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An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600 kg . It is anchored to the sea bottom by a cable. The density of seawater is 1025 kg/m3.Part A What is the buoyant force on the chamber? Express your answer to three significant figures and include the appropriate units.Part B What is the tension in the cable? Express your answer to two significant figures and include the appropriate units.

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Answer:

A)

[tex]8.75\times10^{5} N[/tex]

B)

[tex]1.7\times10^{4} N[/tex]

Explanation:

[tex]d[/tex] = diameter of the spherical chamber = 5.50 m

Volume of the spherical chamber is given as

[tex]V = \frac{4\pi (\frac{d}{2} )^{3}}{3}\\ V = \frac{4(3.14) (\frac{5.50}{2} )^{3}}{3}\\V = 87.07 m^{3}[/tex]

[tex]\rho[/tex] = density of the seawater = 1025 kg m⁻³

Part A)

Force of buoyancy by seawater on the chamber is given as

[tex]F_{b} = \rho V g \\F_{b} = (1025) (87.07) (9.8)\\F_{b} = 8.75\times10^{5} N[/tex]

Part B)

[tex]T[/tex] = tension force in the cable in upward direction

[tex]m[/tex] = mass of the chamber = 87600 kg

Weight of the chamber is given as

[tex]F_{g} = mg \\[/tex]

Using equilibrium of force in the vertical direction, we have

[tex]T + F_{g} = F_{b}\\T + m g = F_{b}\\T + (87600) (9.8) = 8.75\times10^{5}\\T = 16520 N \\T = 1.7\times10^{4} N[/tex]

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