Question

What is the minimum area of the top surface of a slab of ice with a uniform thickness of 0.30 m floating on fresh water that will hold up a car of mass 1100 kg sitting at the center on the top surface? The density of water is 1.0 × 103 kg/m3 and density of ice is 0.92 × 103 kg/m3 .

Answer 1

Answer:

The minimum area of the top surface of a slab of ice is 45.8 m²

Explanation:

Given that,

Thickness = 0.30 m

Mass of car = 1100 kg

Density of water $\rho_{w}=1.0\times10^{3}\ kg/m^3$

Density of ice $\rho_{i}=0.92\times10^{3}\ kg/m^3$

We need to calculate the minimum area of the top surface of a slab of ice

Using Buoyant force force

Total weight of car + ice weight = weight of water displaced

$W_{c}+A\times t\times\rho_{i}\times g=\rho\times A\times t\times g$

Put the value into the formula

$1100\times g+0.30\times A\times0.92\times10^{3}\times g=1.0\times10^{3}\times A\times0.30\times g$

$(1100+A\times0.30\times0.92\times10^{3})g=1.0\times10^{3}\times A\times0.30\times g$

$1100+276 A=300A$

$A=\dfrac{1100}{24}$

$A=45.8\ m^2$

Hence, The minimum area of the top surface of a slab of ice is 45.8 m²