# 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/m^{3}. If the block is placed in the water, what is the buoyant force… Acceleration 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 m^{3}

Density of water (ρ) = 1000 kg/m^{3}

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

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 m^{3}. If specific weight of the liquid is 10 N/m^{3}, 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 m^{3}

Specific weight of the liquid = 10 N/m^{3}

__Wanted :__ Object’s weight in liquid

__Solution :__

Object’s weight in liquid = object’s weight in air – buoyant force

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

**Formula of buoyant force :**

F_{A }= ρ g V

F_{A }= 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/m^{3}

w / V = 10 N/m^{3}

m g / V = 10 N/m^{3}

m (10) / V = 10 N/m^{3}

m / V = 1 kg/m^{3}

ρ = 1 kg/m^{3}

The density of liquid is 1 kg/m^{3}

**The magnitude of buoyant force **:

F_{A }= ρ g V = (1 kg/m^{3})(10 m/s^{2})(1.5 m^{3}) = 15 kg m/s^{2 }= 15 Newton

**Object’s weight in fluid :**

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

Object’s weight in fluid = 85 New ton

3. A ship sailing in the sea enters a wide and deep river. The density of seawater is 1100 kg/m^{3}, the density of river water is 1000 kg/m^{3}. 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 (ρ_{1}) = 1100 kg/m^{3}

Density of river water (ρ_{2}) = 1000 kg/m^{3}

__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 (F_{B}) = weight (w):

Archimedes‘ principle states that the buoyant force acting on an object in fluid

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

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/s^{2} and the density of kerosene is 0.8 gr/cm^{3}, then determine the density of gold.

A. 1.93 gr/cm^{3}

B. 8.65 gr/cm^{3}

C. 19.3 gr/cm^{3}

D. 193 gr/cm^{3}

__Known :__

Mass of gold (m_{gold}) = 193 gram = 0.193 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Buoyant force (F_{A}) = 8000 dyne = 8 x 10^{3} dyne = (8 x 10^{3})(10^{-5 }N) = 8 x 10^{-2 }N = 0.08 Newton

Density of kerosene (ρ) = 0.8 gr/cm^{3 }= 800 kg/m^{3 }

__Wanted :__ Density of gold

__Solution :__

Weight of gold in air :

w = m g = ρ_{b} V g —– Equation 1

*V is volume of gold in kerosene.*

Buoyant force (F_{B}) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) :

w – w’ = F_{A}

w – w’ = ρ_{f} V g —– Equation 2

*V is the volume of gold in kerosene.*

Both equations above can be written again below :

The correct answer is C.