conservation of momentum. mA v A mB v B mA v A mB v B with 2) vB 0 Dividing by mA and using y-components with (mB /m A 10.767 0 (v A ) y 2(vB ) y ( vB ) y ] 10.767e (1) Coefficient of restitution. ( vB ) y ( vB ) y (v A ) y (v A ) y e [(v A ) y e (v A ) y (2) Solving Eqs. (1) and (2) simultaneously with e 0.8 gives (v A ) y ( vB ) y vA vB 2.153 ft/s 6.460 ft/s 2.153 ft/s , 6.460 ft/s (a) Sphere A rises. Use conservation of energy to find h. T1 T1 V1 h 1 mA (v A ) 2 , V1 0, T2 0, V2 mA gh 2 1 T2 V2: m A (v A ) 2 0 0 m A gh 2 (v A ) 2 2g (2.153) 2 (2)(32.2) h 0.0720 ft PROBLEM 13.178 (Continued) (b) Plate B falls and compresses the spring. Use conservation of energy. Let 0 be the initial compression of the spring and impact. In the initial equilibrium state, Fy 0: k 0 be the additional compression of the spring after WB 0 or k 1 k 2 2 0 0 WB (3) Just after impact: T1 1 mB (vB ) 2 , V1 2 T2 0 At maximum deflection of the plate, V2 (V2 ) g T2 V2 (V2 )e WB 1 k( 2 0 )2 Conservation of energy: 1 mB (vB )2 2 T1 V1 1 k 2 2 0 0 WB 1 k 2 2 0 k 0 1 k 2 2 Invoking the result of Eq. (3) gives 1 mB (vB )2 2 Data: mB 1 k 2 2 (4) vB 6.460 ft/s 0.08075 lb s 2 /ft, 3h (3)(0.072) 2 0.216 ft k 72.2 lb/ft k mB (vB ) 2 (0.08075)(6.460) 2 (0.216) 2 Chapter 13, Solution 180 (a) At C: Conservation of total momentum. mA mB m mAv A 5 mi/h mB vB 7.333 ft/s mA v A mB vB vA vB (1) 7.33 vB Work and energy. Car A (after impact): T1 T2 U1 U1 T1 U1 1 m A (v A ) 2 2 2 2 2 1 m A (v A ) 2 2 0 F f (12) k m A g (12 ft) T2 0 (2) (12 ft) (0.3) (32.2 ft/s 2 ) 231.84 ft/s 2 15.226 ft/s k m A g (12) (v A ) 2 vA Car B (after impact): T1 T2 U1 T1 U1 2 2 1 mB (vB ) 2 2 0 k mB g (3) T2 k mB g (3) 1 mB (vB ) 2 2 vB2 ( vB ) 2 vB (2) (3ft) (0.3) (32.2 ft/s 2 ) 57.96 ft 2 /s 2 7.613 ft/s PROBLEM 13.180 (Continued) 7.333 v A2 From (1) vB vA vB 30.2 ft/s 20.6 mi/h 7.333 15.226 7.613 (b) Relative velocities. ( vA vB ) e vB vA ( 7.333 30.2) e e 7.613 15.226 (7.613) 0.2028 (37.53) e 0.203 Chapter 13, Solution 201 Data: mA k 2 kg, mB 0.2, e 0.8, 1 kg, k 20 , 800 N/m, x 0.1 m, d 1.5 m 40 , l 1.0 m Block slides down the incline. Fy N mA g cos 0: 0 N mA g cos (2)(9.81) cos 20 18.4368 N Ff kN (0.2)(18.4368) 3.6874 N Use work and energy. Datum for Vg is the impact point near B. T1 0, (V1 )e 12 k x1 2 1 (800)(0.1) 2 2 4.00 J (V1 ) g mA gh1 U1 2 m A g ( x d )sin Ff ( x d ) 1 2 mA v A 2 (2)(9.81)(1.6)sin 20 (3.6874)(1.6) 5.8998 J 0 10.7367 J T2 1 2 2 (1) (v A ) 1.000 v A V2 2 T1 V1 U1 2 2 vA 2 T2 V2 : 0 4.00 10.7367 5.8998 1.000 v A 0 8.8369 m 2 /s 2 2.9727 m/s 20 vA Impact: Conservation of momentum. Both A and B, horizontal components : 0 m A v A cos mB v B m A v A cos (2)(2.9727) cos 20 2v A cos 20 (1.00)vB (1) PROBLEM 13.201 (CONTINUED) Relative velocities: ( vB ) n vB cos vB cos 20 (v A ) n vA vA e [(vB ) n e [v A 0] (v A ) n ] (0.8)(2.9727) (2) Solving Eqs. (1) and (2) simultaneously, vA vB 1.0382 m/s 3.6356 m/s Sphere B rises: Use conservation of energy. T1 T2 T1 V1 2 v2 1 mB (vB ) 2 V1 0 2 1 2 mB v2 V2 mB gh2 2 1 T2 V2 : mB (vB )2 0 2 (vB ) 2 2 gl (1 cos ) (3.6356) 2 2 mB gl (1 cos ) 1 2 mB v2 2 mB g (1 cos ) (2)(9.81)(1 cos 40 ) 8.6274 m /s 2 (a) Speed of B. (b) Tension in the rope. an Fn v2 2.94 m/s 1.00 m 2 v2 8.6274 1.00 8.6274 m/s 2 mB an : T mB g cos T mB an mB (an g cos ) T 16.14 N (1.0) (8.6274 9.81cos 40 ) ...
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