Question

In: Physics

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 kg woman to be able to stand on it without getting her feet wet?

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 kg woman to be able to stand on it without getting her feet wet?

 

Solutions

Expert Solution

Concepts and reason

The concept required to solve this question is the buoyant force. Initially, write the expression for the buoyant force of the water and the force due to her and the ice weight. Then, rearrange the expression for the volume. Finally, substitute the given values in the expression of the volume.

Fundamentals

A body experiences an upward force when it is immersed in a liquid. This force will be equal to the weight of the fluid displaced by the body when it is partially or fully immersed in the liquid. This upward force exerted by the fluid on the body is known as the Buoyancy force.

 

The expression of the buoyant force FBF_{\mathrm{B}} for the water is, FB=ρwVgF_{\mathrm{B}}=\rho_{\mathrm{w}} V g

Here, ρw\rho_{\mathrm{w}} is the density of water, V\mathrm{V} is the volume, and g\mathrm{g} is the acceleration due to gravity. The total weight FTF_{\mathrm{T}} when the woman standing on the ice slab is given by, FT=ρiceVg+mgF_{\mathrm{T}}=\rho_{\mathrm{ice}} V g+m g

Here, ρice \rho_{\text {ice }} is the density of the ice, V\mathrm{V} is the volume, g\mathrm{g} is the acceleration due to gravity, and m\mathrm{m} is the mass of the

woman. The buoyant force is equal to the total weight and it is given as:

ρwVg=ρiceVg+mg\rho_{\mathrm{w}} V g=\rho_{\mathrm{ice}} V g+m g

The buoyant force depends on the density of the fluid, the volume, and the acceleration due to gravity. For the floating of ice slab on the lake, the buoyant force should be equal to the weight of the object.

 

The buoyant force is equal to the total weight and it is given as:

ρwVg=ρiceVg+mg\rho_{\mathrm{w}} V g=\rho_{\mathrm{ice}} V g+m g

Rearrange the expression for the V.

ρwVgρiceVg=mg(ρwρice)gV=mg(ρwρice)V=mV=m(ρwρice) \begin{array}{c} \rho_{\mathrm{w}} V g-\rho_{\mathrm{ice}} V g=m g \\ \left(\rho_{\mathrm{w}}-\rho_{\mathrm{ice}}\right) g V=m g \\ \left(\rho_{\mathrm{w}}-\rho_{\mathrm{ice}}\right) V=m \\ V=\frac{m}{\left(\rho_{\mathrm{w}}-\rho_{\mathrm{ice}}\right)} \end{array}

Substitute 45.0 kg45.0 \mathrm{~kg} for m,1000 kg/m3\mathrm{m}, 1000 \mathrm{~kg} / \mathrm{m}^{3} for ρw,\rho_{\mathrm{w}}, and 920 kg/m3920 \mathrm{~kg} / \mathrm{m}^{3} for ρice\rho_{\mathrm{ice}} in the above equation.

V=45.0 kg(1000 kg/m3920 kg/m3)V=\frac{45.0 \mathrm{~kg}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}-920 \mathrm{~kg} / \mathrm{m}^{3}\right)}

=0.5625 m3=0.5625 \mathrm{~m}^{3}

The minimum volume for the slab is 0.5625 m30.5625 \mathrm{~m}^{3}.

The Archimedes principle states that the buoyant force on a body is equal to the weight of the fluid displaced by the body.


The minimum volume for the slab is  .

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