Question

One day when you come into physics lab you find several plastic hemispheres floating like boats in a tank of fresh water. Each lab group is challenged to determine the heaviest rock that can be placed in the bottom of a plastic boat without sinking it. You get one try. Sinking the boat gets you no points, and the maximum number of points goes to the group that can place the heaviest rock without sinking. You begin by measuring one of the hemispheres, finding that it has a mass $21g$and a diameter of $8.0cm$. What is the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat?

Short answer

Therefore, the length of the submerging of cylinder is equal to$18cm$

Step-by-step solution

Step :1 Introduction 

The upward force experienced by an object immersed in a fluid is equal to the weight of the fluid displaced by the immersed portion of the body. This force is called as buoyant or up thrust force.

Step :2 Explanation 

The depth to which the cylinder is submerged is equal to the length of the stretching of the spring.

In the case of static equilibrium, the net force acting on the cylinder is equal to zero.

$\mathrm{\Sigma }{F}_y=0$

$F_{\mathrm{B}}+F_{\mathrm{s}}-mg=0$

${\rho }_{\mathrm{w}}Ayg+ky=mg$

$y=\frac{mg}{{\rho }_{\mathrm{w}}Ag+k}$

Here, $F_B$is buoyant force due to liquid of density ${\rho }_{\mathrm{w}}$, $F_S$is spring force, $m$is mass, $g$is acceleration due to gravity,$A$is cross-sectional area of the cylinder, $y$is length of submerging of cylinder, and$K$ is spring constant of the spring.

Step :3 Calculating the length 

Calculate the length of the submerging of cylinder.

Substitute $1.0\mathrm{kg}\text{for}m,9.8\mathrm{m}/{\mathrm{s}}^2,1000\mathrm{kg}/{\mathrm{m}}^3\text{for}{\rho }_{\mathrm{w}},\pi (0.025\mathrm{m}{)}^2\text{for}A\text{, and}35\mathrm{N}/\mathrm{m}\mathrm{for}\mathrm{k}$

in the equation

$y=\frac{\left(1.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\pi (0.025\mathrm{m}{)}^2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+35\mathrm{N}/\mathrm{m}}$

$=0.181\mathrm{m}$

$=0.181\mathrm{m}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)$

$=18.1\mathrm{cm}$