To solve this problem, we need to use the principle of buoyancy, which states that the weight of the water displaced by an object is equal to the weight of the object.

1) The weight of the pontoon in fresh water:

First, we need to calculate the volume of the pontoon submerged in fresh water. This is given by the product of the width, length, and draught of the pontoon, which is 6m * 12m * 1.5m = 108 cubic meters.

The weight of the water displaced by the pontoon is equal to the weight of the pontoon. The density of fresh water is approximately 1000 kg/m^3. So, the weight of the water (and therefore the weight of the pontoon) is the volume of the water displaced multiplied by the density of the water and the acceleration due to gravity (9.81 m/s^2).

Weight of pontoon = Volume * Density * g = 108 m^3 * 1000 kg/m^3 * 9.81 m/s^2 = 1,059,480 N.

2) The draught in sea water:

The weight of the pontoon will remain the same in sea water, but because sea water is denser, the pontoon will not need to displace as much water to float, and therefore the draught will be less.

We can find the new draught by dividing the weight of the pontoon by the product of the width, length, density of sea water, and gravity.

Draught = Weight / (Width * Length * Density * g) = 1,059,480 N / (6m * 12m * 1025 kg/m^3 * 9.81 m/s^2) = 1.44 m.

3) The maximum load that can be carried by the pontoon in sea water:

The maximum load that can be carried by the pontoon is the difference between the weight of the water displaced when the pontoon is fully loaded and the weight of the pontoon itself.

When fully loaded, the pontoon displaces a volume of water equal to the width, length, and maximum allowable draught.

Volume = Width * Length * Max Draught = 6m * 12m * 2m = 144 m^3.

The weight of this volume of sea water is Volume * Density * g = 144 m^3 * 1025 kg/m^3 * 9.81 m/s^2 = 1,445,892 N.

So, the maximum load that can be carried by the pontoon is 1,445,892 N - 1,059,480 N = 386,412 N.

Question

To solve this problem, we need to use the principle of buoyancy, which states that the weight of the water displaced by an object is equal to the weight of the object.

1) The weight of the pontoon in fresh water:

First, we need to calculate the volume of the pontoon submerged in fresh water. This is given by the product of the width, length, and draught of the pontoon, which is 6m * 12m * 1.5m = 108 cubic meters.

The weight of the water displaced by the pontoon is equal to the weight of the pontoon. The density of fresh water is approximately 1000 kg/m^3. So, the weight of the water (and therefore the weight of the pontoon) is the volume of the water displaced multiplied by the density of the water and the acceleration due to gravity (9.81 m/s^2).

Weight of pontoon = Volume * Density * g = 108 m^3 * 1000 kg/m^3 * 9.81 m/s^2 = 1,059,480 N.

2) The draught in sea water:

The weight of the pontoon will remain the same in sea water, but because sea water is denser, the pontoon will not need to displace as much water to float, and therefore the draught will be less.

We can find the new draught by dividing the weight of the pontoon by the product of the width, length, density of sea water, and gravity.

Draught = Weight / (Width * Length * Density * g) = 1,059,480 N / (6m * 12m * 1025 kg/m^3 * 9.81 m/s^2) = 1.44 m.

3) The maximum load that can be carried by the pontoon in sea water:

The maximum load that can be carried by the pontoon is the difference between the weight of the water displaced when the pontoon is fully loaded and the weight of the pontoon itself.

When fully loaded, the pontoon displaces a volume of water equal to the width, length, and maximum allowable draught.

Volume = Width * Length * Max Draught = 6m * 12m * 2m = 144 m^3.

The weight of this volume of sea water is Volume * Density * g = 144 m^3 * 1025 kg/m^3 * 9.81 m/s^2 = 1,445,892 N.

So, the maximum load that can be carried by the pontoon is 1,445,892 N - 1,059,480 N = 386,412 N.

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