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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 protected] ) NOTE: 1. Free body diagram is shown in Fig.( 1(a) ). θ y x F mg e ^ v m d (a) Free body diagram θ 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 Newton’s second law, or ∑ F = m a , we can write the equations as, summationdisplay F x :- F d cos θ = m ¨ x (1) summationdisplay F y :- F d sin θ- mg = m ¨ y (2) As one can see from Fig.( 1(b) ), sin θ and cos θ are simply as, cos θ = ˙ x v sin θ = ˙ 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 α y = 0 =- 1 2 gt 2 + v t sin α + 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 α x = v t cos α Hence, the travel distance is x ( t = 2 . 135) = 20 . 53 [m] Note before solving 4. and 5....
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- Fall '09
- Distance, travel distance