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Can someone explain to me why when we suspended a ball with a density greater than that of water in water by the help of a string why is the weight measured by the spring balance=Weight of beaker + Bouyant force acting on the ball . Why shouldn't it be weight of beaker + Weight of ball - Bouyant force on the ball.

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  • $\begingroup$ What result would you expect if you had exactly the same set-up but without the water, so buoyancy force was effectively zero ? $\endgroup$ – gandalf61 Jan 18 at 6:34
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When the ball is suspended, the weight of the ball is balanced by the tension in the string and the buoyancy force. Here buoyancy force on the ball has an equal and opposite reaction on the water itself thus this extra force must be accounted for.

Why shouldn't it be weight of beaker + Weight of ball - Bouyant force on the ball

This would be the case if the ball was not suspended and is at equilibrium at the bottom of the container . The apparent weight of the ball will be "Weight of ball - Bouyant force on the ball" The apparent weight + the container weight is the result you have arrived to.

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  • $\begingroup$ This helps a lot thanks but one more thing I don't understand is that wouldn't there also be a normal force acting in the block due to water which wil have a equal and opposite reaction as well? $\endgroup$ – Naruto Uchiha Jan 18 at 9:38
  • $\begingroup$ The normal force is equal in magnitude to the apparent weight whose equivalent opposite force increases the reading on the scale. But note we also have to account for the buoyant force on the container thus the overall effect will be the weight of the container +weight of the ball. Should have clarified in the answer mistake on my side. Ping me if anything is still not clear. $\endgroup$ – JustJohan Jan 18 at 10:04
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No matter how much denser the ball is than the water, when held submerged but not sank to the bottom, the weight of the beaker on a scale will only increase by the weight of the volume of water the ball displaces, its buoyant force, per Archimedes principle of buoyancy. The tension on the string will support the extra amount of the ball's weight that is greater than that of the water displaced. You can see that the water level rises when the ball is submerged. It rises by the amount it would rise if the same volume of water as the ball's volume was added instead of the ball.

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Newton's Third Law: action begets reaction.

As the water exerts a buoyant force on the ball, the water exerts an opposite force on the beaker, thus scales. It's a classic part of density determination for irregularly shaped bodies.

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