- #1

marlon

- 3,792

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Let's start with the very base of Newtonian mechanics. It works like this : Suppose we work in two dimensions denoted by a x-axis and an y-axis. You can work in as many dimensions as you want because all you have to do is add a unit vector to the formula's, as you will see.

Starting from the acceleration ,one can calculate the velocity and position by using integrals at any time : r_0 is initial position at t=0, v_0 is initial velocity

[tex]\vec{F} = F_x \vec{e_x} + F_y \vec{e_y} = m(a_x \vec{e_x} + a_y \vec{e_y})[/tex]

Where the e_x and e_y denoted the x and y-direction (ie the unit vectors)

Now, integrating will yield

[tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]

[tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]

Now, the trick really is (and that's the essential part) to apply the same procedure in each direction. The procedure i am talking about is projecting each vector along a direction using the triangle-equalities.

For example : in the x-direction you will have :

[tex]F_x = ma_x [/tex]

[tex]v_x = v_{0x} + a_xt[/tex] and

[tex]x = r_{0x} + v_{0x}t+ a_x \frac{t^2}{2}[/tex]

[tex]v_{0x} = ||\vec{v_0}||cos( \theta)[/tex]

[tex]r_{0x} = ||\vec{r_0}||cos( \theta)[/tex]

the ||.|| denotes the MAGNITUDE of the vector

You see ? the clue is that each vector can be written as a sum of an x and y component [tex]\vec{A} = A_x \vec{e_x} + A_y \vec{e_y}[/tex]

[tex]A_x = ||\vec{A}||cos( \theta)[/tex]

[tex]A_y = ||\vec{A}||sin( \theta)[/tex]

You will need to be carefull with the signs of the x and y components because those depend of the direction of each component with respect to the actual x and y axis.

Some exercises :

1) Derive the equations for both position and velocity when the acceleration is [tex]\vec{a} = 0 \vec{e_x} - g \vec{e_y}[/tex]... Thus a_x = 0 and a_y = -g...This should ring a bell...

2) So, when a force is given, like : F = m(2e_x - 9.81e_y) and the initial position has components r_0 = 2e_x + 6e_y and the initial velocity v_0 = 6e_y, can you write down the equations for both velocity and position in each direction ?

In the end , you must realize that this system is very easy because, nomatter how complicated the force may look, the procedure to determin both position and velocity as a function of time is always the same: Projecting the vectors along the given directions. Once, you have done that, you can do almost anything with these formula's...

regards

marlon

Starting from the acceleration ,one can calculate the velocity and position by using integrals at any time : r_0 is initial position at t=0, v_0 is initial velocity

[tex]\vec{F} = F_x \vec{e_x} + F_y \vec{e_y} = m(a_x \vec{e_x} + a_y \vec{e_y})[/tex]

Where the e_x and e_y denoted the x and y-direction (ie the unit vectors)

Now, integrating will yield

[tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]

[tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]

Now, the trick really is (and that's the essential part) to apply the same procedure in each direction. The procedure i am talking about is projecting each vector along a direction using the triangle-equalities.

For example : in the x-direction you will have :

[tex]F_x = ma_x [/tex]

[tex]v_x = v_{0x} + a_xt[/tex] and

[tex]x = r_{0x} + v_{0x}t+ a_x \frac{t^2}{2}[/tex]

[tex]v_{0x} = ||\vec{v_0}||cos( \theta)[/tex]

[tex]r_{0x} = ||\vec{r_0}||cos( \theta)[/tex]

the ||.|| denotes the MAGNITUDE of the vector

You see ? the clue is that each vector can be written as a sum of an x and y component [tex]\vec{A} = A_x \vec{e_x} + A_y \vec{e_y}[/tex]

[tex]A_x = ||\vec{A}||cos( \theta)[/tex]

[tex]A_y = ||\vec{A}||sin( \theta)[/tex]

You will need to be carefull with the signs of the x and y components because those depend of the direction of each component with respect to the actual x and y axis.

Some exercises :

1) Derive the equations for both position and velocity when the acceleration is [tex]\vec{a} = 0 \vec{e_x} - g \vec{e_y}[/tex]... Thus a_x = 0 and a_y = -g...This should ring a bell...

2) So, when a force is given, like : F = m(2e_x - 9.81e_y) and the initial position has components r_0 = 2e_x + 6e_y and the initial velocity v_0 = 6e_y, can you write down the equations for both velocity and position in each direction ?

In the end , you must realize that this system is very easy because, nomatter how complicated the force may look, the procedure to determin both position and velocity as a function of time is always the same: Projecting the vectors along the given directions. Once, you have done that, you can do almost anything with these formula's...

regards

marlon

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