0
$\begingroup$

Assume that density of the body is less than the fluid.

If we put a body in a given fluid the buoyant force equalise the weight of the body. If we increase weight, the buoyant force increase?

If it does So it must be self-adjusting force, is it have any maximum value just like friction as?

$\endgroup$
3
  • $\begingroup$ @5 Dots buoyant force by a particular liquid will increase only if the volume of the body inside the liquid is increased. $\endgroup$
    – Ankit
    Sep 21, 2020 at 13:06
  • $\begingroup$ @Ankit, Imagine a floating boat. The fact that it's floating means that the buoyant force on it is equal to its weight. Then, imagine you drop a sack of sand into the boat. Now it weighs more. but is it still floating? If so, then then the buoyant force must have increased. It must have increased by exactly the same as the weight of the sack of sand. "Yes but,..." I hear you say, "that's because more of the boat now is 'inside the liquid.'" Uh Huh! And that equilibrium between its weight, and how much of it is "in the liquid" is precisely what the OP is trying to understand. $\endgroup$ Sep 21, 2020 at 21:30
  • $\begingroup$ @Solomon Slow oo now I get what OP is looking out for. $\endgroup$
    – Ankit
    Sep 22, 2020 at 2:38

3 Answers 3

8
$\begingroup$

Like you said, the bouyant force is equal to the weight of the displaced water. Making the object heavier will sink it further until either a new equilibrium is reached by displacing more water to counteract the increase in weight or the entire object is submerged, but the equilibrium is still not reached. The maximum value would be the weight of the water with the same volume as the object.

$\endgroup$
2
$\begingroup$

Increasing the weight of the object doesn't change the buoyant force. The buoyant force only depends on the submerged volume of the object and the density of the fluid it is in. Changing only the weight of the object will just determine if the object will float or sink given the constant buoyant force.

$\endgroup$
18
  • 1
    $\begingroup$ Why the down vote? $\endgroup$ Sep 21, 2020 at 12:47
  • 1
    $\begingroup$ If it's floating (which it is because the density is less than the fluid's density), increasing the weight makes it go down which increases the submerged volume... that's kinda important to mention... $\endgroup$
    – user253751
    Sep 21, 2020 at 15:05
  • $\begingroup$ @user253751 nothing in my answer contradicts that. Only changing the weight doesn't change the buoyant force. I didn't say increasing the weight and then allowing things to change accordingly doesn't change the resulting buoyant force. Furthermore, the OP's problem doesn't say the object is partially submerged (you can hold any object under a fluid completely), and the OP doesn't say after we increase the weight we allow for anything else to happen. I read the problem literally, not adding in my own assumptions. $\endgroup$ Sep 21, 2020 at 16:02
  • 1
    $\begingroup$ Well lots of things are possible if you stop time. If you only make the incline steeper, does the ball roll faster? No! Why not? Because you didn't give it time to accelerate! If you only pour water on the floor, does the floor get wet? No! Why not? Because the water hasn't fallen out of the container yet! This is extremely silly pedantry. $\endgroup$
    – user253751
    Sep 21, 2020 at 16:20
  • 1
    $\begingroup$ @BioPhysicist Agree. $\endgroup$
    – Bob D
    Sep 21, 2020 at 21:30
1
$\begingroup$

If we put a body in a given fluid the buoyant force equalise the weight of the body. If we increase weight, the buoyant force increase?

Correct. As long as the density of the object is less than or equal to the density of the liquid.

So it must be self adjusting force, is it have any maximum value just like friction as??

Not exactly sure what you mean by "self adjusting", but the buoyant force does have to increase in order for a heavier object to float. But there is a maximum buoyant force which occurs when the density of the object equals the density of the liquid. I don't see this as analogous to a maximum friction force, by which I assume you mean static friction force. The static friction force does match the applied force to prevent relative motion between the surfaces up to a maximum possible static friction force. Clearly, the buoyant force does not prevent relative motion between the surface of the object and the liquid because the heavier object will submerge further than the lighter object until equilibrium is established.

In any case, in order for an object to float, the upward buoyant force must equal the weight of the object. In other words, in order to float the following equation must be satisfied for all cases where the density of the object is less than or equal to the density of the liquid.

$$V_{o}ρ_{o}g=V_{l}ρ_{l}g$$

The left side of the equation is the weight of the object. $V_o$ is the total volume of the object, which is not necessarily the submerged volume, and $ρ_{o}$ is the density of the object.

The right side of the equation is the buoyant force, which is the weight of the volume of liquid $V_l$ that is displaced by the object times the density of the liquid, $ρ_{l}$.

The greater the weight of the object on the left side of the equation the greater the buoyant force has to be on the right in order for the object to float. The maximum buoyant force occurs when the density of the object equals the density of the liquid, $ρ_{o}=ρ_{l}$, meaning the volume of water displaced equals the volume of the object, $V_{l}=V_{o}$ and the object floats completely submerged. Any further increase in weight results in the object sinking and no further displacement of water, so the buoyant force remains constant. Adding weight simply causes the object to sink faster (assuming the density of the liquid is constant).

Hope this helps.

$\endgroup$
1
  • $\begingroup$ Got it, thanks mate $\endgroup$
    – 5 Dots
    Sep 22, 2020 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.