what is buoyant force?
Answers
Answer:
Explanation: What does buoyant force mean?
Have you ever dropped your swimming goggles in the deepest part of the pool and tried to swim down to get them? It can be frustrating because the water tries to push you back up to the surface as you're swimming downward. The name of this upward force exerted on objects submerged in fluids is the buoyant force.
So why do fluids exert an upward buoyant force on submerged objects? It has to do with differences in pressure between the bottom of the submerged object and the top. Say someone dropped a can of beans in a pool of water. [Not again!]
Because pressure (P_{gauge}=\rho gh)(P
gauge
=ρgh)left parenthesis, P, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h, right parenthesis increases as you go deeper in a fluid, the force from pressure exerted downward on the top of the can of beans will be less than the force from pressure exerted upward on the bottom of the can.
Essentially it's that simple. The reason there's a buoyant force is because of the rather unavoidable fact that the bottom (i.e. more submerged part) of an object is always deeper in a fluid than the top of the object. This means the upward force from water has to be greater than the downward force from water.
[Hold on..what if?]
Knowing conceptually why there should be a buoyant force is good, but we should also be able to figure out how to determine the exact size of the buoyant force as well.
We can start with the fact that the water on the top of the can is pushing down F_{down}F
down
F, start subscript, d, o, w, n, end subscript, and the water on the bottom of the can is pushing up F_{up}F
up
F, start subscript, u, p, end subscript. We can find the total upward force on the can exerted by water pressure (which we call the buoyant force F_{buoyant}F
buoyant
F, start subscript, b, u, o, y, a, n, t, end subscript) by simply taking the difference between the magnitudes of the upward force F_{up}F
up
F, start subscript, u, p, end subscript and downward force F_{down}F
down
F, start subscript, d, o, w, n, end subscript.
F_{buoyant} =F_{up} - F_{down}F
buoyant
=F
up
−F
down
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, F, start subscript, u, p, end subscript, minus, F, start subscript, d, o, w, n, end subscript
We can relate these forces to the pressure by using the definition of pressure P=\dfrac{F}{A}P=
A
F
P, equals, start fraction, F, divided by, A, end fraction which can be solved for force to get F=PAF=PAF, equals, P, A . So the force exerted upward on the bottom of the can will be F_{up}=P_{bottom}AF
up
=P
bottom
AF, start subscript, u, p, end subscript, equals, P, start subscript, b, o, t, t, o, m, end subscript, A and the force exerted downward on the top of the can will be F_{down}=P_{top}AF
down
=P
top
AF, start subscript, d, o, w, n, end subscript, equals, P, start subscript, t, o, p, end subscript, A. Substituting these expressions in for each FFF respectively in the previous equation we get,
F_{buoyant} =P_{bottom} A - P_{top}AF
buoyant
=P
bottom
A−P
top
AF, start subscript, b, u, o, y, a, n, t, end subscript, equals, P, start subscript, b, o, t, t, o, m, end subscript, A, minus, P, start subscript, t, o, p, end subscript, A
We can use the formula for hydrostatic gauge pressure P_{gauge}=\rho ghP
gauge
=ρghP, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h to find expressions for the upward and downward directed pressures. The force from pressure directed upward on the bottom of the can is P_{bottom}=\rho gh_{bottom}P
bottom
=ρgh
bottom
P, start subscript, b, o, t, t, o, m, end subscript, equals, rho, g, h, start subscript, b, o, t, t, o, m, end subscript and the force from pressure directed downward on the top of the can is P_{top}=\rho gh_{top}P
top
=ρgh
top
P, start subscript, t, o, p, end subscript, equals, rho, g, h, start subscript, t, o, p, end subscript . We can substitute these into the previous equation for each pressure respectively to get,
F_{buoyant} =(\rho gh_{bottom}) A - (\rho gh_{top})AF
buoyant
=(ρgh
bottom
)A−(ρgh
top
)AF, start subscript, b, u, o, y, a, n, t, end subscript, equals, left parenthesis, rho, g, h, start subscript, b, o, t, t, o, m, end subscript, right parenthesis, A, minus, left parenthesis, rho, g, h, start subscript, t, o, p, end subscript, right parenthesis, A
Notice that each term in this equation contains the expression \rho g AρgArho, g, A. So we can simplify this formula by pulling out a common factor of \rho g AρgArho, g, A to get,
F_{buoyant} =\rho gA(h_{bottom} -h_{top})F
buoyant
=ρgA(h
bottom
−h
top
)
Answer:
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