I am stuck with a question. Please help:

A metal block is weighed in the air, the spring balance shows 20 N. Then a beaker with water is taken weighing 40 N. The metal block is immersed in the liquid, the apparent weight being 16 N. Thus the buoyancy recorded is 4N. So, what is the weight of beaker with water and block now?

NOTE added for clarification: While floating, the metal block remains attached to the spring balance in such a way that the spring continues to carry a load of 16 N.

In some books i have seen that they say to add the buoyant force to the original weight of beaker. But, I argue if Two blocks of different masses but same volume sink in water in separate beakers of equal weight. Then, would the weight of the beakers be the same?


The buoyant force is the amount of force that the liquid is exerting on the object. It takes 4 N less to keep the object from falling down when it is in the liquid, so the buoyant force is 4 N.

Due to the nature of "equal and opposite reactions", this means the mass is also exerting 4 N of force on the fluid. That 4 N force on the liquid will make the beaker weight 4 N more (only 4 N of the weight is actually pushing down on the liquid).

Therefore the beaker will have the weight original weight plus the 4 N buoyant force; for 44 N total.

It's different if the objects aren't being held by a spring or rope. If they can sink then the beaker would weigh 20 N more, since the entire mass will be pushing down on the beaker. In this case we measured the force on the beaker to be only 4 N of those 20 N.

| cite | improve this answer | |
  • $\begingroup$ This is incorrect. The weight of the beaker plus block is obviously 60N. The entire mass of the block is pushing down on the beaker. Buoyant fluid is not an anti-gravity device... $\endgroup$ – Pirx Feb 8 '17 at 17:49
  • $\begingroup$ @Pirx The block is attached to a spring balance that reads 20 N before being submerged and 17 N after. If the spring is still supporting 17 N of the weight; why would the beaker also have the full weight of the block? That would be true if the block were in the beaker and not supported partially by a spring balance; which is not the case here. $\endgroup$ – JMac Feb 8 '17 at 18:10
  • $\begingroup$ JMac, I assumed he removed the weight from the balance before placing it in the beaker. Question is unclear. $\endgroup$ – Jake Watrous Feb 8 '17 at 18:38
  • $\begingroup$ @JakeWatrous It's not unclear. It says it is attached to a spring scale and you are taking measurements from it the whole time; this means it will have to stay attached if you're using the spring scale and getting a reading that isn't 0. If it was sinking or floating on it's own the spring scale would be resisting nothing; and the weight measured for the beaker would be the block and beaker weight. The spring scale is resisting some of the weight though; only 20% of the weight is being held up by buoyant forces; so it only adds that much weight. $\endgroup$ – JMac Feb 8 '17 at 18:43
  • 1
    $\begingroup$ @Pirx I wouldn't say extremely poorly. It doesn't explicitly state that you're measuring the apparent weight with the same spring scale, that is a cause of confusion. It is quite obviously implied though, as you are still getting measurements about it's weight somehow; and they only use the term immersed instead of sunken or dropped. $\endgroup$ – JMac Feb 8 '17 at 19:21

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.