 Ch 2. Particle Force and Acceleration Multimedia Engineering Dynamics Rect.Coord. Normal/Tang. Coord. PolarCoord. Orbital Mechanics Computational Mechanics
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll DYNAMICS - CASE STUDY SOLUTION Force Diagram - Vectors Force Diagram - Components Begin with a free-body diagram. The velocity has been included on the free-body diagram for reference. To apply Eqs. 5, the x and y components of the total force on the projectile must be determined. Since v/|v| is a unit vector, D can be written as The external forces are weight and drag, The x and y components of the total force are Now, consider the case when C = 0.002, and let Δt = 0.1. At the initial time to      to = 0      vx(to) = 400 ft/s           vy(to) = 400 ft/s Using the four basic equations for position and velocity (Eqs. 5) the following is determined, x - position      x(to + Δt) = x(to) + vx(to) Δt      x(0 + 0.1) = 0 + 400 (0.1)                     = 40 ft y - position      y(to + Δt) = y(to) + vy(to) Δt      y(0 + 0.1) = 0 + 400(0.1)                     = 40 ft x - velocity                                   vx(to + Δt) = vx(to) + (1/m) ΣFx(to, x(to), y(to), vx(to), vy(to))Δt                                   vx(0 + 0.1) = vx(0) - (C/m) vx(0) (vx 02 + vy 02)0.5 Δt                                   vx(0.1) = 400 - (0.002/100) (400) (4002 + 4002)0.5 (0.1)                                              = 399.6 ft/s                                x - velocity                                    vy(to + Δt) = vy(to) + (1/m) ΣFy(to, x(to), y(to), vx(to), vy(to))Δt                                    vy(0 + 0.1) = vy(0) - g Δt - (C/m) vx (0) (vx 02 + vy 02)0.5 Δt                                    vy(0 + 0.1) =  400 - 32.2(0.1) - (0.002/100) (400) (4002 + 4002)0.5 (0.1)                                                      = 396.3 ft/s  Results Plot Continuing this way, the results for the first five time-steps can be obtained as shown in the table at the left. These values can then be plotted to obtain an approximate solution for the trajectory of the projectile: When there is no drag (C = 0), the closed-form solution for the trajectory can be obtained and compared with the numerical solution. The graph below presents this comparison using Δt = 3.5s, Δt = 1.0s, and Δt = 0.1s. Notice that the numerical solution with Δt = 0.1s is the closest to the exact solution. This indicates that as the time-step is decreased, the numerical solution becomes more accurate. Practice Homework and Test problems now available in the 'Eng Dynamics' mobile app
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