Let the total length of the rod be $l$, mass be $m$, and cross-sectional area be $A$. The relative density of the wood is given as $\gamma = \frac{\rho_{rod}}{\rho_{water}}$.

The forces acting on the rod are:

• Weight ($m\vec{g}$): Acting downwards at the center of mass (distance $l/2$ from the top end). Magnitude $mg$.
• Tension ($\vec{T}$): Acting upwards at the top end of the rod.
• Buoyant Force ($\vec{B}$): Acting upwards at the center of buoyancy. If $l’$ is the length of the rod submerged, the center of buoyancy is at a distance $l’/2$ from the bottom end, or $l – l’/2$ from the top end.

The magnitude of the buoyant force is given by the weight of the displaced water:

$$ B = (A l’) \rho_{water} g $$

We can express $m$ in terms of density: $m = (A l) \rho_{rod} = (A l) \gamma \rho_{water}$. Substituting $\rho_{water} A = \frac{m}{l \gamma}$, we get:

$$ B = \frac{m}{l \gamma} l’ g = \frac{mg}{\gamma} \left(\frac{l’}{l}\right) $$

While the rod is being pulled out slowly, it maintains rotational equilibrium. We take the torque about the point of suspension (the top end where $T$ acts). The torque due to Tension is zero.

Let $\theta$ be the angle the rod makes with the horizontal (or vertical, the $\sin\theta$ term cancels out regardless). The condition $\sum \vec{\tau} = 0$ gives:

$$ \tau_{gravity} = \tau_{buoyancy} $$ $$ mg \left( \frac{l}{2} \sin\theta \right) = B \left( l – \frac{l’}{2} \right) \sin\theta $$

Canceling $\sin\theta$ and substituting the expression for $B$:

$$ mg \frac{l}{2} = \left[ \frac{mg}{\gamma} \frac{l’}{l} \right] \left( l – \frac{l’}{2} \right) $$ $$ \frac{l}{2} = \frac{l’}{\gamma l} \left( \frac{2l – l’}{2} \right) $$ $$ \gamma l^2 = l'(2l – l’) $$ $$ l’^2 – 2l l’ + \gamma l^2 = 0 $$

Solving this quadratic equation for $l’$:

$$ l’ = \frac{2l \pm \sqrt{4l^2 – 4\gamma l^2}}{2} = l (1 \pm \sqrt{1 – \gamma}) $$

Since the submerged length $l’$ must be less than $l$, we take the negative root:

$$ l’ = l(1 – \sqrt{1 – \gamma}) $$

Crucial Conclusion: As long as the rod is tilted (partially submerged and not vertical), the submerged length $l’$ remains constant, independent of the height $h$.

From the translational equilibrium in the vertical direction ($\sum F_y = 0$):

$$ T + B = mg \implies T = mg – B $$

Substitute $B = \frac{mg}{\gamma} \frac{l’}{l}$ and the derived value of $l’$:

$$ T = mg \left[ 1 – \frac{1}{\gamma} \frac{l(1 – \sqrt{1-\gamma})}{l} \right] $$ $$ T = mg \left[ 1 – \frac{1 – \sqrt{1-\gamma}}{\gamma} \right] $$

Given $\gamma = 3/4$:

$$ T = mg \left[ 1 – \frac{1 – \sqrt{1 – 3/4}}{3/4} \right] = mg \left[ 1 – \frac{1 – 1/2}{3/4} \right] = mg \left[ 1 – \frac{1/2}{3/4} \right] $$ $$ T = mg \left[ 1 – \frac{2}{3} \right] = \frac{mg}{3} $$

Range for Phase 1: This phase lasts until the rod becomes vertical. In the critical vertical position, the submerged length is still $l’ = l/2$. The height of the top end $h$ is the length of the non-submerged part: $h = l – l’ = l – l/2 = l/2$. Thus, this applies for $0 \le h \le l/2$.

Once $h > l/2$, the rod hangs vertically. The submerged length is no longer constant but depends on $h$ as $l_{sub} = l – h$.

The buoyant force becomes:

$$ B = \frac{mg}{\gamma} \frac{l_{sub}}{l} = \frac{mg}{\gamma} \frac{l – h}{l} = \frac{mg}{\gamma} \left( 1 – \frac{h}{l} \right) $$

The tension is:

$$ T = mg – B = mg \left[ 1 – \frac{1}{\gamma} \left( 1 – \frac{h}{l} \right) \right] $$

Substituting $\gamma = 3/4$:

$$ T = mg \left[ 1 – \frac{4}{3} \left( 1 – \frac{h}{l} \right) \right] $$

This phase continues until the rod is completely out of the water, i.e., $h = l$. So, $l/2 \le h \le l$.

When $h \ge l$, the rod is fully in the air. $B = 0$.

$$ T = mg $$