Determine the tension in the cable

A spherical navigation buoy is tethered to the lake floor by a vertical cable . The outside diameter of the buoy is 1.00m. The interior of the buoy consists of an aluminum shell 1.0cm thick, and the rest is solid plastic. The density of aluminum is 2700kg/m^3 and the density of the plastic is 280kg/m^3. The buoy is set to float exactly halfway out of the water.

Determine the tension in the cable

Thank you!

Volume (V) of a sphere of radius R is given by:
V = 4πR³/3

The radius of the buoy = 1.00 / 2 = 0.500 m, so the volume of the buoy, Vb, will be:
Vb = 4π * 0.500³ / 3
Vb = 0.524 m³

The plastic sphere filling the aluminium shell has a radius 1 cm less than the buoy, so its volume (Vp) will be:

Vp = 4π * 0.490³ / 3
Vp = 0.493 m³

The volume of the aluminium shell (Va) is the difference between the two, so
Va = (Vb - Vp)
Va = 0.524 - 0.493
Va = 0.0308 m³

Mass = Volume * Density

Mass of aluminium shell = Va * 2700 = 83.16 kg
Mass of plastic filler = Vp * 280 = 138 kg

Total mass of buoy = 83.16 + 138 = 221.2 kg
Weight of buoy = mg = 221.2 * 9.81 = 2170 N {Taking g = 9.81 m/s²}

The buoy floats half way out of the water, so it displaces a volume of water (Vw)
Vw = Vb / 2
Vw = 0.524 / 2 = 0.262 m³
Taking the density of water as 1000 kg/m³
Mass of water displaced = 0.262 * 1000 = 262 kg
Weight of water displaced = mg = 262 * 9.81 = 2570 N

From Archimedes' Principle, buoyant force = weight of water displaced = 2570 N

The buoy is in equilibrium, so downward forces = upward forces
Tension force + weight force = buoyant force
Tension force + 2170 N = 2570 N
Tension = 2570 - 2170 = 400 N