Question
Oil having a density of 921 kg/m^3 floats on water. A rectangular block of wood 3.07 cm high and with a density of 973 kg/m^3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block? Answer in units of m.
Answers
Answered by
GPT-4o mini
To solve the problem, we need to analyze the forces acting on the block of wood, which is floating between the oil and the water.
Let's denote:
- \( h_{w} \): the height of the block submerged in water (m).
- \( h_{o} \): the height of the block submerged in oil (m).
- \( H \): the total height of the block (3.07 cm = 0.0307 m).
- \( \rho_{w} \): the density of water (about 1000 kg/m³).
- \( \rho_{o} \): the density of oil (921 kg/m³).
- \( \rho_{b} \): the density of wood (973 kg/m³).
Since the block is floating, the weight of the block is equal to the buoyant force acting on it from both the oil and the water.
1. Calculate the volume of the block:
\[
V = \text{Area} \times H
\]
Since the area is not explicitly given, we can work in terms of area \( A \).
2. The weight of the block (\( W \)):
\[
W = \rho_{b} \cdot V = \rho_{b} \cdot A \cdot H
\]
3. The buoyant force from the oil and water:
\[
F_{b} = \text{buoyant force from oil} + \text{buoyant force from water}
\]
\[
F_{b} = \rho_{o} \cdot V_{o} + \rho_{w} \cdot V_{w}
\]
where:
- \( V_{o} = A \cdot h_{o} \) (volume submerged in oil)
- \( V_{w} = A \cdot h_{w} \) (volume submerged in water)
4. The volume of the block is related to \( h_{o} \) and \( h_{w} \) by:
\[
h_{o} + h_{w} = H
\]
Substituting \( V_{w} = A \cdot h_{w} \) and \( V_{o} = A \cdot h_{o} \), we have:
\[
F_{b} = \rho_{o} \cdot A \cdot h_{o} + \rho_{w} \cdot A \cdot h_{w}
\]
This can be simplified by dividing by \( A \):
\[
F_{b} = \rho_{o} \cdot h_{o} + \rho_{w} \cdot h_{w}
\]
5. Setting the weight equal to the buoyant force:
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot h_{w}
\]
6. Now insert \( h_{w} = H - h_{o} \) into the equation to eliminate \( h_{w} \):
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot (H - h_{o})
\]
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot H - \rho_{w} \cdot h_{o}
\]
7. Rearranging gives:
\[
\rho_{b} \cdot H - \rho_{w} \cdot H = (\rho_{o} - \rho_{w}) \cdot h_{o}
\]
\[
H \cdot (\rho_{b} - \rho_{w}) = h_{o} \cdot (\rho_{o} - \rho_{w})
\]
Now solve for \( h_{o} \):
\[
h_{o} = \frac{H \cdot (\rho_{b} - \rho_{w})}{\rho_{o} - \rho_{w}}
\]
8. Substituting known values:
- \( \rho_{b} = 973 \, \text{kg/m}^3 \)
- \( \rho_{o} = 921 \, \text{kg/m}^3 \)
- \( \rho_{w} = 1000 \, \text{kg/m}^3 \)
- \( H = 0.0307 \, \text{m} \)
Now plug in the values:
\[
h_{o} = \frac{0.0307 \cdot (973 - 1000)}{921 - 1000}
\]
\[
= \frac{0.0307 \cdot (-27)}{-79}
\]
\[
= 0.0307 \cdot \frac{27}{79}
\]
Calculating this gives:
\[
h_{o} \approx 0.0104 \, \text{m}
\]
9. Finally, calculate the depth of the block below the interface:
Since the total height \( H = 0.0307 \, \text{m} \):
\[
\text{Depth below interface} = h_{o} = 0.0104 \, m
\]
Thus, the bottom of the block is approximately **0.0104 m** below the interface between the two liquids.
Let's denote:
- \( h_{w} \): the height of the block submerged in water (m).
- \( h_{o} \): the height of the block submerged in oil (m).
- \( H \): the total height of the block (3.07 cm = 0.0307 m).
- \( \rho_{w} \): the density of water (about 1000 kg/m³).
- \( \rho_{o} \): the density of oil (921 kg/m³).
- \( \rho_{b} \): the density of wood (973 kg/m³).
Since the block is floating, the weight of the block is equal to the buoyant force acting on it from both the oil and the water.
1. Calculate the volume of the block:
\[
V = \text{Area} \times H
\]
Since the area is not explicitly given, we can work in terms of area \( A \).
2. The weight of the block (\( W \)):
\[
W = \rho_{b} \cdot V = \rho_{b} \cdot A \cdot H
\]
3. The buoyant force from the oil and water:
\[
F_{b} = \text{buoyant force from oil} + \text{buoyant force from water}
\]
\[
F_{b} = \rho_{o} \cdot V_{o} + \rho_{w} \cdot V_{w}
\]
where:
- \( V_{o} = A \cdot h_{o} \) (volume submerged in oil)
- \( V_{w} = A \cdot h_{w} \) (volume submerged in water)
4. The volume of the block is related to \( h_{o} \) and \( h_{w} \) by:
\[
h_{o} + h_{w} = H
\]
Substituting \( V_{w} = A \cdot h_{w} \) and \( V_{o} = A \cdot h_{o} \), we have:
\[
F_{b} = \rho_{o} \cdot A \cdot h_{o} + \rho_{w} \cdot A \cdot h_{w}
\]
This can be simplified by dividing by \( A \):
\[
F_{b} = \rho_{o} \cdot h_{o} + \rho_{w} \cdot h_{w}
\]
5. Setting the weight equal to the buoyant force:
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot h_{w}
\]
6. Now insert \( h_{w} = H - h_{o} \) into the equation to eliminate \( h_{w} \):
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot (H - h_{o})
\]
\[
\rho_{b} \cdot H = \rho_{o} \cdot h_{o} + \rho_{w} \cdot H - \rho_{w} \cdot h_{o}
\]
7. Rearranging gives:
\[
\rho_{b} \cdot H - \rho_{w} \cdot H = (\rho_{o} - \rho_{w}) \cdot h_{o}
\]
\[
H \cdot (\rho_{b} - \rho_{w}) = h_{o} \cdot (\rho_{o} - \rho_{w})
\]
Now solve for \( h_{o} \):
\[
h_{o} = \frac{H \cdot (\rho_{b} - \rho_{w})}{\rho_{o} - \rho_{w}}
\]
8. Substituting known values:
- \( \rho_{b} = 973 \, \text{kg/m}^3 \)
- \( \rho_{o} = 921 \, \text{kg/m}^3 \)
- \( \rho_{w} = 1000 \, \text{kg/m}^3 \)
- \( H = 0.0307 \, \text{m} \)
Now plug in the values:
\[
h_{o} = \frac{0.0307 \cdot (973 - 1000)}{921 - 1000}
\]
\[
= \frac{0.0307 \cdot (-27)}{-79}
\]
\[
= 0.0307 \cdot \frac{27}{79}
\]
Calculating this gives:
\[
h_{o} \approx 0.0104 \, \text{m}
\]
9. Finally, calculate the depth of the block below the interface:
Since the total height \( H = 0.0307 \, \text{m} \):
\[
\text{Depth below interface} = h_{o} = 0.0104 \, m
\]
Thus, the bottom of the block is approximately **0.0104 m** below the interface between the two liquids.