Question

3) A piece of metal with a mass of 3.8 kg and density of 1808 kg/m3 is suspended from a string and then completely immersed in a container of water. The density of water is 1000 kg/m3 .
(a) Determine the volume of the piece of metal.
(b) Determine the tension in the string after the metal is immersed in the container of water.

SHOW ALL WORK AND FORMULAS! THANKS!

Answer 1

(a) volume=mass/density

given, mass of metal piece =3.8kg

density of metal piece =1808 kg/m³

therefore, volume of metal piece = 3.8/1808

= 2.1017x10^-3 m³

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(b) When a body isTotally Submerged in a fluid, an upward force acts on that body known as buoyant force

Buoyant force = (ρ_fluidV_fluidg) = W_fluid .........................(1)

But downward force acting from the body is its silf weight. Given by

Self weight = mg = ρ_objectV_object g .................................(2)

But, For completely submerged object:
V_object = V_{fluid displaced} .........................................(3)

Given that the metal piece is suspended from a string and then completely immersed in a container of water.Then the tension in the string after the metal is immersed in the container of water is the net force acting on the body.

Net force = tension in string = self weight - buoyant force

= (2)_equation - (1)_equation

= ( ρ_objectV_object g) - (ρ_fluid V_fluid g)

= V_object g (  ρ_{object -} ρ_fluid ) ...................( from (3) )

= 2.1017x10^-3 x 9.8 x ( 1808 -1000 )

Hence, Tension in the string = 16.642 N ........................................( N- Newtons )