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Unformatted text preview: ME 240: Introduction to Dynamics and Vibrations Mechanical Engineering Department The University of Michigan Fall 2009 Computer Assignment #1 Solution October 9, 2009 Prepared by Joosup Lim ( email@example.com ) NOTE: 1. Free body diagram is shown in Fig.( 1(a) ). y x F mg e ^ v m d (a) Free body diagram y x v . . (b) Relation between x , y and v Figure 1: Free body diagram and velocities in x and y components 2. Using Cartesian coordinates, and Newtons second law, or F = m a , we can write the equations as, summationdisplay F x :- F d cos = m x (1) summationdisplay F y :- F d sin - mg = m y (2) As one can see from Fig.( 1(b) ), sin and cos are simply as, cos = x v sin = y v 1 Hence, the equations of motion are shown as, summationdisplay F x :- F d x v = m x (3) summationdisplay F y :- F d y v- mg = m y (4) 3. Assuming no aerodynamic drag force ( F d = 0 ), from equation of motions, one can simply get x = 0 y =- g First, one can find the time, t , when the stone hits the ground ( y = 0 ) from y =- g , y =- g y =- gt + v sin y = 0 =- 1 2 gt 2 + v t sin + 1 . 82 One may get t =- . 174 , 2 . 135 , and take positive value. Hence the travel time is t = 2 . 135 [sec]. Horizontal travel distance ( x ) with respect to time t can be obtained as, x = 0 x = v cos x = v t cos Hence, the travel distance is x ( t = 2 . 135) = 20 . 53 [m] Note before solving 4. and 5....
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This note was uploaded on 10/23/2009 for the course MECHENG me 240 taught by Professor Hulbert during the Fall '09 term at University of Michigan.
- Fall '09