Just need an answer checked on a problem

[AryRezvani]

Homework Statement

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.

Homework Equations

Both work and equations are below.

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.

 

[PeterO]

Homework Statement

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.

Homework Equations

Both work and equations are below.

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.

Is this buoy supposed to be floating 1/3 out of the water or 1/2 out of the water [see red above]