{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework5-solutions

# homework5-solutions - ME 104 Homework 5 Solutions Prof Karl...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ) ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online