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A beaker of water resting on a scale registers a weight of 0.61 N. Lowering a steel cube completely into the water (without making any contact with the beaker) causes the reading on the scale to increase to 0.69 N. The steel cube has density $7800 ~kg/m^3$ is replaced with a wooden block with exactly the same dimensions, but a density of only $850 ~kg/m3.$ I need help understanding why the scale still shows 0.69 N.

I thought the weight would be between 0.61 N and 0.69 N because you replace the steel cube with something of the same volume but less dense, and I understand this will have a mass less than that of the cube since $$m=\rho\times v$$. A decrease in mass should cause the scale reading to be lowered. Why is it not so here?

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Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object according to Archimedes. Considering that there is something to cancel out the weight of the body(a rope for e.g.)and since the object is fully immersed and is rest implies you are pushing it with a force equal to the buoyant force. That force is being read by the scale. Now since the two objects displaces same volume of water, according to the principle stated they experience the same upward force, hence you should push each of them with same force. Hence the same reading.

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    $\begingroup$ There shouldn't be any need for you to "push" the objects at all - gravity provides more than enough force to "pull" them into the liquid. $\endgroup$
    – Brionius
    Dec 30, 2015 at 17:41
  • $\begingroup$ gravity provides more than enough force to "pull" them into the liquid Not true if its an old wood with enough air pockets. Well, the observation is true, edited the answer for more clarity. $\endgroup$
    – Sathyam
    Dec 30, 2015 at 17:53
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    $\begingroup$ You're right in the case of the wood, I was thinking of the steel cube. $\endgroup$
    – Brionius
    Dec 30, 2015 at 18:14
  • $\begingroup$ When the iron cube sinks to the bottom wouldn't the scale read the weight of water plus the weight of iron cube? The iron cube accelerates down when it is immersed while the wooden piece does not. So when the iron cube rests on the base it shoukd be the sum of the weights right? $\endgroup$
    – N.S.JOHN
    Jan 2, 2016 at 4:02
  • $\begingroup$ @N.S.JOHN The question concerns lowering the steel cube, not dropping it. In the latter case your question is relevant. $\endgroup$
    – Sathyam
    Jan 2, 2016 at 11:28
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The reason this result seems mysterious is because as stated, you've glossed over the "without making any contact with the beaker" part. To suspend a denser-than-water cube in water without sinking to the bottom, you'll need a rope holding up the cube. The tension in that rope will provide whatever force is necessary to maintain static equilibrium. The tension plus the force of buoyancy therefore balances the force of gravity. According to Newton's 3rd law, the buoyant force from the water on the cube (which is the only force from the water on the cube) is equal to the force of the cube on the water.

Thus, the only force the cube puts on the water, and therefore the beaker, and therefore the scale, is equal in magnitude to the force of buoyancy on the cube, which is dependent only on volume, not on density, and is therefore equal for the steel and wooden cube.

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