1. A cylindrical disk of wood weighing 45.0N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g/cm3. The cylinder of oil is 75.0 cm deep and has the same diameter of the wood. (a) What is the gauge pressure at the top of the oil column? Suppose now someone puts a weight of 83.0N on the wooden disk and no oil seeps through. (b) what is the change in pressure at the top of the oil [3 pts] and at the bottom of the oil column?
∆p=p-p_o=p_g W/A=W/(πr^2 )
p_g=(45 N)/(π(.150m)^2 )=636 Pa
∆p=(83 N)/(π(.150〖m)〗^2 )=1170 Pa
(a) Well, it seems like we have a floating disk made of wood on a cylinder of oil. Let's dive into it!
First, we need to find the volume of the oil. Since the cylinder has the same diameter as the wood disk, we can use the formula for the volume of a cylinder: V = πr^2h, where r is the radius of the cylinder and h is the height.
Given that the diameter of the wood disk is 30.0 cm, the radius would be half of that, which is 15.0 cm or 0.15 m. The height of the cylinder of oil is 75.0 cm or 0.75 m.
Using the formula for the volume of a cylinder, we can calculate the volume of the oil:
V = π(0.15)^2(0.75) = 0.053 m^3
Next, we need to find the weight of the oil. We can use the formula: weight = density x volume x gravity.
Given that the density of the oil is 0.850 g/cm^3, which is 850 kg/m^3, and the volume of the oil is 0.053 m^3, we can calculate the weight of the oil:
weight = 850 x 0.053 x 9.8 = 429.95 N
Now, let's find the weight of the wood disk. It's given as 45.0 N.
The gauge pressure at the top of the oil column is the pressure exerted by the weight of the oil pushing down on the top surface:
pressure = weight of oil / area
The area of the top surface is the same as the area of the cylindrical wood disk, which is πr^2. So, the gauge pressure at the top can be calculated as:
pressure = weight of oil / (πr^2)
Substituting the values:
pressure = 429.95 / (π(0.15)^2) = 6125.38 Pa
(b) Now, someone adds a weight of 83 N on the wood disk. The change in pressure at the top of the oil is simply the weight added divided by the area of the wood disk:
change in pressure = weight added / (πr^2)
Given that the weight added is 83 N, we can calculate the change in pressure:
change in pressure = 83 / (π(0.15)^2) = 1187.5 Pa
As for the change in pressure at the bottom of the oil column, it remains the same because the additional weight does not affect the oil below it. So there you have it, the change in pressure at the top of the oil column is 1187.5 Pa and the change in pressure at the bottom remains the same.
To solve this problem, we will use the concepts of buoyancy and pressure.
(a) Calculating the gauge pressure at the top of the oil column:
First, we need to find the volume of the wood disk.
The diameter of the wood disk is 30.0 cm, so the radius would be 15.0 cm (or 0.15 m).
The volume of the disk (V_disk) can be calculated using the formula for the volume of a cylinder:
V_disk = π * r^2 * h = π * (0.15 m)^2 * 0.001 m (since 1 cm = 0.01 m)
V_disk = 0.001 * π * 0.0225 m^3
Next, let's find the volume of oil supporting the wood disk.
The height of the oil column is given as 75.0 cm, which is 0.75 m.
The volume of the oil (V_oil) can be calculated using the same formula for the volume of a cylinder:
V_oil = π * r^2 * h = π * (0.15 m)^2 * 0.75 m
V_oil = 0.75 * π * 0.0225 m^3
Now, we can find the mass of the oil that supports the wood disk.
The density of the oil is given as 0.850 g/cm^3, which is 850 kg/m^3.
The mass of the oil (m_oil) can be calculated by multiplying the density by the volume:
m_oil = 850 kg/m^3 * 0.75 * π * 0.0225 m^3
m_oil = 55.972 kg
Finally, we can find the weight of the oil (W_oil) that is supporting the wood disk:
W_oil = m_oil * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2)
W_oil = 55.972 kg * 9.8 m/s^2
W_oil = 548.1716 N
The gauge pressure at the top of the oil column is equal to the weight of the oil divided by the area of the top surface of the oil column:
P_top = W_oil / A_oil_top
The area of the top surface of the oil column (A_oil_top) can be calculated using the formula for the area of a circle:
A_oil_top = π * r^2 = π * (0.15 m)^2
P_top = 548.1716 N / (π * 0.0225 m^2)
P_top ≈ 9252.7 Pa
Therefore, the gauge pressure at the top of the oil column is approximately 9252.7 Pa.
