Question

A pontoon is used on a river to transfer materials to a barge. The pontoon is constructed from high density polyethylene and has a mass of 40 tonnes. The dimensions of the pontoon are 19 m long x 10m wide x 0.75m high.

a) What is the freeboard of the pontoon when it is floating in the river?

b) What additional mass (tonnes) can the pontoon carry if a maximum draught of 0.45m is available?

c) If the additional mass is to be carried as sand (density 1600kg/m3) in a container 6 m long and 5m wide, how deep a layer of sand can be carried? If instead of sand, the pontoon is to carry bentonite at a density of 2600kg/m3, what depth can be carried?

Answer 1

When a body is immersed in a fluid, the fluid exerts an upward for on the body which is equal to the weight of fluid displaced by the immersed body. This upward force is called buoyant force.

If the weight of body is less than the upward force, then the body floats in the fluid.

Pontoons are working with the same principle. The material that is used to make pontoon is having density less than that of water.

Here the given details are:

Mass of pontoon = 40000 kg

Volume of pontoon = 19×10×0.75

                                 = 142.5 m3

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Part a

As the mass of pontoon floating in the water is 40000 kg , the mass of water displaced will be equal to 40000 kg.

Volume of water displaced will be equal to the volume of body immersed. Let dd be the immersed depth.

Density of water = 1000 kg/m3

Mass of water displaced = Mass of pontoon

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Part b

If, maximum draught is 0.45 m, Volume of body immersed = 19×10×0.45

                                                     Volume of body immersed = 85.5 m3

Mass of water displaced = Volume of body immersed×1000

                                           = 85.5×1000

Mass of water displaced = 85500 kg

Mass of pontoon + Maximum additional mass = Mass of water displaced

40000 + Maximum additional mass = 85500

Maximum additional mass = 45500 kg

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Part c

Density of sand = 1600 kg/m3

Depth of sand in the container = d

Volume of sand in the container = 6×5×d

                                                    = 30d

Mass of sand = Density of sand × Volume of sand

                      = 1600 ×30d

                      = 48000d

Mass of sand = Maximum additional mass

48000d = 45500

d = 0.948 m

Therefore, depth of sand = 0.948 m

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If the material was Bentonite instead of sand,

Mass of Bentonite = Density of Bentonite × Volume of sand

                      = 2600 ×30d

                      = 78000d

Mass of Bentonite = Maximum additional mass

78000d = 45500

d = 0.948 m

Therefore, depth of Bentonite = 0.583 m

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