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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.

 

     
     
   
 
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