Buoyant force – problems and solutions

1. A block of wood with length = 2.5 m, width = 0.5 m and height = 0.4 m. The density of water is 1000 kg/m3. If the block is placed in the water, what is the buoyant forceAcceleration due to gravity is 10 N/kg.

Known :

Volume of the block (V) = length x width x height = 2.5 x 0.5 x 0.4 = 0.5 m3

Density of water (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 10 N/kg

Wanted : The magnitude of the buoyant force

Solution :

Formula of buoyant force :

F = ρ g V

F = buoyant force, ρ = density of water, g = acceleration due to gravity, V = volume

F = (1000)(10)(0.5) = (1000)(5) = 5000 Newton

Read :  Momentum – problems and solutions

2. Weight of an object in air is 100 N. The object is placed in a liquid. Increase in volume of liquid is 1.5 m3. If specific weight of the liquid is 10 N/m3, what is the weight of the object in liquid.

Known :

Object’s weight in air (w) = 100 Newton

Increase in volume of liquid = volume of the object in liquid (V) = 1.5 m3

Specific weight of the liquid = 10 N/m3

Wanted : Object’s weight in liquid

Solution :

Object’s weight in liquid = object’s weight in airbuoyant force

Object’s weight in liquid = 100 Newton – buoyant force

Formula of buoyant force :

FA = ρ g V

FA = buoyant force = the force exerted by the liquids on the object in water

ρ = density of liquid

g = acceleration due to gravity

V = object’s volume in liquid

Specific weight :

Specific weight of liquid = 10 N/m3

w / V = 10 N/m3

m g / V = 10 N/m3

m (10) / V = 10 N/m3

m / V = 1 kg/m3

ρ = 1 kg/m3

The density of liquid is 1 kg/m3

The magnitude of buoyant force :

FA = ρ g V = (1 kg/m3)(10 m/s2)(1.5 m3) = 15 kg m/s2 = 15 Newton

Object’s weight in fluid :

Object’s weight in fluid = 100 Newton – 15 Newton

Object’s weight in fluid = 85 Newton

3. A ship sailing in the sea enters a wide and deep river. The density of seawater is 1100 kg/m3, the density of river water is 1000 kg/m3. Determine comparison of the volume of the object is in seawater and in river water.

A. 11 : 10

B. 10 : 11

C. 121 : 100

D. 1 : 1

Known :

Density of seawater (ρ1) = 1100 kg/m3

Density of river water (ρ2) = 1000 kg/m3

Wanted: Comparison of the volume of the object is in seawater and in river water. comparison of the volume of the object is in seawater and in river water.

Solution :

If the object is floating then buoyant force (FB) = weight (w):

Buoyant force – problems and solutions 1

Archimedes‘ principle states that the buoyant force acting on an object in fluid (water) is equal to the weight of the fluid (water) it displaces. The volume of the object in fluid (water) is equal to the volume of fluid (water) moved.


Comparison of the volume of the object in seawater and in river water:

Buoyant force – problems and solutions 2

The correct answer is B.

4. Gold, whose mass is 193 grams is in kerosene having an upward force of 8000 dynes. If the acceleration due to gravity is 10 m/s2 and the density of kerosene is 0.8 gr/cm3, then determine the density of gold.

A. 1.93 gr/cm3

B. 8.65 gr/cm3

C. 19.3 gr/cm3

D. 193 gr/cm3

Known :

Mass of gold (mgold) = 193 gram = 0.193 kg

Acceleration due to gravity (g) = 10 m/s2

Buoyant force (FA) = 8000 dyne = 8 x 103 dyne = (8 x 103)(10-5 N) = 8 x 10-2 N = 0.08 Newton

Density of kerosene (ρ) = 0.8 gr/cm3 = 800 kg/m3

Wanted : Density of gold

Solution :

Weight of gold in air :

w = m g = ρb V g —– Equation 1

V is volume of gold in kerosene.

Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) :

w – w’ = FA

w – w’ = ρf V g —– Equation 2

V is the volume of gold in kerosene.

Both equations above can be written again below :

Buoyant force – problems and solutions 3

Buoyant force – problems and solutions 4

The correct answer is C.