Approach

The problem involves two identical elastic cords with negligible relaxed lengths. This implies that the cords behave like ideal springs with natural length zero. The tension in such a cord is directly proportional to its total length ($T = k \cdot \text{length}$). By using vector mechanics, we can find the resultant force exerted by the two cords on the particle at point P and equate it to the external holding force.

1. Force Analysis with Vectors

Let the position of the origin be $\mathbf{O} = (0,0)$. The nails are at positions $\mathbf{A}$ and $\mathbf{B}$. Since they are equidistant from the origin, if $\mathbf{A} = -\mathbf{d}$, then $\mathbf{B} = +\mathbf{d}$. Therefore, $\mathbf{A} + \mathbf{B} = 0$.

Let the ball be at position $\mathbf{P}$. The cords behave as zero-length springs with stiffness $k$. The force exerted by a spring with one end at $\mathbf{X}$ and the other at $\mathbf{P}$ is $\mathbf{F}_{\text{elastic}} = -k(\mathbf{P} – \mathbf{X})$.

The total elastic restoring force $\mathbf{F}_{\text{res}}$ on the ball is the vector sum of the forces from both cords: $$ \mathbf{F}_{\text{res}} = -k(\mathbf{P} – \mathbf{A}) – k(\mathbf{P} – \mathbf{B}) $$ $$ \mathbf{F}_{\text{res}} = -k(2\mathbf{P} – (\mathbf{A} + \mathbf{B})) $$

Since $\mathbf{A} + \mathbf{B} = 0$: $$ \mathbf{F}_{\text{res}} = -2k\mathbf{P} $$ This means the net restoring force is always directed towards the origin with magnitude $2k|\mathbf{OP}|$.

2. Equilibrium Calculation

To hold the ball in equilibrium, the external force $F$ must balance this restoring force. $$ F = |\mathbf{F}_{\text{res}}| = 2k |\mathbf{OP}| $$

We are given:

• Coordinates of P: $(4, 3)$ m
• External Force Magnitude: $F = 1000$ N

First, calculate the distance $|\mathbf{OP}|$: $$ |\mathbf{OP}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m} $$

Now substitute into the force equation: $$ 1000 = 2k(5) $$ $$ 1000 = 10k $$ $$ k = 100 \text{ N/m} $$