Solution to Question 12
1. System Analysis and Stiffness
The thread has a natural length of approximately zero (negligible relaxed length) and a force constant $k$. For such a spring, the tension $T$ is directly proportional to its current length $L$: $$T = kL$$ Let the total length of the thread be $l$. The thread is stretched between points $A$ (origin) and $B$. Let the coordinates of $A$ be $(0,0)$. Since the thread length is $l$ and it makes an angle $\theta$ with the horizontal, the coordinates of $B$ are: $$x_B = l \cos \theta, \quad y_B = l \sin \theta$$
When the spider is at position $P(x,y)$, it divides the thread into two segments. Let $\alpha$ be the fraction of the thread’s material on the left side (AP). The stiffness of a spring is inversely proportional to its natural length (or amount of material). Thus, the stiffness of the two segments are: $$k_1 = \frac{k}{\alpha} \quad \text{and} \quad k_2 = \frac{k}{1-\alpha}$$
2. Force Equilibrium
The forces acting on the spider are its weight $mg$ and the tensions $T_1$ and $T_2$ from the two segments of the thread.
Since the relaxed length is zero, the tension vector behaves like $\vec{T} = -k_{segment} \vec{r}_{segment}$.
Force exerted by segment AP (towards A):
$$\vec{F}_1 = -k_1 (x \hat{i} + y \hat{j}) = -\frac{k}{\alpha} (x \hat{i} + y \hat{j})$$
Force exerted by segment PB (towards B):
$$\vec{F}_2 = k_2 ((x_B – x) \hat{i} + (y_B – y) \hat{j}) = \frac{k}{1-\alpha} ((l \cos \theta – x) \hat{i} + (l \sin \theta – y) \hat{j})$$
Horizontal Equilibrium ($x$-direction)
The net horizontal force must be zero: $$-\frac{k}{\alpha} x + \frac{k}{1-\alpha} (l \cos \theta – x) = 0$$ Dividing by $k$ and rearranging: $$\frac{x}{\alpha} = \frac{l \cos \theta – x}{1-\alpha}$$ $$x(1-\alpha) = \alpha(l \cos \theta – x)$$ $$x – x\alpha = \alpha l \cos \theta – \alpha x$$ $$x = \alpha l \cos \theta$$ From this, we find the fraction $\alpha$ in terms of $x$: $$\alpha = \frac{x}{l \cos \theta}$$ This implies the horizontal position is directly proportional to the fraction of the string length.
Vertical Equilibrium ($y$-direction)
The net vertical force, including gravity, must be zero: $$-\frac{k}{\alpha} y + \frac{k}{1-\alpha} (l \sin \theta – y) – mg = 0$$
3. Deriving the Trajectory Equation
Substitute $\alpha = \frac{x}{l \cos \theta}$ into the vertical equilibrium equation.
Note that $\frac{k}{\alpha} = \frac{k l \cos \theta}{x}$ and $\frac{k}{1-\alpha} = \frac{k l \cos \theta}{l \cos \theta – x}$.
Substituting these into the force equation:
$$-\left( \frac{k l \cos \theta}{x} \right) y + \left( \frac{k l \cos \theta}{l \cos \theta – x} \right) (l \sin \theta – y) = mg$$
Divide the entire equation by $k l \cos \theta$:
$$-\frac{y}{x} + \frac{l \sin \theta – y}{l \cos \theta – x} = \frac{mg}{k l \cos \theta}$$
Rearrange to solve for $y$:
$$\frac{l \sin \theta – y}{l \cos \theta – x} = \frac{y}{x} + \frac{mg}{k l \cos \theta}$$
Multiply by $(l \cos \theta – x)$:
$$l \sin \theta – y = \left( \frac{y}{x} + \frac{mg}{k l \cos \theta} \right) (l \cos \theta – x)$$
$$l \sin \theta – y = y \frac{l \cos \theta}{x} – y + \frac{mg}{k} – \frac{mg}{k l \cos \theta} x$$
Cancel $-y$ from both sides:
$$l \sin \theta = y \frac{l \cos \theta}{x} + \frac{mg}{k} – \frac{mg}{k l \cos \theta} x$$
Isolate the term with $y$:
$$y \frac{l \cos \theta}{x} = l \sin \theta – \frac{mg}{k} + \frac{mg}{k l \cos \theta} x$$
Multiply by $\frac{x}{l \cos \theta}$:
$$y = x \frac{\sin \theta}{\cos \theta} – \frac{mg x}{k l \cos \theta} + \frac{mg x^2}{k l^2 \cos^2 \theta}$$
$$y = x \tan \theta – \frac{mg}{k l \cos \theta} x + \frac{mg}{k l^2 \cos^2 \theta} x^2$$
Grouping the $x$ terms gives the final parabolic equation.