The weight of an object under Water!

Well, the calculation has always been a trending one in the field of Mathematics and Science as no theory can be proved without a proof…

Here we go providing you with a solution for calculating the Weight of an Object in the Water!

The weight of the submerged body of an object can be calculated by subtracting the weight of the volume of water which is been displaced from its air weight.

Taking into consideration two facts:

• Fresh Water weighs 62.366 pounds/cubic foot
• Saline Water weighs 64.043 pounds/cubic foot

Hence, the weight of an Object in Water is:

Ws = Wa – Ww

Where

Ws: Submerged Weight of Object

Wa: Weight of Object in Air

Ww: Weight of Water Displaced

Since Weight can be demonstrated in terms of volume and density. Hence, we substitute the known values of the density…

1. For Fresh Water:

Ws/sea = Wa-(62.366)*Vo                                     (Equation 1)

Where

Ws/sea: Submerged Weight of Object in Seawater in pounds

Wa: Weight of Object in pounds

Vo: Volume of Object in cubic feet

2. For Saline Water:

Ws/sea = Wa-(64.0430)*Vo                         (Equation 2)

Where,

Ws/sea: Submerged Weight of Object in Seawater in pounds

Wa: Weight of object in pounds

Vo:  Volume of an object in cubic feet

Now comparing the two equations, 1 and 2 you can observe that an Object which is submerged in fresh water would lose less of its Air Weight to the Buoyant Force of the Fresh Water than the Saline water. Well, the weight of the object would be more when placed in the fresh water. For material with greater density, for example, Lead the difference is of diminutive, wherein one ton of Lead would only gain 4 pounds/ton in a Fresh Water. Talking about the concrete Lead it would gain approx. Of 50 pounds/ton in the Freshwater.

Experiment more with science and you will gain more!!!

Cheers!