# Splitting up a force into horizontal and vertical components?

My Bedford and Fowler textbook (4th edition) has a chapter on numerical solutions. I'm having trouble understanding how the textbook splits up the components of force in the $x$ and $y$ directions to model the drag on a projectile:

The aerodynamic frag force on the projectile is of magnitude $C|{v^2}|$, where $C$ is a constant.

We must determine the $x$ and $y$ components of the total force on the projectile. Let $D$ be the drag force. Because $v/|v|$ is a unit vector in the direction of $v$, we can write the drag force as:

$$D = -C{|v|^2} \frac{v}{|v|} = C|v|v.$$

The external forces on the projectile are its weight and the drag so we have:

$$\sum F = -mgj - C|v|v,$$

and the total components of the force are:

$$\sum F_x = C\sqrt{v_x^2 + v_y^2}\quad v_x$$

$$\sum F_y = -mg- C\sqrt{v_x^2 + v_y^2}\quad v_y.$$

I don't understand how the individual components of the drag force $D = C|v|^2$ (acting in opposition to the projectile's movement) becomes something like this: $$-C\sqrt{v_x^2 + v_y^2}* v_x~ ?$$

I understand that the absolute value of something can be represented as: $$|a|=\sqrt{a^2}.$$

And the magnitude of velocity from its $x$ and $y$ components can be calculated as $$V=\sqrt{v_x^2+v_y^2}.$$

But I don't get how the non-absolute $v$ term magically becomes $v_x$ or $v_y$ while the absolute term doesn't become $|v_x|$ or $|v_y|$ too.

What about equations without absolute values in them like the drag force equation? $F_D = \frac{1}{2}pv^2AC_D$, where $p,A, C_D$ are constants. Would the horizontal and vertical components of forces then be:

$$F_x = -\frac{1}{2}pv_x^2AC_D$$ and $$F_y = -mg -\frac{1}{2}pv_y^2AC_D~ ?$$

• The term is $\left| v \right| \vec v = \sqrt{v_x^2 + v_y^2} \big( \vec e_x v_x + \vec e_y v_y \big)$. Then you just split this up into components. The scalar multiplier will be the same for both components. Sep 6, 2015 at 12:48
• @SebastianRiese You should post this as an answer. Sep 6, 2015 at 13:22
• It's always better to write vector equations with vector symbols to avoid confusion like you have. You can't automatically do symbol substitution without thinking about what the concept of the symbol is. Also, you have a typo in the first equation with the $\Sigma$ sign. Sep 6, 2015 at 13:24
• My confusion probably stems from the fact that the textbook page that I wrote this from does not have any vector symbols above the appropriate variables. Sep 6, 2015 at 13:33
• @user155876 Do they use bold-face for vectors? Normally this is explained in an introductory chapter. Sep 6, 2015 at 13:45

I find this type of question is always easier if you draw a diagram. The drag force $F$ acts in the opposite direction to the velocity so it looks like:

and the components of the drag force are:

\begin{align} F_x &= F \cos\theta \\ F_y &= F \sin\theta \end{align}

$\cos\theta$ is $v_x/v$ and $v = \sqrt{v_x^2 + v_y^2}$ so we get:

$$F_x = F \frac{v_x}{\sqrt{v_x^2 + v_y^2}}$$

Since $F = -Cv^2$ we get:

$$F_x = -Cv^2 \frac{v_x}{\sqrt{v_x^2 + v_y^2}}$$

and because $v = \sqrt{v_x^2 + v_y^2}$ this becomes:

\begin{align} F_x &= -C \left(v_x^2 + v_y^2\right) \frac{v_x}{\sqrt{v_x^2 + v_y^2}} \\ &= -C v_x \sqrt{v_x^2 + v_y^2} \end{align}

And likewise for $F_y$.

# Review of vectors

A vector is a quantity with a magnitude and a direction, which makes it somewhat different from how normal quantities, which are just magnitudes. One huge difference is that vectors "transform" a certain way when we rotate our coordinates, for example. They "transform" just the same way that an arrow in space transforms, if we do not care about where the arrow is located but only how the arrow is pointing. So when you get really technical, this will become the definition of a vector; we'll just say that a vector is anything which "transforms like a vector" and give the example case of displacement vectors in space to define how we want them to transform.