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Suppose I have a rod that has a density of $X <1$. If I were to submerge that rod in water (density 1), I would expect $X$ of the rod to be below water and $1-X$ of it to be above water (simple buoyancy).

But here's the catch. Suppose I attach one corner of the rod to a frictionless hinge, suspended close to the water. How will that affect what percent of the rod is above water in equilibrium now? My teacher told me it is NOT simply $1-X$. He also didn't specify how high above the water the hinge was, so I suppose that doesn't matter. Please tell me if more info is needed.

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Here's a solution assuming a "thin rod"; i.e. assuming that the free end of the rod is fully submerged and "the stretch where the rod begins to dive and is only partially submerged" is negligibly small compared to the stretch (from the end attached to the hinge) that is lifted completely above the water as well as compared to the fully submerged stretch. (Without this assumption or approximation the problem may be considered and even solved as well, but the solution seems too unwieldy to be presentable.)

Let's call $H$ the height above water level of the end with the hinge. According to the stated assumptions, the "stretch above water" therefore has a length of $H / \text{Sin[} \, \phi \, \text{]}$, where $\phi$ is the angle the rod makes with the horizontal water surface, and the remaining fully submerge stretch has a length of $L - H / \text{Sin[} \, \phi \, \text{]}$.

This configuration is stable

  • if the hinge is sufficiently supported to hold the rod end in height $H$. (That's supposed to be satisfied by construction. See the calculation of the required supporting force below.) And:

  • if the torque ("around the hinge") due to the weight of the rod, which acts to turn/pull the rod down towards the vertical (i.e. towards increasing $\phi$) is balanced by the torque due to the force/buoyancy/lift acting on the submerged stretch of the rod:

$0 = \int_0^L \, \mathbf{r \cdot dF}_{\text{weight}} - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{r \cdot dF}_{\text{lift}}$,

$0 = \int_0^L \, \mathbf{r \cdot g} \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{r \cdot g} \, \rho_{\text{water}} \, A \, dl$,

$0 = \int_0^L \, l \, \text{Cos[} \, \phi \, \text{]} \, g \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, l \, \text{Cos[} \, \phi \, \text{]} \, g \, \rho_{\text{water}} \, A \, dl$,

where $A$ is the (otherwise considered negligible) cross-sectional area of the rod. Let's assume that it is constant along the length of the rod; therefore it cancels. Also assume that the rod is of uniform density. This leads to:

$0 = \rho_{\text{rod}} \, \int_0^L \, l \, dl - \rho_{\text{water}} \, \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, l \, dl$,

$\rho_{\text{rod}} \, L^2 \, / 2 = \rho_{\text{water}} \, (L^2 \, / 2 - (H / \text{Sin[} \, \phi \, \text{]})^2 \, / 2).$

Solving for the length of the stretch of rod above water:

$H / \text{Sin[} \, \phi \, \text{]} = L \, \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }$,

or following the terminology given in the question:

$H / \text{Sin[} \, \phi \, \text{]} = L \, \sqrt{ 1 - X }$.

If the height of the hinge reaches or exceeds the value $L \, \sqrt{ 1 - X }$ then the rod simply hangs vertically (dipping in the water as long as the height of the hinge doesn't exceed $L$, of course). If the height of the hinge approaches $0$ it also approaches the (physically inevitable) actual thickness of the rod and the initially made assumption breaks down; the rod then swims more or less horizontally, with $1 - X$ of its volume above water.

Finally, the force supported by the hinge has to compensate the weigth and buoyancy of the rod:

$\mathbf{F}_{\text{hinge}} = $
$\int_0^L \, \mathbf{dF}_{\text{weight}} - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{dF}_{\text{lift}} = $
$\int_0^L \, \mathbf{g} \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{g} \, \rho_{\text{water}} \, A \, dl = $
$\mathbf{g} \, \left( \rho_{\text{rod}} \, A \, L - \rho_{\text{water}} \, A \, (L - H / \text{Sin[} \, \phi \, \text{]}) \right)$.

Inserting the result from above:

$\mathbf{F}_{\text{hinge}} = $
$\mathbf{g} \, \left( \rho_{\text{rod}} \, A \, L - \rho_{\text{water}} \, A \, (L - L \, \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $
$\mathbf{g} \, \left( m_{\text{rod}} - \rho_{\text{water}} \, A \, L \, (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $
$\mathbf{g} \, \left( m_{\text{rod}} - m_{\text{rod}} \, \rho_{\text{water}} / \rho_{\text{rod}} \, (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $
$\mathbf{g} \, m_{\text{rod}} \, \left( 1 - \rho_{\text{water}} / \rho_{\text{rod}} (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) \equiv $
$\mathbf{g} \, m_{\text{rod}} \, \left( 1 - 1/X (1 - \sqrt{ 1 - X }) \right) $,

which is less than $\mathbf{g} \, m_{\text{rod}}$ for $X \lt 1$.

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