I have some doubts on the following derivation of the EOM of a stationary string.
Let $F_x, F_y$ be horizontal and vertical tension of the string
$\mu$ be the mass per unit length of the string [kg/m]
$$\frac{F_{2y}}{F_{2x}} = \tan\theta_+ = y'(x + dx)$$ $$\frac{F_{1y}}{F_{1y}} = \tan\theta_- = y'(x)$$ $$dm = \mu dx$$ No net force, $$F_{1x} = F_{2x} = F$$ $$F_{2y} - F_{1y} = gdm = \mu gdx$$ $$F(y'(x + dx) - y'(x)) = \mu gdx$$ $$F\frac{d^2y}{dx^2} = \mu g$$
Solving gives $$y = \frac{\mu g}{2F}x^2 + Ax + B$$
I am new to calculus, so I have some questions, please explain in a way that a beginner can understand, thanks!:
- why do we use $dm = \mu dx$ instead of $dm = \mu \sqrt{1 + (\frac{dy}{dx})^2} dx$ here? because it seems to me that $\mu \cdot dx$ doesn't describe the mass at some point on a curve, but $\mu \cdot ds$ does.
- Isn't a string held at both end called catenary which is not a parabola? then why in this case it is a parabola? What are the difference between a catenary and this graph?
I might be making some very basic conceptual mistakes because I am mostly self learning now. So please help me clear my doubt.