(b) The change in pressure at the top of the oil column and the bottom of the oil column when a weight of 83.0 N is added to the wood disk:
The pressure at the top of the oil column would remain the same because the oil is incompressible. Therefore, the change in pressure at the top is zero.
At the bottom of the oil column, the pressure would increase due to the added weight. The change in pressure can be calculated by dividing the weight added by the area of the bottom surface of the oil column:
P_bottom = (W_oil + 83.0 N) / A_oil_bottom
The area of the bottom surface of the oil column (A_oil_bottom) can be calculated using the formula for the area of a circle:
A_oil_bottom = π * r^2 = π * (0.15 m)^2
P_bottom = (548.1716 N + 83.0 N) / (π * 0.0225 m^2)
P_bottom ≈ 11862.7 Pa
Therefore, the change in pressure at the bottom of the oil column is approximately 11862.7 Pa.
To solve this problem, we need to use concepts of buoyancy and fluid pressure.
(a) To find the gauge pressure at the top of the oil column, we need to consider the pressure due to the weight of the oil column and the atmospheric pressure.
1. Convert the density of the oil to kg/m^3:
The density of oil is given as 0.850 g/cm^3. To convert it to kg/m^3, we divide by 1000 (since there are 1000 grams in 1 kilogram) and multiply by 100^3 (since there are 100 centimeters in 1 meter):
Density of oil = 0.850 g/cm^3 * (1 kg / 1000 g) * (100 cm / 1 m)^3
= 850 kg/m^3
2. Calculate the weight of the oil column:
The weight of the oil column is equal to the weight of the disk floating on it. Given that the weight of the disk is 45.0 N, the weight of the oil column is also 45.0 N.
3. Calculate the pressure due to the weight of the oil column:
Pressure = density * gravitational acceleration * height
Here, density = 850 kg/m^3
gravitational acceleration = 9.8 m/s^2 (standard value)
height = 75 cm = 0.75 m (converted from centimeters to meters)
Pressure = 850 kg/m^3 * 9.8 m/s^2 * 0.75 m
= 6172.5 Pa
4. Calculate the atmospheric pressure:
Atmospheric pressure is typically around 101325 Pa (standard value).
5. Finally, find the gauge pressure at the top of the oil column:
Gauge pressure = Pressure at the top - Atmospheric pressure
= 6172.5 Pa - 101325 Pa
= -95025.5 Pa
Therefore, the gauge pressure at the top of the oil column is approximately -95025.5 Pa.
(b) To calculate the change in pressure at the top and bottom of the oil column when a weight of 83.0 N is added:
1. Calculate the new weight of the oil column:
Since the wooden disk now has a weight of 45.0 N added to its weight, the total weight of the disk and the added weight is 45.0 N + 83.0 N = 128.0 N.
2. Find the new pressure due to the weight of the oil column:
We use the same formula as before: Pressure = density * gravitational acceleration * height.
Here, density = 850 kg/m^3
gravitational acceleration = 9.8 m/s^2 (standard value)
height = 75 cm = 0.75 m (converted from centimeters to meters)
Pressure = 850 kg/m^3 * 9.8 m/s^2 * 0.75 m
= 6172.5 Pa
3. Calculate the change in pressure at the top of the oil column:
Change in pressure = New pressure - Old Pressure
= 6172.5 Pa - (-95025.5 Pa)
= 101198 Pa
4. Calculate the pressure at the bottom of the oil column:
The pressure at the bottom of the oil column is the same as the pressure at the top plus the additional pressure due to the added weight.
Pressure at the bottom = New pressure + Pressure due to added weight
= 6172.5 Pa + 83.0 N / (30.0 cm / 100 m)
= 6172.5 Pa + 27667 Pa
= 33839.5 Pa
Therefore, the change in pressure at the top of the oil column is approximately 101198 Pa, and the pressure at the bottom of the oil column is approximately 33839.5 Pa.