Physics Asked on April 9, 2021
We know that $text{density}= frac{text{mass}}{text{volume}}$, and density of water is $1000, text{kg/m}^3$.
If I want to float a $100 ,text{kg}$ metal block then using the above formula of density I get $text{volume}= 100/ 1000 = 0.1,text{m}^3$
Now if the dimension of the block is $50 ,text{cm }× 50 ,text{cm} × 50 ,text{cm}$, then the volume of the block comes out to be $0.125,text{m}^3$.
So does this means that block with the given dimensions will float on water as the volume of the block comes out to be greater than that required for a $100, text{kg}$ object to float.
Question to your answer depends that whether the block is hollow or solid.
If it's solid then obviously any metal block i.e block of any dimensions will sink. I am neglecting surface tension and only considering buoyant force and gravity.
But if the block is hollow then the density of material is different from the density of the block . We tend to take combine density of air plus the metal.
Answered by Danny LeBeau on April 9, 2021
So does this means that block with the given dimensions will float on water as the volume OF the block comes out to be greater than that required for a 100 kg object to float.
If the block is solid and homogeneous, and it has a density equal to or less than the water, it doesn't matter what its dimension are. It will float.
It will float partially above water if its density is less than water, float completely submerged if its density exactly equals the water, and sink if its density is greater than water. All according to the equation:
$$V_{o}ρ_{o}g=V_{l}ρ_{l}g$$
The left side of the equation is the total weight of the object. $V_o$ is the total volume of the object, not necessarily the submerged volume of the object.
The right side of the equation is the upward buoyant force exerted by the liquid on the object. $V_l$ is the volume of fluid displaced by the submerged volume of the object.
From the equation, if the density of the object equals the density of the liquid, it is completely submersed but floats, meaning $V_o$ = $V_l$. That's because the upward buoyant force exactly equals the weight of the object. If it is less than the fluid density, the object will float partially submerged, meaning the volume of the object is greater than the volume of the displaced liquid. If it is greater than to fluid density, it will sink because its weight will be greater than the buoyant force.
Hope this helps.
Answered by Bob D on April 9, 2021
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