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 Newton

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 (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:

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.

**What is the fundamental principle behind the buoyant force?****Answer**: The fundamental principle behind buoyant force is Archimedes’ Principle. It states that an object submerged in a fluid experiences an upward force called the buoyant force that is equal in magnitude to the weight of the fluid displaced by the object.

**Why does an object feel lighter when it’s submerged in water?****Answer**: When an object is submerged in water, it displaces a certain volume of water. The weight of this displaced water exerts an upward buoyant force on the object, which counteracts some of the object’s weight, making it feel lighter.

**If an object floats on water, how does the weight of the object compare to the buoyant force acting on it?****Answer**: If an object floats, the buoyant force acting on it is equal to the weight of the object. That’s why the object remains at equilibrium without sinking or rising.

**Does buoyant force act on objects in the air?****Answer**: Yes, buoyant force acts on objects in any fluid, including air. However, because air is much less dense than liquids like water, the buoyant force in air is much smaller and often negligible for everyday objects.

**Why do some objects sink in water while others float?****Answer**: Whether an object sinks or floats depends on the relationship between its weight and the buoyant force. If the buoyant force (due to the displaced fluid) is greater than the object’s weight, it floats. If the weight is greater, it sinks.

**If you were to take a hollow metal ball and a solid metal ball of the same volume and material, and submerge them in water, which one would experience a greater buoyant force?****Answer**: The buoyant force depends on the volume of fluid displaced and not on the mass of the object. Since both balls displace the same volume of water, they would experience the same buoyant force.

**Why do ships made of steel, which is much denser than water, float?****Answer**: Ships are designed with large hollow spaces inside, which means their overall average density is less than that of water. The water displaced by the ship exerts a buoyant force that can support the ship’s weight, allowing it to float.

**If you push a floating ball deeper into water and then release it, what will happen?****Answer**: If you push a floating ball deeper into the water, you’re increasing the volume of water it displaces, which increases the buoyant force. When you release it, the increased buoyant force will push it upwards until it returns to its equilibrium position.

**What happens to the buoyant force on a submerged object if you move it from freshwater to saltwater?****Answer**: Saltwater is denser than freshwater. Therefore, for the same volume displaced, the buoyant force in saltwater will be greater than in freshwater. So, an object might float higher in saltwater than in freshwater.

**Is it possible for an object to be stable in one orientation (e.g., vertical) but unstable in another (e.g., horizontal) while floating in water?**

**Answer**: Yes, the stability of a floating object depends on the distribution of its weight relative to the buoyant force. If the center of gravity is directly above the center of buoyancy, the object will be stable. However, if you change the orientation, the relationship between the centers might change, leading to instability in that particular orientation.