Question 3.1: An unloaded pontoon being used in a river estuary to transpo......
An unloaded pontoon being used in a river estuary to transport construction equipment has a mass of 15 tonnes (1 tonne = 1000 kg). In plan the pontoon is 8m long by 5 m wide. It is rectangular in section and has sides 1.5 m high. The water is saline with a density of 1025 kg/m³. (a) What is the depth of immersion of the unloaded pontoon? (b) What weight can be carried by the pontoon while still maintaining a freeboard of 0.5 m?
(a) A floating body displaces its own mass of water, in this case 15 tonnes or 15000 kg. Remembering that mass density = mass/volume, then: Volume of water displaced by the pontoon, V = mass/density of saline water = 15000/1025
V = 14.63 m³
Depth of immersion, h = volume (V ) of water displaced/plan area of pontoon (A) = 14.63/(8 × 5)
h = 0.37 m
(b) If the freeboard is 0.5 m then the depth of immersion h = (1.5 – 0.5) = 1.0 m. Therefore volume of water now displaced, V = h × A = 1.0 × (8 × 5) = 40 m³. Thus the maximum amount of water that can be displaced by the loaded pontoon is 40 m³. The mass of the displaced saline water = 40 × 1025 = 41000 kg or 41 tonnes. The pontoon’s mass is 15 tonnes, so it can carry a mass of (41 – 15) = 26 tonnes. Thus the weight that the pontoon can carry = 26000 × 9.81 = 255.06 × 10³ N