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Q62P

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Physics For Scientists & Engineers
Found in: Page 444

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Short Answer

Review: The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measurement in air, however, is disturbed by buoyant forces. An object of volume V is weighed in air on an equal-arm balance with the use of counterweights of density ρ . Representing the density of air as ρair and the balance reading as Fg' , show that the true weight Fg is

Fg=Fg'+(v-Fg'ρg)ρgair

Hence it is proved that:

Fg=Fg'+(v-Fg'ρg)ρgair

See the step by step solution

Step by Step Solution

Identification of the given data

The given data can be listed below as,

⦁ The volume of object is, V.

⦁ The density of object is, ρ .

⦁ The density of air is, ρair .

⦁ The balance reading weight is, F'g .

⦁ The true weight is, Fg .

Significance of the buoyant force

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

B=ρfluidgVfluid

Determination of the true weight  Fg

When both weight and the body are in balance, the force relation by using a above concept of step (2) is given by:

Fg-B=Fg'-B'Fg-ρairgv=Fg'-ρairgv'

The value v' is expressed as,

v'=m'ρ

And

m'=Fg'g

Substitute all the value in the above equation,

Fg-ρairgv=Fg'-ρairgm'ρFg-ρairgv=Fg'-ρairgFg'ρgFg=Fg'-ρairgFg'ρg+ρairgvFg=Fg'+ρairg-Fg'ρg+vFg=Fg'+ρairgv-Fg'ρg

Hence the true weight Fg is, Fg=Fg'+ρairgv-Fg'ρg

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