Homework9-solutions2 - ME 104 Homework 9 Solutions 2 Prof Karl Hedrick University of California Berkeley Chapter 16 Solution 136 Kinematics

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Unformatted text preview: ME 104: Homework 9 Solutions 2 Prof. Karl Hedrick: University of California, Berkeley Chapter 16, Solution 136 Kinematics: Velocity AB 0 vA (0.25 ft)(6 rad/s) 1.5 ft/s 2 vB 0.75 ft 2 rad/s vB BC 1.5 m/s 2 0.75 ft BC Acceleration aA aA aB aBC aBC a AB (0.25 ft)(6 rad/s) 2 9 ft/s 2 (0.75)(3 rad/s 2 ) 3 ft/s 2 (0.375 ft)(2 rad/s) 2 1.5 ft/s 2 1 (a A 2 aB ) 1 (9 2) 2 a AB aA 9 ft/s 2 6 ft/s 2 aB 3 ft/s (0.5 ft) 2 AB AB (0.5 ft) AB 12 rad/s 2 PROBLEM 16.136 (Continued) Kinetics: I AB I AB 1 1 4 lb m AB ( AB)2 (0.5 ft) 2 12 12 32.2 2.588 103 lb ft/s 2 Rod BC: Since BC 0, a 0 MC 0 yields Bx 0 Rod AB: Fx ( Fx )eff : Ax 0 MA ( M A )eff : B y (0.5 ft) WAB (0.25 ft) I AB AB m AB a AB (0.25 ft) 0.5By (4 lb)(0.25 ft) (2.588 10 3 lb ft/s 2 )(12 rad/s) 0.5By 4 lb (6 ft/s 2 )(11.25 ft) 32.2 1 0.03106 0.1863 0.8447 1.689 lb m AB a AB 4 lb (6 ft/s 2 ) 32.2 1.565 lb B 1.689 lb 0.5By By Fy ( Fy )eff : Ay WAB Ay By 4 lb + 1.689 lb Ay Ay 1.565 lb Chapter 16, Solution 144 Kinematics: We resolve the acceleration of G into the accel. of the cart and the accel. of G relative to A: aR aR aG aC aA aG/A aG/A Where Kinetics: Cart and rod aG/A 1 L 2 WR WC L IR aG/A Fx ( Fx )eff : (WC WR )sin 25 30 lb 40 lb 3 ft 1 mR L2 12 1 (3) 1.5 2 ( mC mR )a C mR aG/A cos 25 (40 30)sin 25 70aC aC 1 [(40 30)aC 30(1.5 ) cos 25 ] 32.2 70(32.2)sin 25 (4500 cos 25 ) (32.2)sin 25 45 cos 25 70 (1) Rod MA ( M A )eff : 0 I ( mR aG/R ) L 2 ( mR aC cos 25 ) L 2 1 30 2 (3) 12 g 30 (1.5 )(1.5) g 22.5 30 aC cos 25 (1.5) g 67.5 (45cos 25 )aC 0 0 (0.5cos 25 )aC (2) PROBLEM 16.144 (Continued) (a) Acceleration of the cart. Substitute for from (2) into (1): aC (32.2)sin 25 45 cos 25 (0.5cos 25 ) aC 10 aC 32.2sin 25 2.5 1 210 cos 25 18.490 ft/s 2 aC 18.49 ft/s 2 25° (b) Angular acceleration. From (2): (0.5cos 25 )(18.490) 8.379 rad/s 2 8.38 rad/s 2 Chapter 16, Solution 149* Kinematics: Assume α AB α BC and AB BC 0 Kinetics: Bar BC MB ( M B )eff : PL I BC (maBC ) BC L 2 AB P Fx ( Fx )eff : P Bx m2 L 12 1 mL 2 maBC mL 1 mL 3 L 2 BC L 2 AB BC (1) P Bx mL AB 1 L 2 BC (2) Bar AB: PROBLEM 16.149* (Continued) MA ( M A )eff : Bx L I AB (ma AB ) L 2 Bx m2 L 12 1 mL 3 4 mL 3 5 mL 6 5 AB AB m L 2 AB L 2 AB (3) 1 mL 2 1 mL 6 Add (2) and (3): Subtract (1) from (4) P 0 BC AB BC (4) AB BC (5) AB Substitute for BC in (1): P 1 mL 2 1 mL( 5 3 AB ) 7 mL 6 AB AB 6P 7 mL (6) 30 P 7 mL Eq. (5) Data: BC 5 6P 7 mL BC (7) 20 N L AB 500 mm 6P 7 mL 0.5 m, m 3 kg, P 6 20 N 7 (3 kg)(0.5 m) 11.249 rad/s 2 BC AB 11.43 rad/s 2 30 P 7 mL 30 20 N 7 (3 kg)(0.5 m) BC 57.143 rad/s 2 57.1 rad/s 2 ...
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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 University of California, Berkeley.

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