How to work out the density?

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How to work out the density?

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Answer 1 · 2017-06-07T13:29:24

$ρ_{o b j e c t} = 517 \frac{kg}{m^{3}}$ $rho_(object)=517" ""kg"/"m"^3$

Explanation:

Oh Buoy! Buoyancy is a force exerted by a fluid on an object that is partially or fully immersed.
Archimedes' principle states that this buoyancy force is equal to the weight of the displaced fluid.

The weight of an object or displaced fluid is a force (F) and equals its mass times the gravitational field strength:

$F = m \cdot g$ $F=m*g$

The buoyancy force is the weight of the displaced fluid and is found using the above formula and replacing mass with density times volume:

$F_{b u o y} = m \cdot g = ρ_{o i l} \cdot V_{d} \cdot g$ $F_(buoy)=m*g=rho_(oil)*V_d*g$

$V_{d} = volume of displaced fluid$ $V_d="volume of displaced fluid"$
$ρ_{o i l} = density of fluid$ $rho_(oil)="density of fluid"$
$g = 9.81 \frac{m}{s^{2}}$ $g=9.81" ""m"/("s"^2)$

Because the object is floating and not moving, the magnitude of the downward force due to gravity (the weight of the object) must be equal to the buoyancy force. I.e. the net force is zero:

$F_{b u o y} = F_{o b j e c t}$ $F_(buoy)=F_(object)$

We know the expressions for the two forces and set them equal to each other. The $g$ $g$ 's cancel out so we can solve for the mass of the object.

$ρ_{o i l} \cdot V_{d} \cdot g = m_{o b j e c t} \cdot g$ $rho_(oil)*V_d*g=m_(object)*g$

$ρ_{o i l} \cdot V_{d} = m_{o b j e c t}$ $rho_(oil)*V_d=m_(object)$

The density of the object is mass divided by volume and we can now find its mass and volume. The volume is the sum of the immersed part and the part above the surface:

$V_{o b j e c t} = 2.31 \cdot 10^{- 4} + 3.94 \cdot 10^{- 4} = 6.25 \cdot 10^{- 4}$ $V_(object)=2.31*10^-4+3.94*10^-4=6.25*10^-4$ $m^{3}$ $" ""m"^(3)$

Finally, we substitute in the expressions for mass and volume to solve for the density of the object:

$ρ_{o b j e c t} = \frac{m_{o b j e c t}}{V_{o b j e c t}} = \frac{ρ_{o i l} \cdot V_{d}}{V_{o b j e c t}}$ $rho_(object)=m_(object)/V_(object)=(rho_(oil)*V_d)/V_(object)$

$ρ_{o b j e c t} = \frac{820.0 \cdot 3.94 \cdot 10^{- 4}}{6.25 \cdot 10^{- 4}} = 517 \frac{kg}{m^{3}}$ $rho_(object)=(820.0*3.94*10^-4)/(6.25*10^-4)=517" ""kg"/"m"^3$

Doing a quick logic check: this makes sense as the density of the plastic is less than that of the oil, so it floats.

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