An undersea research chamber is spherical with an external diameter of 5.20 m. The mass of the chamber, when occupied, is 74,400 kg. It is anchored to the sea bottom by a cable. What is (a) the buoyant force on the chamber, and (b) the tension in the cable?

Buoyancy is an upward force exerted by the fluid in the upward direction that opposes the weight of a fully or partially submerged body. It is equal to the weight of the water displaced. It is expressed in Newtons.

The external diameter of the research chamber is \({D_{{\rm{chamber}}}} = 5.20\;{\rm{m}}\).

The mass of the chamber is \({m_{{\rm{chamber}}}} = 74400\;{\rm{kg}}\).

Let us understand the question better with the help of the diagram shown below:

There are three forces on the chamber:

The buoyant force \({F_{\rm{B}}}\), the tension in the cable \({F_{\rm{T}}}\), and the weight of the chamber \(mg\).

Now, we know that the buoyant force is the weight of water displaced by the chamber.

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= {\rho _{{\rm{water}}}}{V_{{\rm{chamber}}}}g\\{F_{{\rm{buoyant}}}} &= {\rho _{{\rm{water}}}}\frac{4}{3}\pi R_{{\rm{chamber}}}^3g\end{aligned}\)

Here \({\rho _{{\rm{water}}}}\) is the density of seawater, which is \(1.025 \times {10^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}\)(standard), \({V_{{\rm{chamber}}}}\) is the volume of the chamber (equal to the volume of water displaced), and \({R_{{\rm{chamber}}}}\) is the radius of the chamber \(\left( {{R_{{\rm{chamber}}}} = \frac{{{D_{{\rm{chamber}}}}}}{2} = \frac{{5.20}}{2} = 2.60\;{\rm{m}}} \right)\).

Substitute the values in the above formula for buoyant force.

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= \left( {1.025 \times {{10}^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}} \right) \times \frac{4}{3}\pi \left( {2.60\;{{\rm{m}}^3}} \right)\left( {9.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right)\\{F_{{\rm{buoyant}}}} &= 7.3953 \times {10^5}\;{\rm{N}}\\{F_{{\rm{buoyant}}}} &\approx 7.40 \times {10^5}\;{\rm{N}}\end{aligned}\)

Hence, the buoyant force on the spherical undersea research chamber is \(7.40 \times {10^5}\;{\rm{N}}\).

To find the tension, we will be using the equilibrium condition for the chamber, i.e.,

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= mg + {F_T}\\{F_{\rm{T}}} &= {F_{{\rm{buoyant}}}} - mg\end{aligned}\)

Substitute the values in the above formula for buoyant force.

\(\begin{aligned}{F_{\rm{T}}} &= 7.3953 \times {10^5}\;{\rm{N}}\; - \;\left( {7.44 \times {{10}^4}\;{\rm{kg}}} \right)\left( {9.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right)\left( {\frac{{1\;{\rm{N}}}}{{1\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}}}} \right)\\{F_{\rm{T}}} &= 1.0 \times {10^4}\;{\rm{N}}\end{aligned}\)

Hence, the tension in the cable is \(1.0 \times {10^4}\;{\rm{N}}\).