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 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 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
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 air – buoyant 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 New ton
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):
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/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 :
The correct answer is C.