Buoyancy and Archimedes Principle
Objects in a fluid seem to weigh less than they do because of a concept called buoyancy. Buoyancy occurs because the pressure of a fluid increases with depth. When an object is in a fluid, the force pushing up on the bottom of the object is larger than the force pushing down from the top. The buoyant force can be calculated by finding the difference between the force pushing on the bottom of the object (F2) and the force pushing on the top of the object (F1):
FB = F2 – F1
F1, or the force pushing down on the top of the object, is equal to the pressure of the fluid multiplied by the area of the top of the object, or:
F1 = PA
The pressure of the fluid on the object can be determined by multiplying the density of the fluid by gravity by the height of the fluid from the top of the object to the surface:
P = rf g h1
F1 =rf g h1A
The pressure of the fluid pushing up on the bottom of the object can be calculated in much the same way, except the height used when calculating the pressure is the distance from the bottom of the object to the surface (h2):
P = rf g h2
F2 = rf g h2 A
The buoyant force = FB = F2 – F1,
FB = rf g h2 A – rf g h1 A
We can simplify the equation:
FB = rfgh2A – rfgh1A = rfgA(h2-h1) (h2-h1= h, the height of the object)
FB = rfghA
And Ah = V, so:
FB = rfgV
To continue manipulating the equation, we can use the density equation ρ = m/V to determine that:
FB = rfgV = mfg, because rV = m
In plain words, this equation says that the buoyant force on an object in a fluid is equal to the weight of the fluid that object displaces (we will define this weight as the weight of the fluid that would occupy the volume of the object submerged in it). This concept is Archimedes’ Principle, and was discovered by the famous scientist when he determined the density of a crown by comparing its weight when submerged to its weight when in the air.
These concepts can be applied when dealing with floating objects as well. Generally, an object will float on a fluid if the fluid has a greater density than the object. An object floats at equilibrium when the buoyant force acting on it equals its weight. It is this principle that allows a submarine or fish to hover at the same depth in water, or a hot air balloon to float in the atmosphere.
A) What is the buoyant force on a block of gold with a volume of .025m3 submerged in a tank of water (density 1.0 x 103 kg/m3)?
Use the buoyant force equation FB = rfgV
Plug in the values FB = (1.0 x 103 kg/m3)(9.8 m/s2)(.025m3)
Solve for the buoyant force FB = 245 N
Often it is helpful to see what others have to say on the subject. Here are a few good sites to check out:
Buoyancy basics in simple terms
Good overall introduction to Archimedes' Principle, with exercises
Basic definition of the Principle with relevant links
All-around site on Archimedes, including biography and the story of the golden crown
The NTNU Virtual Physics Laboratory provides several excellent applets that demonstrate principles of Physics. Click Here for an an interactive introduction to buoyant forces, and here for an applet that shows how buoyant forces work with different densities, volumes, and masses.
For Practice Problems, Try: Giancoli Multiple Choice Practice Questions (It will be a few lessons before all of this is covered)