A uniform cylinder of length L and mass M hav | vedclass

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Question

(C) At equilibrium,the forces acting on the cylinder are the spring force $kx_0$ (upwards),the buoyant force $F_B$ (upwards),and the gravitational for

Vedclass · Accepted Answer

$\frac{Mg}{k}\left( 1 - \frac{LA\sigma}{2M} \right)$

Vedclass · Answer

(C) At equilibrium,the forces acting on the cylinder are the spring force $kx_0$ (upwards),the buoyant force $F_B$ (upwards),and the gravitational force $Mg$ (downwards).The equation of equilibrium is: $kx_0 + F_B = Mg$.The buoyant force $F_B$ is equal to the weight of the displaced liquid: $F_B = V_{submerged} \cdot \sigma \cdot g$.Since the cylinder is half-submerged,the submerged volume is $V_{submerged} = A \cdot \frac{L}{2}$.Thus,$F_B = \left( A \cdot \frac{L}{2} \right) \sigma g = \frac{LA\sigma g}{2}$.Substituting this into the equilibrium equation:$kx_0 + \frac{LA\sigma g}{2} = Mg$$kx_0 = Mg - \frac{LA\sigma g}{2}$$x_0 = \frac{Mg - \frac{LA\sigma g}{2}}{k} = \frac{Mg}{k} \left( 1 - \frac{LA\sigma}{2M} \right)$.

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