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The condition for flotation of a body a little bit above the surface of the liquid is that the upthrust force must be greater than its weight. Why do we in numericals take upthrust equal to the weight of the body?

In my college textbook, it says that a body which floats partially submerged has Fb>W(buoyant force>weight of the body)but everywhere else as in numericals I see that since the body is floating Fnet=0 and body is at equilibrium so Fb=W.Isn't it contradicting?

Please clear my doubt asap as I have many sleepless nights.enter image description here

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Floating actually means to rest or cause to rest on the surface of a fluid or in a fluid or space without sinking.

That means the object should be on the surface or to come up of the surface after putting some external force on the body and leaving it [edit added:] That is, the body when underwater experiences net force upwards, but once the net forces are zero, it floats with a certain portion above the water's surface [end edit] The case for which,

$F_b = Mg$

it means that the body is floating and is at the rest at the same time.

And for

$F_b > Mg$

Here, the net force in the upward direction

This means the object is willing to come at the $surface$ until both the forces becomes equal. Which can also be categorized as a case of floating.

But on the other hand if

$Mg > F_b$

That means Net force is in the direction the gravitational force.

Then the object will always tends to sink it will never come out by itself on the surface even though if we pull it out by the force applied by an external agent it will still tend to sink. that's just the case.

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  • $\begingroup$ So do you mean that as the body rises the upthrust reduces and eventually Fb=W?If so can you explain a bit further? $\endgroup$
    – AJknight
    Jan 14, 2022 at 3:14
  • $\begingroup$ One more thing at the surface since volume immersed is reduced, is that why the upthrust reduces and becomes equal to weight? $\endgroup$
    – AJknight
    Jan 14, 2022 at 3:45
  • $\begingroup$ Yes yes, just like that the upthrust force is equal to $Vρg$ which means as the immersed volume gets decreased with the time (area is constant and height of the immersed volume is decreasing) hence the force will also get decreased and it will keep decreasing until it gets reduced to the magnitude equal to the gravitational force acting on the whole body. $\endgroup$ Jan 14, 2022 at 4:05
  • $\begingroup$ For additional information I must tell you that if in case the object is let's say a $Cone$ immersed vertically inverted (the pointed end is inside the fluid) the case of decreasing volume will be different here as the area will also be decreased with the height as the object moves upwards. If the cube is immersed facing one whole side of it inside the fluid then the area will remain constant as discussed above in my answer (that's the simple case in which the cube in immersed) only height will be decreasing. $\endgroup$ Jan 14, 2022 at 4:13
  • $\begingroup$ ok so my textbook kinda gave an incomplete explanation right? $\endgroup$
    – AJknight
    Jan 14, 2022 at 6:31

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