Question #85447

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Question #85447

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Answer 1 · 2017-06-23T22:09:48

You must remember that the buoyant force or upthrust is equal to the weight of the fluid displaced and NOT equal to the weight of the object.

Explanation:

What you are asking about is Buoyant force and Floating bodies.

We know that in any given fluid, pressure increases with depth. When an object is immersed in a fluid, the upward force on the bottom of an object is greater than the downward force on the top of the object. this gives rise to Buoyant Force.
A floating body will displace its own weight of fluid such that there is no vertical resultant force on the body. The volume of the floating object that is below the surface will depend on both the density of the object and density of the fluid in which it is floating.

We can calculate the buoyant force on an object by adding up the forces exerted on all of sides of object. Lets us consider a cylindrical object which has top surface area $= A$ $=A$ and is at depth $d_{1}$ $d_1$ . The pressure at this depth is

$P_{1} = d_{1} ρ g$ $P_1=d_1rhog$
where $ρ$ $rho$ is the density of the fluid and $g = 9.81 m s^{- 1}$ $g=9.81ms^-1$ is the acceleration due to gravity.

The downward force on the top surface is

$F_{1} = Pressure \times Area = d_{1} ρ g A$ $F_1="Pressure"xx"Area"=d_1rhogA$

Similarly, the upwards force acting on the bottom surface of object at depth $d_{2}$ $d_2$ is

$F_{2} = d_{2} ρ g A$ $F_2=d_2rhogA$

It is a cylindrical object, the net force on its sides is zero, as the forces on different parts of sides are equal and opposite to each other. Thus, the net upward force on the object is

$F_{net} = F_{2} - F_{1} = ρ g A (d_{2} - d_{1})$ $F_"net"=F_2-F_1=rhogA(d_2-d_1)$
This is Force of buoyancy $F_{B}$ $F_B$

For an object of mass $m$ $m$ to float (or to be in equilibrium)

$F_{B} = m g$ $F_B=mg$

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Question #85447

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