Solving Numerical problems on Archimedes’ Principle and Buoyancy

In this post, we will quickly revise Archimedes’ Principle and its legendary background. Then we will solve a few numerical problems in physics using this principle.


Archimedes’ Principle and Buoyancy

If an object is submerged in a liquid, the object displaces a volume of the liquid equal to the volume of the submerged object. There is a Legend that Archimedes made this observation when he sat in a bathtub that was filled to the top of the tub. The volume of water that overflowed was equal to his own volume.

The forces exerted by the fluid on the sides of the submerged object are balanced. However, the forces exerted by the fluid on the top and bottom of the object are not equal.

The force exerted by the liquid below the object is greater than the force exerted by the liquid above it; as a result, the liquid exerts a net upward force on the submerged or floating object. This force is called buoyancy, and its magnitude is equal to the weight of the displaced water.

 Archimedes’ Principle states that the buoyant force is equal to the weight of the displaced liquid.

Numerical Problems on Archimedes’ Principle

Numerical Problem 1 

The density of steel is 9000. kg/m3 and the density of water is 1000. kg/m3. If a cube of steel that is 0.100 m on each side is placed in a tank of water and weighed while under water, what is the apparent weight of the cube?

Solution:

Here, the cube of steel has 0.100 m on each side

The volume of the cube is 0.00100 m3.

The density of steel is 9000. kg/m3

The mass of the cube is = volume x density = 9.00 kg.

The weight of the cube when not submerged in water = (9.00 kg)(9.80 m/s2) = 88.2 N

The mass of water displaced by the cube =volume of displaced water x density of water = 0.00100 m3 x 1000 kg/m3 = 1.00 kg

The weight of the water displaced by the cube = mass of water x g= 9.80 N

The buoyant force on the steel cube = 9.80 N

Apparent weight of cube under water = 88.2 N – 9.80 N = 78.4 N

Numerical Problem

A hollow metal cube 1.00 m on each side has a mass of  600. kg.  How deep will this cube sink when placed in a vat of water?

Solution:

Since the weight of the cube is 5880 N, it will need to displace 5880 N of water in order to float.

Say the cube deeps by d meter.

Volume of submerged portion of cube =(1.00 m)(1.00 m)(d m)=d m3

Mass of water displaced = Volume of submerged portion of cube x density of water = d x1000 kg = 1000d kg

Weight of water displaced =1000d x 9.8 = 9800d N

So, 9800 d = 5880

d=0.600 m

The cube will sink such that 0.6 m are underwater.

Exercise

  1. A cylinder with a radius of 11 cm and a height of 3.4 cm has a mass of 10.0 kg.
    1. What is the weight of this cylinder?
    2. What is the weight of this cylinder when it is submerged in water?
  2. A wooden raft is 2.00 m wide, 3.00 m long, and 0.200 m deep.  The raft and its occupants have a mass of 700. kg.  How deep will the raft sink below the water when floating?
  3. For the raft in problem #2, how many 50. kg people can be added to the raft before it sinks completely under water?
  4. The density of gold is 19,320 kg/m3 and the density of mercury is 13,500 kg/m3.  If a cube of gold that is 0.100 m on each side is placed in a tank of mercury and weighed while under the surface, what is the apparent weight of the cube?

Summary

  • If an object is submerged in a liquid, the object will displace a volume of the liquid equal to the volume of the submerged object.
  • The forces exerted by the fluid on the sides of the submerged object are balanced, but the forces exerted by the fluid on the top and bottom of the object are not equal.
  • The liquid exerts a net upward force on the submerged or floating object, called buoancy.
  • The magnitude of buoancy is equal to the weight of the displaced water.
  • Archimedes’ Principle states that the buoyant force is equal to the weight of the displaced liquid.
Solving Numerical problems on Archimedes’ Principle and Buoyancy
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