What does 'BUOYANT force' mean?
Have you ever dropped your swimming goggles in the deepest part of the pool and tried swimming down to get them? It can be difficult as the water tries to push you back to the surface as you swim down. The force above which is applied to objects submerged in liquids is the name of the victorious force.
So why do liquids have such a strong effect on underwater objects? It relates to differences in weight between the bottom of the underwater object and the top. Say someone dropped a can of beans in a basin of water.
Since weight (P_ {diameter} = \ rho GH) (Pgauge = ρgh) is a poppy of the left, P, beginning subscript, g, a, u, g, e, end subscript, co-equal, rho, g, h, right hooks increase as you go deeper into a liquid, the force from a weight placed on top of a can of beans will be less than the force from a weight that is applied up high at the bottom of the canal.
In fact, it is that simple. The reason for the existence of an influential force is that it is inevitable that the base (i.e. a more submerged part) of an object is always deeper in the liquid than the top of the object. This means that the upward force from the water must be greater than the force down from the water.
We know conceptually why there should be a strong force, but we should be able to figure out how to determine exactly what the size of the victorious force is as well.
We can start with the water on top of the canister pushing down F_ {down} FdownF, start subscript, d, o, w, n, end subscript, and the water at the bottom of the canal push up F_ {up} FupF, start a subscription, u, p, end subscription. We get the total force upon the water-pressure canister (called the victorious force F_ {buoyant} FbuoyantF, start subscription, b, u, o, y, a, n, t, end of membership) by just take the difference between the force sizes up to F_ {up} FupF, subscribe start, u, p, subscribe end and downforce F_ {down} FdownF, start to subscribe, d, o, w, n, end subscription.
F_ {buoyant} = F_ {up} - F_ {down} Fbuoyant = Fup - FdownF, start subscription, b, u, o, y, a, n, t, end subscription, equal, F, subscription start, u, p, subscribe end, minus, F, subscribe start, d, o, w, n, subscribe end
We can relate these forces to the weight using the weight definition P = \ frac {F} {A} P = AFP, equivalent, starting fraction, F, divided by, A, end fraction to release to gain force. F = PAF = PAF, equivalent, P, A. So the force that will be placed above the bottom of the canister is F_ {up} = P_ {bottom} AFup = PbottomAF, start a subscription, u, p, end subscribe, equal, P, start to subscribe, b, o, t, t, o, m, end subscribe, A and the force applied to the top of the canister F_ {down} = P_ {top} AFdown = PtopAF, start to subscribe, d, o, w, n, end subscribe, equal, P, start to subscribe, t, o, p, end subscribe, A. Insert these expressions for each individual FFF in the preceding equation we receive,
F_ {buoyant} = P_ {bottom} A - P_ {top} AFbuoyant = PbottomA - PtopAF, subscription start, b, u, o, y, a, n, t, subscription end, equal, P , start subscription, b, o, t, t, o, m, end subscription, A, minus, P, start subscription, t, o, p, end subscription, A.
We can use the formula for hydrostatic gas pressure P_ {diameter} = \ rho ghPgauge = ρghP, initial subscript, g, a, u, g, e, termination subscript, equivalent, rho, g, h to find phrases for the upside and weights directed downwards. The force from weight directed upwards at the bottom of the canister is P_ {bottom} = \ rho gh_ {bottom} P bottom = ρghbottomP, start subscript, b, o, t, t, o, m, end subscript, equal, rho, g, h, initial subscript, b, o, t, t, o, m, termination subscript and is the force from downward directed pressure on top of canister P_ {top} = \ rho gh_ {top} Ptop = ρghtopP, start a subscription, t, o, p, end subscription, equal, rho, g, h, start a subscription, t, o, p, end of a subscription. We can add these to the previous equation to get each individual weight,
F_ {buoyant} = (\ rho gh_ {bottom}) A - (\ rho gh_ {top}) AFbuoyant = (ρghbottom) A− (ρghtop) AF, start to subscribe, b, u, o, y, a, n, t, end subscript, equivalent, left poppy, rho, g, h, start subscript, b, o, t, t, o, m, end subscript, right poppy, A, minus, left poppy, rho, g, h, start a subscription, t, o, p, end subscription, correct parenting, A.
