Steel Ball Density: Impact of Water Filling on Mercury Surface

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When a steel ball is floating on mercury and the container is filled with water, the ball does not sink deeper into the mercury as expected. Instead, it remains partially submerged but less deeply than before. This phenomenon is attributed to the buoyant force exerted by the water, which counteracts some of the weight of the ball. The discussion highlights confusion regarding the mechanics of buoyancy and pressure in this scenario. Ultimately, the interaction between the water and the ball influences its level of submersion in the mercury.
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Homework Statement


Consider a steel ball floating on the surface of mercury in a half-filled container. What happens when the rest of the container is filled with water?


The Attempt at a Solution



Wouldn't the ball sink a little farther into the mercury with the added weight of the water on top of the ball? Or would it just stay where it's at? The mercury is obviously going to stay on bottom, and the steel ball is going to float in the water.
 
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You are correct. The steel ball should sink a little bit farther into the mercury due to the pressure from the water which will be above it.
 
Well apparently the answer was the ball remains partially submerged in mercury but not as deeply as before. I'm not sure I understand why that is, but in case anyone else is wondering, that's the way it is heh.
 
Hmm. Maybe it has to do with the buoyancy of the ball. The buoyant force of the water on the ball may cancel with some of the weight, thus causing less submergence in the mercury. Maybe something like this is the case? I don't see how this can be the case is the ball is on top of the mercury and no water is beneath the ball though...I won't say anything for sure, since I am not sure if this is the case.
 
Yeah, i don't get it either.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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