# Archimedes Principle – Example 1

Example 1:
Determine the upthrust acted on the objects immerse in the water below.
a.

b.

c.

Answer:
a. Upthrust = Weight of the displaced water = 15N

b. Upthrust = Weight of the displaced water = 32N

c. Upthrust = Weight of the displaced water = 20N

Example 2:
An iron block which has volume 0.3m³ is immersed in water. Find the upthrust exerted on the block by the water. [Density of water = 1000 kg/m³]

Answer:

Density of water, ρ = 1000 kg/m³
Volume of water, V = 0.3 m³
Gravitational Field Strength, g = 10 N/kg
Upthrust, F = ?

F = ρVg
F = (1000)(0.3)(10)
F = 3000N

Example 3:

The figure above shows an empty boat floating at rest on the water. Given that the mass of the boat is 150kg. Find

1. the upthrust acting on the boat.
2. The mass of the water displaced by the boat.
3. The maximum mass that the boat can load safely if the volume of the boat at the safety level is 3.0 m³.

Answer:
a. According to the principle of flotation, the upthrust is equal to the weight of the boat.

Upthrust,
F = Weight of the boat
= mg
= (150)(10)
= 1500N

b. According to the Archimedes’ Principle, the weight of the water displaced = Upthrust

Weight of the displaced water,
W = mg
(1500) = m(10)
m = 150kg

c.
Maximum weight can be sustained by the boat

F=ρVg
F=(1000)(3)(10)
=30,000N

Maximum weight of the load
= Maximum weight sustained – Weight of the boat
= 30,000 – 1,500 = 28,500N

Maximum mass of the load
= 28500/10 = 2850 kg

Example 4:

In figure above, a cylinder is immersed in water. If the height of the cylinder is 20cm, the density of the cylinder is 1200kg/m³ and the density of the liquid is 1000 kg/m³, find:
a. The weight of the object
b. The buoyant force

Answer:
a.
Volume of the cylinder, V  = 50 x 20 = 1000cm³ = 0.001m³
Density of the cylinder, ρ = 1200 kg/m³
Gravitational Field Strength, g = 10 N/kg
Weight of the cylinder, W = ?

W=ρVg W=(1200)(0.001)(10)=12N
b.
Volume of the displaced water = 50 x 12 = 600cm³ = 0.0006m³
Density of the water, ρ = 1000 kg/m³
Upthrust, F = ?
F=ρVg F=(1000)(0.0006)(10)=6N

Example 5
The density and mass of a metal block are 5.0×103 kg m-3 and 4.0kg respectively. Find the upthrust that act on the metal block when it is fully immerse in water.
[ Density of water = 1000 kgm-3 ]

Answer:
In order to find the upthrust, we need to find the volume of the water displaced. Since the block is fully immersed in water, hence the volume of the water displaced = volume of the block.

Volume of the block,
V= m ρ = 4 5.0× 10 3 =0.0008 m 3
Upthrust acted on the block,
F=ρVg F=(1000)(0.0008)(10)=8N

Example 6:

A metal block that has volume of 0.2 m³ is hanging in a water tank as shown in the figure to the left. What is the tension of the string? [ Density of the metal = 8 × 10³ kg/m³, density of water = 1 × 10³ kg/m³]

Answer:
Let,
Tension = T
Weight = W
Upthrust = F

Diagram below shows the 3 forces acted on the block.

The 3 forces are in equilibrium, hence
T+F=W T=W−F T= ρ 1 Vg− ρ 2 Vg T=( ρ 1 − ρ 2 )Vg =(8000−1000)(0.2)(10)=14,000N

Video

Example 7:
A wooden sphere of density 0.9 g/cm³ and mass 180 g, is anchored by a string to a lead weight at the bottom of a vessel containing water. If the wooden sphere is completely immersed in water, find the tension in the string.

Answer:
Let’s draw the diagram that illustrate the situation:

We need to determine the volume of the displaced water to find the upthrust.
Let the volume of the wooden sphere = V
V= m ρ = 180 0.9 =200c m 3
Let,
Tension = T
Weight = W
Upthrust = F

All the 3 forces are in equilibrium, hence
T+W=F T=F−W T= ρ 1 Vg− ρ 2 Vg T=( ρ 1 − ρ 2 )Vg =(1000−800)(0.0002)(10)=0.4N

Example 8:

Figure above shows a copper block rest on the bottom of a vessel filled with water. Given that the volume of the block is 1000cm³. Find the normal reaction acted on the block.
[Density of water = 1000 kg/m³; Density of copper = 3100 kg/m³]

Answer:
Volume of the block, V = 1000cm³ = 0.001m³

Let,
Normal Reaction = R
Weight = W
Upthrust = F

Diagram below shows the 3 forces acted on the block.
All the 3 forces are in equilibrium, hence
R+F=W R=W−F R= ρ 1 Vg− ρ 2 Vg R=( ρ 1 − ρ 2 )Vg =(3100−1000)(0.001)(10)=21N