Questions: An iron casting containing a number of cavities weighs 6.75 E+3 N in air and 4.18 E+3 N in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is 7.87 g / cm^3.

Solution Steps

The buoyant force is the difference in weight of the iron casting in air and in water. This can be calculated as:

\[ F_b = W_{\text{air}} - W_{\text{water}} = 6.75 \times 10^3 \, \text{N} - 4.18 \times 10^3 \, \text{N} = 2.57 \times 10^3 \, \text{N} \]

The buoyant force is equal to the weight of the water displaced by the casting. Using the relationship \( F_b = \rho_{\text{water}} \cdot V \cdot g \), where \(\rho_{\text{water}} = 1000 \, \text{kg/m}^3\) and \(g = 9.81 \, \text{m/s}^2\), we can solve for the volume \(V\):

\[ V = \frac{F_b}{\rho_{\text{water}} \cdot g} = \frac{2.57 \times 10^3 \, \text{N}}{1000 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2} \]

\[ V = 0.2621 \, \text{m}^3 \]

The total volume of the casting, including the cavities, is the volume of the displaced water. The volume of the cavities is the difference between this total volume and the volume of the solid iron part. The volume of the solid iron part can be calculated using its weight in air and the density of iron:

\[ V_{\text{iron}} = \frac{W_{\text{air}}}{\rho_{\text{iron}} \cdot g} = \frac{6.75 \times 10^3 \, \text{N}}{7.87 \times 10^3 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2} \]

\[ V_{\text{iron}} = 0.0873 \, \text{m}^3 \]

The volume of the cavities is:

\[ V_{\text{cavities}} = V - V_{\text{iron}} = 0.2621 \, \text{m}^3 - 0.0873 \, \text{m}^3 = 0.1748 \, \text{m}^3 \]

Final Answer