Answer to Question #140618 in Molecular Physics | Thermodynamics for Adeola

Question

A body of 3kg and volume 4×10m³ is hung from a balance of graduated in newtons what is the reading of the balance when the body is immersed in water is (a) in air (b) fully immersed in water (c) ⅓ immersed in water (d) fully immersed in paraffin. ( Density of water = 10³kgm³, density of paraffin = 800kgm³
(ii) find the mass of an iceberg which floats in sea water displacing 45m³( density of sea water = 1030kgm³
(iii) what is the tension in the cable holding a mine of mass 600kg and volume 0.8m³ to the sea bed, so that it is totally suerged.

Answer 1

(i) Use the equation for buoyancy force: the force that acts upward ("pulls" bodies out of liquid) is the acceleration due to gravity times liquid density times volume of the submerged part of the object. While this force acts upward, the force of gravity does not change and acts downward. The weight is what scales measure, weight is force of gravity minus buoyancy force.

Thus, we have the following for each situation.

(a) in air:

$$W=F-B=mg-\rho_{air} gV=\\ =3\cdot9.8-1.29\cdot9.8\cdot4000=-50539\text{ N}.$$

This body not simply floats in air but rises up.

(b) fully immersed in water

$$W=F-B=mg-\rho_{water} gV=\\ =3\cdot9.8-1000\cdot9.8\cdot4000=-39200000\text{ N}.$$

Huge buoyancy force acts on the body fully immersed in water.

(c) ⅓ immersed in water

$$W=F-B=mg-\rho_{water} gV_{1/3}=\\ =3\cdot9.8-1.29\cdot9.8\cdot4000\cdot\frac{1}{3}=-13067000\text{ N}.$$

Huge buoyancy force acts on the body 1/3 immersed in water, although it is smaller than in the previous case.

(d) fully immersed in paraffin

$$W=F-B=mg-\rho_{paraf.} gV=\\ =3\cdot9.8-800\cdot9.8\cdot4000=-31400000\text{ N}.$$

(ii) What body displaces is the volume of the submerged part of body. If it floats, then the force of gravity equals the buoyancy force:

$$mg=\rho gV,\\ m=\rho V=1030\cdot45=46350\text{ kg}.$$

(iii) The tension in the cable is caused by the difference between the buoyancy force and the force of gravity:

$$T=|F-B|=|mg-\rho g V|=\\ =|600\cdot9.8-1030\cdot0.8\cdot9.8|=2062\text{ N}$$