74.9k views
0 votes
A semicircular plate 2 feet in diameter sticks straight down into freshwater (weight - density 62.4lb/ft ^ 3) with the diameter along the surface. Find the force exerted by the water on one side of the plate.

User ZeroCho
by
8.0k points

1 Answer

3 votes

The force exerted by the water on one side of the plate, found using Archimedes principle is about 130.69 lb

How can the Archimedes principle be used to find the force exerted on the plate?

Archimedes principle states that the force exerted by the semicircular plate on the water is equivalent to the weight of the water displaced

The diameter of the semicircular plate is along the surface, therefore, the volume of the freshwater displaced is equivalent to the volume of the semicircular plate

Volume of the semicircular plate = (2/3) × π × r³

Where the radius, r = Diameter/2

The diameter of the semicircular plate = 2 feet

Therefore, the radius of the semicircular plate is; 2 feet/2 = 1 ft

Volume of the semicircular plate is; (2/3) × π × 1³ = (2/3) × π ft³

Weight of the freshwater displaced is; 62.4 lb/ft³ × (2/3) × π ft³ ≈ 130.69 lb

Force exerted by the water in one side of the plate = Weight of the freshwater displaced

Therefore; Force exerted by the water in one side of the plate ≈ 130.69 lb

User Jonathan W
by
7.7k points