Solution: Thermal Elasticity and Buoyancy
1. Problem Setup
We have an elastic cord with relaxed length $l = 80 \text{ cm} = 0.8 \text{ m}$ and room-temperature stiffness $k = 100 \text{ N/m}$.
The total height of the hook above the block is $H = 100 \text{ cm} = 1.0 \text{ m}$.
When cold water (height $h$) is added, the cord is divided into two sections with different temperatures and stiffnesses:
- Section 1 (Above Water): Length $l_1$ (relaxed). Stiffness $k$.
- Section 2 (In Water): Length $l_2 = l – l_1$ (relaxed). Stiffness increases to $4k$ (due to $0^\circ \text{C}$).
2. Equilibrium Condition
The block is just about to leave the bottom. This means the normal force is zero, and the elastic force $F$ balances the apparent weight of the block in water.
$$ F = W_{app} = 40 \text{ N} $$
Since the cord is massless, tension $F$ is uniform throughout.
3. Equations for Extension
Upper Part (Air):
Geometric length = $H – h$.
Stiffness of this part: $k_1 = \frac{k l}{l_1}$.
Hooke’s Law: $F = k_1 (Extension) = \frac{k l}{l_1} (H – h – l_1)$.
Lower Part (Water):
Geometric length = $h$.
Stiffness of this part (using $4k$ base stiffness): $k_2 = \frac{4k l}{l – l_1}$.
Hooke’s Law: $F = k_2 (Extension) = \frac{4k l}{l – l_1} (h – (l – l_1))$.
4. Solving the System
Given: $k l = 100 \times 0.8 = 80$, $F = 40$, $H = 1$.
From Upper Part Equation:
$$ 40 = \frac{80}{l_1} (1 – h – l_1) $$
$$ \frac{1}{2} l_1 = 1 – h – l_1 \implies 1 – h = 1.5 l_1 \implies l_1 = \frac{2(1 – h)}{3} \quad \dots(1) $$
From Lower Part Equation:
$$ 40 = \frac{4(80)}{0.8 – l_1} (h – 0.8 + l_1) $$
$$ \frac{40}{320} = \frac{h – 0.8 + l_1}{0.8 – l_1} \implies \frac{1}{8} (0.8 – l_1) = h – 0.8 + l_1 $$
$$ 0.1 – 0.125 l_1 = h – 0.8 + l_1 \implies h – 0.9 + 1.125 l_1 = 0 \quad \dots(2) $$
Substitute (1) into (2):
$$ h – 0.9 + 1.125 \left( \frac{2 – 2h}{3} \right) = 0 $$
$$ h – 0.9 + 0.375 (2 – 2h) = 0 $$
$$ h – 0.9 + 0.75 – 0.75h = 0 $$
$$ 0.25 h – 0.15 = 0 $$
$$ h = \frac{0.15}{0.25} = \frac{15}{25} = 0.6 \text{ m} $$
Final Answer: The minimum height of water required is 60 cm.
We have an elastic cord with relaxed length $l = 80 \text{ cm} = 0.8 \text{ m}$ and room-temperature stiffness $k = 100 \text{ N/m}$. The total height of the hook above the block is $H = 100 \text{ cm} = 1.0 \text{ m}$. When cold water (height $h$) is added, the cord is divided into two sections with different temperatures and stiffnesses:
- Section 1 (Above Water): Length $l_1$ (relaxed). Stiffness $k$.
- Section 2 (In Water): Length $l_2 = l – l_1$ (relaxed). Stiffness increases to $4k$ (due to $0^\circ \text{C}$).
The block is just about to leave the bottom. This means the normal force is zero, and the elastic force $F$ balances the apparent weight of the block in water. $$ F = W_{app} = 40 \text{ N} $$ Since the cord is massless, tension $F$ is uniform throughout.
Upper Part (Air):
Geometric length = $H – h$.
Stiffness of this part: $k_1 = \frac{k l}{l_1}$.
Hooke’s Law: $F = k_1 (Extension) = \frac{k l}{l_1} (H – h – l_1)$.
Lower Part (Water):
Geometric length = $h$.
Stiffness of this part (using $4k$ base stiffness): $k_2 = \frac{4k l}{l – l_1}$.
Hooke’s Law: $F = k_2 (Extension) = \frac{4k l}{l – l_1} (h – (l – l_1))$.
Given: $k l = 100 \times 0.8 = 80$, $F = 40$, $H = 1$.
From Upper Part Equation: $$ 40 = \frac{80}{l_1} (1 – h – l_1) $$ $$ \frac{1}{2} l_1 = 1 – h – l_1 \implies 1 – h = 1.5 l_1 \implies l_1 = \frac{2(1 – h)}{3} \quad \dots(1) $$
From Lower Part Equation: $$ 40 = \frac{4(80)}{0.8 – l_1} (h – 0.8 + l_1) $$ $$ \frac{40}{320} = \frac{h – 0.8 + l_1}{0.8 – l_1} \implies \frac{1}{8} (0.8 – l_1) = h – 0.8 + l_1 $$ $$ 0.1 – 0.125 l_1 = h – 0.8 + l_1 \implies h – 0.9 + 1.125 l_1 = 0 \quad \dots(2) $$ Substitute (1) into (2): $$ h – 0.9 + 1.125 \left( \frac{2 – 2h}{3} \right) = 0 $$ $$ h – 0.9 + 0.375 (2 – 2h) = 0 $$ $$ h – 0.9 + 0.75 – 0.75h = 0 $$ $$ 0.25 h – 0.15 = 0 $$ $$ h = \frac{0.15}{0.25} = \frac{15}{25} = 0.6 \text{ m} $$