Question

A spherical buoy of diameter 0.5 m and mass 35 kg is attached to the seabed by a mooring rope and floats fully submerged as shown above. Calculate the tension in the mooring rope. The density of sea water is 1020 kgm−3.

Answer 1

Let’s consider the free-body diagram in the picture above. From the FBD we can see

that the buoyant force tends to pull the buoy upward while the force of gravity (or

weight of the buoy) tends to pull the buoy downward. So, we can write the tension in

the mooring rope as follows:

$𝑇 = 𝐹_𝐵 − 𝑚𝑔$

here, 𝐹_𝐵 is the buoyant force, 𝑚𝑔 is the force of gravity (or weight of the buoy).

By the definition, the buoyant force is equal to the weight of the sea water displace:

$F_B= \rho_{sea water}V_{sea water}g$

here, 𝜌𝑠𝑒𝑎 𝑤𝑎𝑡𝑒𝑟 is the density of the sea water, 𝑉𝑠𝑒𝑎 𝑤𝑎𝑡𝑒𝑟 = 𝑉𝑏𝑢𝑜𝑦 is the volume of the

sea water displaced that is equal to the volume of the buoy, 𝑔 is the acceleration due to

gravity.

We can find the volume of the spherical buoy from the formula:

$V_{bouy}= \frac{4}{3}\pi R^3_{buoy}$

$So, T=F_B-mg=\rho_{sea water} \frac{4}{3} \pi R^3_{buoy} g-mg=(\rho_{sea water} \frac{4}{3} \pi R^3_{buoy}-m)g$

$T=(1020 \times \frac{4}{3} \times \pi \times 0.25^3-35)0.98=311.24N$