Problem 17: Effect of Acceleration on Submerged Depth
Initial State ($a=0$):
The cylinder acts under Gravity ($mg$), Buoyancy ($F_B$), and Spring tension ($kx_0$). Since the spring is extended, it pulls down. $$ F_B = mg + kx_0 $$ $$ \rho_w S h_{sub} g = mg + kx_0 \quad \dots(1) $$ where $h_{sub}$ is the submerged length.
Accelerating State (Upward acceleration $a$):
In the frame of the vessel, we apply a pseudo force $ma$ downwards. Effective gravity becomes $g’ = g+a$. Let the new extension be $x’$ and new submerged length be $h’_{sub}$. Note: If the spring extends by an additional $\Delta x$ (cylinder moves up), the submerged portion decreases by $\Delta x$ (since water moves with the vessel). $$ x’ = x_0 + \Delta x $$ $$ h’_{sub} = h_{sub} – \Delta x $$
New Equilibrium Equation: $$ F_B’ = W’ + F_{spring}’ $$ $$ \rho_w S h’_{sub} (g+a) = m(g+a) + k x’ $$ Substituting expressions in terms of $\Delta x$: $$ \rho_w S (h_{sub} – \Delta x) (g+a) = m(g+a) + k (x_0 + \Delta x) $$
From equation (1), we know $m = \rho_w S h_{sub} – \frac{k x_0}{g}$. Let’s rearrange (1) as $\rho_w S h_{sub} = m + \frac{k x_0}{g}$. Actually, simpler method: Multiply eq (1) by $\frac{g+a}{g}$: $$ \rho_w S h_{sub} (g+a) = m(g+a) + k x_0 \frac{g+a}{g} $$ Compare this with the acceleration equation expanded: $$ \rho_w S h_{sub} (g+a) – \rho_w S \Delta x (g+a) = m(g+a) + k x_0 + k \Delta x $$ Substitute the left term from our modified eq (1): $$ [ m(g+a) + k x_0 (1 + \frac{a}{g}) ] – \rho_w S \Delta x (g+a) = m(g+a) + k x_0 + k \Delta x $$ Cancel $m(g+a)$ and $k x_0$: $$ k x_0 \frac{a}{g} – \rho_w S \Delta x (g+a) = k \Delta x $$ $$ k x_0 \frac{a}{g} = \Delta x [ k + \rho_w S (g+a) ] $$
Since $\Delta x$ is positive (cylinder moves up relative to the base), the submerged length decreases by this amount.