Sorry this might seem like a dumb question, but I'm having trouble understanding the concept behind buoyant force and scale readings.

Suppose I have a beaker filled with water, and the beaker is placed on a measuring scale.

a) If I then place a ball into the beaker, the ball floats, what will be scale reading be? I understand that Mg=Buoyant force, since the ball floats, they cancel out. Does that mean there's no change in scale reading?

b) If I place a ball and the ball submerges completely under the water, but does not touch the base of the beaker, what will the scale reading be? Now that Mg>Buoyant force, does the scale reading increase by the difference of the two? Mg-Buoyant force?

c) If I place a ball and the ball completely sinks under the water and touches the base of the beaker, what will be scale reading be? I understand that Mg>>>Fb, but why does Fb play no effect in the scale reading in this case?


1 Answer 1


Buoyant force has no effect on the interaction between the beaker and the scales it stands on.

The scales will register a weight which is the sum of the weights of the beaker, water and ball.

Consider that if the ball is at least partially immersed so that some part is below the level of water, it has displaced some water and so raised the level of the water in the beaker.

The water pressure at the bottom is a function of the height of the water column above it and so is a higher pressure. Since pressure acts normally to the interface between water and beaker base it presses down on the base of the beaker which transmits this force to the scales.

  • $\begingroup$ Thanks for the answer! So for all three cases I do not need to be considering the Buoyant force and I can just simply add Mg+Wbeaker+water to get the scale reading? I'm kind of confused because I also came across this physics.stackexchange.com/questions/73095/… $\endgroup$
    – J.Y.Y
    Oct 13, 2015 at 9:30
  • $\begingroup$ Wow thank you so much! So for case 1 and 2, I would get the scale reading by adding the Weight of the beaker+water in addition to the buoyant forces, but for case 3, since the ball is in direct contact to the measuring device and transmitting its weight directly, I would also have to add the weight of the ball? So for case 3 the scale reading would be Weight of the beaker+water + Fb + Weight of the ball? $\endgroup$
    – J.Y.Y
    Oct 13, 2015 at 9:40
  • $\begingroup$ @J.Y.Y In all three cases, to calculate the force on the scales, you don't need to worry about buoyancy of objects in equilibrium inside the beaker. It is an irrelevant distraction. The scales weigh the sum of the weights of the objects ultimately resting on it. It doesn't matter if the contents are floating, suspended on springs, magnetically levitating etc. $\endgroup$ Oct 13, 2015 at 10:22
  • $\begingroup$ @J.Y.Y Your teacher may be wanting you to analytically arrive at the conclusion I have stated. You can do so by calculating all the forces on all the objects involved (once they have arrived at a stable state) I expect you will find that when you calculate the force exerted on the scales by the beaker that any buoyant forces you have accounted for are cancelled out. $\endgroup$ Oct 13, 2015 at 10:30
  • $\begingroup$ Is it alright if i assume i can ignore the link above then? Thank you very much for your help! $\endgroup$
    – J.Y.Y
    Oct 13, 2015 at 10:37

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