(a) What is the minimum area of a block of ice \(0.305 \mathrm{~m}\) thick floating on water that will hold up an automobile of mass \(1120 \mathrm{~kg} ?(b)\) Does it matter where the car is placed on the block of ice? The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\).

Firstly, we need to calculate the volume of ice required to hold up the automobile. We know from Archimedes' principle that the upthrust force equals the weight of fluid displaced by the submerged part of the object. Hence, we equate the weight of the car (which is its mass multiplied by gravitational acceleration, i.e. \(1120 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2}\)) to the weight of the water displaced (which is the density of the water multiplied by the volume of ice submerged and the gravitational acceleration i.e. \(1000 \mathrm{~kg/m^{3}} \times V \times 9.8 \mathrm{~m/s^{2}}\)). So, \( 1120 \mathrm{~kg} \times 9.8 \mathrm{~m/s^{2}} = 1000 \mathrm{~kg/m^{3}} \times V \times 9.8 \mathrm{~m/s^{2}}\). Simplifying, we find the volume of required ice, \(V\), to be \(11.2 \mathrm{~m^{3}}.\)

Knowing the volume, we can find the required area of the ice block, \(A\), using the known thickness of the ice, \(0.305 \mathrm{~m}\). Since \( V = A \times \text{thickness} \), we find that \( A = V / \text{thickness} = 11.2 \mathrm{~m^{3}} / 0.305 \mathrm{~m} = 36.7 \mathrm{~m^{2}}.\)

Given the the ice block is uniform and the water underneath it is also uniform, the placement of the car on the ice block doesn't matter. As long as the total weight does not exceed that of the water displaced by the volume of submerged ice block, the ice block will float, regardless of where the car is situated. However, distributing the weight evenly would reduce the risk of the ice cracking.