Saturday, 1 February 2020

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



Answer



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:



Drag


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 $F_y$.



No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...