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Troublshooting your pc for dummies - Bourg D.M.

Bourg D.M. Troublshooting your pc for dummies - Wiley publishing , 2002. - 350 p.
ISBN 0-596-00006-5
Download (direct link): fordummiestroublesho2005.pdf
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Particle Kinetics in 3D
As in kinematics, extending the equations of motion for a particle to three dimensions is easy to do. You simply need to add one more component and will end up with three equations as follows:
Y^Fx = max = may
= maz
The resultant force and acceleration vectors are now
a = ax i + ayj + a7k a = Ja2 + a2 + a2
EF=EF-i+E^j+LF=k E^/E^+E^+E^
To hammer these concepts home, I want to present another example.
Let’s go back to the cannon program discussed in Chapter 2. In that example I made some simplifying assumptions so that 1 could focus on the kinematics of the problem without complicating it too much. One of the more significant assumptions I made was that there was no drag acting on the projectile as it flew through the air. Physically, this would be valid only if the projectile were moving through a vacuum, which of course, is unlikely here on the earth. Another significant assumption I made was that there was no wind to act on the projectile, affecting its course. These two considerations, drag and wind, are important in real-life projectile problems, so to make this example a little more interesting, and more challenging to the user if this were an actual game, I’ll now go ahead and add these two considerations.
First, assume that the projectile is a sphere and that the drag force acting on it as it flies through the air is a function of some drag coefficient and the speed of the projectile.
Particle Kinetics in 3D [ 75
This drag force can be written as follows:
Fd = -Qy y
Fj = —Cjvxi — CjVyj — CdVz k !
where Cd is the drag coefficient, v is the velocity of the projectile (v*, vy, and vz are its components), and the minus sign means that this drag force opposes the projectile’s motion. Actually, I’m cheating a bit here, since in reality the fluid dynamic drag would be more a function of speed squared. I’m doing this here to facilitate a closed-form solution.
Second, assume that the projectile is subjected to a blowing wind and that the force of this wind on the projectile is a function of some drag coefficient and the wind speed. This force can be written as follows:
F w = Fw —
where Cw is the drag coefficien t, vw is the wind speed, and the minus sign means that this force opposes the projectile’s motion when the wind is blowing in a direction opposite of the projectile’s direction of motion. When the wind is blowing with the projectile, say, from behind it, then the wind will actually help the projectile along instead of impede its motion. In general, C-& is not necessarily equal to Cj shown in the drag formula. Referring to Figure 2-3, I’ll define the wind direction as measured by the angle y. The x-and ^-components of the wind force can now be written in terms of the wind direction, y, as follows:
. F?X — Fw cos y = "(CuWu,) cos y
Fwz = Fu, cos y = “(Cu,vw) sin y
Finally, let’s apply a gravitational force to the projectile instead of specifying the effect
of gravity as a constant acceleration, as was done in Chapter 2. This allows you to
include the force due to gravity in the equations of motion. Assuming that the projectile is relatively close to sea level, the gravitational force can be written as
F, - -mgj
where the minus sign indicates that it acts in the negative y-direction (pulling the projectile toward the earth), and g on the right side of this equation is the acceleration due to gravity at sea level.
Now that all of the forces have been identified, you can write the equations of motion in each coordinate direction:
? ^2 = ~F»* ~~ Fdx = m(dvx/dt)
'22,Fy = -Fdy - Fsy ?? m(dvy/dt)
^2 F* = ~F»z ~ Fdz = m{dvjdt)
76 | Chapter 4: Kinetics
Note here that I already made the substitution dv/dt for acceleration in each equation. Following the same procedure shown in the previous section, you now need to integrate each equation of motion twice: once to find an equation for velocity as a function of time and again to find an equation for displacement as a function of time. As before, I’ll show you how this is done component by component.
You might be asking yourself now, “Where’s the thrust force from the cannon that propels the projectile in the first place?” In this example I’m looking specifically at the motion of the projectile after it has left me muzzle of the cannon where there is no longer a thrust force acting on the projectile (it isn’t self-propelled). To account for the effect of the cannon thrust force, which acts over a very short period of time while the projectile is within the cannon, you have to consider the muzzle velocity of the projectile when it initially leaves the cannon. The components cf the muzzle velocity in the coordinate directions will become initial velocities in each direction, and they will be included in the equations of motion once they have been integrated. The initial velocities will show up in the velocity and displacement equations just as they did in the example in Chapter 2. You’ll see this in the following sections.
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