**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

- the upthrust acting on the boat.
- The mass of the water displaced by the boat.
- 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 5.0×10

^{3}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