A small ferryboat is 3.90 m wide and 5.70 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.30 cm into the river. What is the weight of the truck? kN
Question
A small ferryboat is 3.90 m wide and 5.70 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.30 cm into the river. What is the weight of the truck? kN
Solution
To solve this problem, we need to use the principle of buoyancy, which states that the weight of the water displaced by the boat is equal to the weight of the boat (or in this case, the additional weight of the truck).
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First, we need to find the volume of the water displaced by the boat when the truck is loaded onto it. The boat sinks an additional 4.30 cm (or 0.043 m) into the river, so the volume of the water displaced is the area of the boat multiplied by this additional depth.
Volume = length * width * height Volume = 5.70 m * 3.90 m * 0.043 m = 0.90261 m^3
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The weight of the water displaced is then the volume of the water multiplied by the density of the water and the acceleration due to gravity. The density of water is typically 1000 kg/m^3 and the acceleration due to gravity is approximately 9.81 m/s^2.
Weight = volume * density * gravity Weight = 0.90261 m^3 * 1000 kg/m^3 * 9.81 m/s^2 = 8852.6 N
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The weight of the truck is then this weight converted to kilonewtons (kN), since 1 kN = 1000 N.
Weight = 8852.6 N / 1000 = 8.85 kN
So, the weight of the truck is approximately 8.85 kN.
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