Physics on the rocks

Since the ice cube is floating, it should be in static equilibrium. Hence, we should analyse the forces acting on the ice cube, while knowing that the vector sum of forces on it should be zero.

We realise that there are two forces acting on the ice cube: first, the gravitational force exerted by the Earth on the ice cube, and second, the buoyant force exerted by the water on the ice cube.

We can express the weight of the ice cube (a.k.a. the gravitational force exerted on the ice cube) as

\begin{equation} F {\mbox{weight}}=\rho {\mbox{ice}}V_{\mbox{ice}}g \end{equation}

To find the buoyant force exerted on the ice cube, we can use Archimedes' Principle, which states that the buoyant force exerted on any object immersed in a fluid is equal to the weight of the fluid displaced.

Hence, we can express the buoyant force as

\begin{equation} F {\mbox{buoyancy}}=\rho {\mbox{water}}V_{\mbox{displaced}}g \end{equation}

Now that we have knowledge of the two forces involved, we can write the following equation for the vector sum of forces:

\begin{equation} F {\mbox{weight}}-F {\mbox{buoyancy}}=\rho {\mbox{ice}}V {\mbox{ice}}g-\rho {\mbox{water}}V {\mbox{displaced}}g=0 \end{equation}

Now, we can easily manipulate the above equation and substitute some values which can be derived from the information given in the question:

\begin{equation} \mbox{Ratio}=\frac{\rho {\mbox{ice}}}{\rho {\mbox{water}}}=\frac{V {\mbox{displaced}}}{V {\mbox{ice}}}=\frac{2-0.166}{2}=0.917 \end{equation}

Given: Side of cube, consequently it's Volume; Density of water; Height of cube above water

To Find: Relative density of ice

Solution: let mass of block of ice be m1, mass of water displaced be m2, volume of block be v1, volume of water be v2, density of block be d1, density of water be d2(=1 g cm^-3).

Also, let surface area of block be a; height of ENTIRE block h1(=2), height of block under surface=1.834 (2-0.166)

As the block of ice is in a stable equilibrium state the archimedes principle can be applied to it.

Weight of block = Weight of water displaced

m1 g = m2 g

v2 d2 = v1 d1 [mass=volume*density]

a 1.834 1=a 2 d1 1.834=2*d1 d1=1.834/2

d_{ice}=0.917

d {ice}/d {water} = 0.917/1 = 0.917

Using Archimedes' principle, we'll get the following equation:

$\frac { \text{density of ice} }{ \text{density of water} }$ = $\frac { \text{weight of displaced ice} }{ \text{total weight} }$

Note that:

density = $\frac { \text{mass} }{ \text{volume} }$ = $\frac { \text{weight / g} }{ \text{volume} }$ ; g = gravitational constant

Hence, weight is proportional to the change of volume. A (Area) of the ice is constant:

$\frac { \text{density of ice} }{ \text{density of water} }$ = $\frac { \text{weight of displaced ice} }{ \text{total weight} }$

$\frac { \text{density of ice} }{ \text{density of water} }$ = $\frac { \text{height of displaced ice} }{ \text{total height} }$ = $\frac { 2 - 0.166 }{ 2 }$ = 0.917

Thus, the ratio of the density of ice to the density of water is 0.917.

We start of by assuming that there is no movement of the cube inside the water, which implies that the total acceleration is zero. The gravitational force exerted on a cube is m g where m can be represented by the product of density and the total volume, that is m g = "rho"(ice) * a^3*g, where a is the side of a cube (2 cm). The force opposing to the gravitational force is the buoyancy force due to the water, but not exerted on a total volume, just on the part of the cube that is under water. Horizintal side of the cube remains the same and it is a=2 cm, but vertical is equal to (a-x). If we put these into the same equation we end up with:

"rho"(ice) a^3 g = "rho"(water) a^2 (a-x)*g

after cancellations of g's we have

"rho"(ice) a^3 = "rho"(water) a^2*(a-x)

and from this equation we have the ratio od densities:

"rho"(ice)/"rho"(water) = (a-x)/a = 0.917

The ice is floating, so the system is in equilibrium. Therefore the bouyant force of the water on the ice and the force of gravity on the ice must be equal and opposite,

$F_{b}=\rho_w g V_{disp}=m_i g = \rho_i V_i g$ ,

where $\rho$ denotes the density of the water or ice depending on subscript. This allows us to solve for $\rho_i/\rho_w=V_{disp}/V_i=((2-0.166) \times 4)/(2 \times 4)=0.917$ .

Volume under water=Vs ; Total volume = V V* ice density = Vs * Water density Ice density/ Water density= Vs/V There is te answer