Note that the term \ rho g AρgArho, g, A. is in every term in this equation. So we can simplify this formula by extracting a common feature of \ rho g AρgArho, g, A available,
F_ {buoyant} = \ rho GA (h_ {bottom} -h_ {top}) Fbuoyant = ρgA (bottom-top) F, start a subscription, b, u, o, y, a, n, t, end of the membership, equivalent, rho, g, A, left poppy, h, start subscript, b, o, t, t, o, m, end subscription, minus, h, starter subscription, t, o, p, end subscript, correct suffix
Now, this term is h_ {bottom} -h_ {top} bottom - though, start a subscription, b, o, t, t, o, m, end subscription, minus, h, start a subscription, t, o, p, end of subscription is important and something interesting is going to happen because of it. The difference between the bottom depth of the can h_ {bottom} bottom, start subscript, b, o, t, t, o, m, end subscript and top depth of the can h_ {top} stop, a start is underwritten, t, o, p, end subscript exactly equal to the height of the canal.
So we can replace (h_ {bottom} -h_ {top}) (h bottom - h top) poppy left, h, start a subscription, b, o, t, t, o, m, end of the membership, minus, h, subscript start, t, o, p, end subscript, a correct popup in the previous formula with the height of the can h_ {can} change, start a subscription, c, a, n, termination subscription available,
F_ {buoyant} = \ rho gAh_ {can} Fbuoyant = ρgAhcanF, start a subscription, b, u, o, y, a, n, t, end of the subscription, equal, rho, g, A, h, the start of the subscription, c, a, n, end of the subscription
Here’s the interesting part. Since A \ times hA × hA, times, is equal to the size of a cylinder, we can start the term Ah_ {can} AhcanA, h, subscript end, c, a, n, end subscript with VVV volume. Perhaps the first instinct was to link this book to the size of a can. But note that this volume will also be equal to the amount of water discharged from the can. By soaking, we mean the amount of water that once existed in the amount of space that the language now has.
So we're definitely going to replace the term AhAhA, h with VVV volume, but should we write this book as the size of a can or the volume of the diffused stream? This is important because the two books may be different if the object is only partially submerged in the stream. The short answer is that we need to use the volume of the output current V_ {fluid} VfluidV, start a subscription, f, l, u, I, d, end subscript in the formula because the dispersed liquid is the factor that determines the force effect.
\ Large F_ {buoyant} = \ rho gV_ {f} Fbuoyant = ρgVfF, subscription start, b, u, o, y, a, n, t, subscription end, equal, rho, g, V , subscription start, f, end subscription.
This equation, when put into words, is called Archimedes' principle. Archimedes' principle is the statement that the force of a force is equal to the weight of the current in which the object is moving. The simplicity and power of this idea are amazing. If you want to know the effect of an object, you only need to determine the pressure of the stream that issued the object.
The fact that simple and beautiful ideas (but not obvious) result from the logical advancement of basic physics principles is part of why people see physics as useful, powerful, and interesting. And the fact that it was discovered by Archimedes of Syracuse 2000 years ago, before Newton's laws, is remarkable, to say the least.
What is confusing about the victorious force and the Archimedes principle?
Sometimes people forget that the density \ rhoρrho is in the formula for an effective force F_b = \ rho V_ {f} Fb = ρVfgF, start subscript, b, end subscript, equal, rho, V, initial subscription, f, membership termination, g and refers to the density of the diffuse current, not the density of the underwater object.
People often forget that the volume in the sugar formula refers to the volume of the diffuse stream (or volume of water under the water), and may not be the total volume of the sugar. which is.
Sometimes people think that the force of influence increases as an object is taken deeper and deeper into a liquid. But the victorious force does not depend on depth. It depends solely on the size of the V_fVfV mobile stream, start a subscription, f, end subscription, stream density \ rhoρrho, and the acceleration due to GGG gravity.
Many people, when asked to apply Archimedes' principle, usually look for upset harassment before launching into a fluid conversation about people jumping naked out of bathtubs. "Everything is turned upside down by a force equal to the weight of the stream in which the object is moving. . "
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