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Chapter 9: Problem 92

A block of wood, with density \(780 \mathrm{kg} / \mathrm{m}^{3},\) has a cubic shape with sides \(0.330 \mathrm{m}\) long. A rope of negligible mass is used to tie a piece of lead to the bottom of the wood. The lead pulls the wood into the water until it is just completely covered with water. What is the mass of the lead? [Hint: Don't forget to consider the buoyant force on both the wood and the lead.]

Short Answer

Expert verified
Answer: The mass of the lead is approximately 8.238 kg.

Step by step solution

01

Calculate the volume and mass of the wooden cube.

First, we need to calculate the volume of the wooden cube. The wooden cube has a side length of 0.330 m. Therefore, the volume (V_wood) of the wooden cube can be calculated as: V_wood = \(0.330^3\) = \((0.330\,\mathrm{m})^3\) = \(0.035937\,\mathrm{m^3}\) Next, we will calculate the mass of the wooden cube using the density. The mass (m_wood) of the wooden cube can be calculated as: m_wood = density_wood × V_wood = \(780\,\mathrm{kg/m^3} \times 0.035937\,\mathrm{m^3}\) = \(28.03086\,\mathrm{kg}\)
02

Calculate the total buoyant force.

As the wooden cube is fully submerged in water, the buoyant force (F_B) will be equal to the weight of the displaced water. We know that the buoyant force is given by F_B = m_water × g, where m_water is the mass of the displaced water and g is the acceleration due to gravity (approximately \(9.81\,\mathrm{m/s^2}\)). The density of water (ρ_water) is \(1000\,\mathrm{kg/m^3}\), so we can calculate the mass of the displaced water (m_water) as: m_water = ρ_water × V_wood = \(1000\,\mathrm{kg/m^3} \times 0.035937\,\mathrm{m^3}\) = \(35.937\,\mathrm{kg}\) Now, we can calculate the total buoyant force (F_B) as: F_B = m_water × g = \(35.937\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2}\) = \(352.5933\,\mathrm{N}\)
03

Calculate the weight of the lead.

As the wooden cube and lead are in equilibrium, the total downward force (weight of the wooden cube and lead) must be equal to the total upward force (buoyant force). Therefore, we can write the equation as: W_wood + W_lead = F_B Where W_wood is the weight of the wooden cube given by m_wood × g and W_lead is the weight of lead that we are looking for. Now, we can solve for W_lead as: W_lead = F_B - W_wood = \(352.5933\,\mathrm{N} - (28.03086\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2})\) = \(80.81861\,\mathrm{N}\)
04

Calculate the mass of the lead.

Finally, we can determine the mass of the lead (m_lead) using the weight calculated in the previous step. As weight = mass × gravity, we can write: m_lead = W_lead / g = \(80.81861\,\mathrm{N} / 9.81\,\mathrm{m/s^2}\) = \(8.238\,\mathrm{kg}\) The mass of the lead is approximately 8.238 kg.

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