1. The problem statement, all variables and given/known data I'd like to mention beforehand this is not my work, i'm simply just trying to understand it. I believe the answer is incorrect, could someone kindly indicate why? A spherical navigation buoy is tethered to the sea floor by a vertical cable as shown in the following figure. The outside diameter of the buoy is 1.00 m. The interior of the buoy consists of an aluminum shell 1.0 cm thick, and the rest is solid plastic. The density of aluminum is 2700 kg/m3 and the density of the plastic is 200 kg/m3. The buoy is set to float exactly one-third out of the water. Determine the tension in the cable. 2. Relevant equations Both work and equations are below. 3. The attempt at a solution Buoy Quiz 1) The density of water is 1 Ton / Cubic Meter or 1000 Kg / Cubic Meter. 2) The law of buoyancy is Volume of object submerged is equal to Volume of liquid displaced. Now, calculate the volume of hemisphere. V = (4/6) x Pi x (R^3) , where R = 1/2 of Diameter = 0.50m Pi = 3.1416 V = 0.262 cu.meter Calculate the volume of aluminium shell : V(Al) = (0.01) x 2 x Pi x R^2 V(Al) = 0.016 cu.meter Therefore, the volume of plastic is : V(Plastic) = Vol. of hemisphere - Vol. of aluminium V(Plastic) = (0.262 - 0.016) = 0.246 cu.meter The Buoyant Force is : Bf = Volume of Hemisphere x Density of Water = (0.262 m^3) x (1000 Kg/m^3) Bf = 262 Kgs. The Weight of Object is : W = Volume of Sphere x Density of Materials W = ((0.016 m^3) x (2700 Kg/m^3) x 2) + ((0.246 m^3) x (200 Kg/m^3) x 2) W = 184.80 Kgs. The Tension (T) in the Cable is: T = Bf - W T = 262 kgs. - 184.80 Kgs. T = 77.20 Kgs.