Question
What is the formula to calculate the maximum tension a string can support when tied to a cylindrical object filled with water, with a cross-sectional radius of 3 cm and a height of 10 cm, filled with water to a height of 5 cm?
Answers
Answered by
GPT-4o mini
To determine the maximum tension a string can support when tied to a cylindrical object filled with water, we need to consider the weight of the water in the cylinder. The tension is essentially equal to the weight of this water.
First, we can calculate the volume of water in the cylinder:
1. **Calculate the volume (V)**:
The volume of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height of the water.
In this case:
- The radius \( r = 3 \) cm = 0.03 m (converting to meters for SI units)
- The height of the water \( h = 5 \) cm = 0.05 m
So, the volume of the water is:
\[
V = \pi (0.03 \, \text{m})^2 (0.05 \, \text{m})
\]
Calculating this:
\[
V = \pi (0.0009 \, \text{m}^2) (0.05 \, \text{m}) \approx \pi (0.000045 \, \text{m}^3) \approx 0.00014137167 \, \text{m}^3
\]
2. **Calculate the weight (W)**:
The weight of the water can be calculated using the formula:
\[
W = V \times \rho \times g
\]
where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
Now, substituting the values:
\[
W = 0.00014137167 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2
\]
\[
W \approx 0.00014137167 \times 9810 \approx 1.38643927 \, \text{N}
\]
The maximum tension \( T \) in the string, which is equal to the weight of the water, is approximately:
\[
T \approx 1.39 \, \text{N}
\]
Thus, the formula to calculate the maximum tension a string can support in this case is:
\[
T = \pi r^2 h \times \rho \times g
\]
where \( r \) is the radius of the cylinder, \( h \) is the height of the water, \( \rho \) is the density of water, and \( g \) is the acceleration due to gravity.
### Final Result
The maximum tension the string can support when tied to the cylindrical object filled with water to a height of 5 cm is approximately **1.39 N**.
First, we can calculate the volume of water in the cylinder:
1. **Calculate the volume (V)**:
The volume of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height of the water.
In this case:
- The radius \( r = 3 \) cm = 0.03 m (converting to meters for SI units)
- The height of the water \( h = 5 \) cm = 0.05 m
So, the volume of the water is:
\[
V = \pi (0.03 \, \text{m})^2 (0.05 \, \text{m})
\]
Calculating this:
\[
V = \pi (0.0009 \, \text{m}^2) (0.05 \, \text{m}) \approx \pi (0.000045 \, \text{m}^3) \approx 0.00014137167 \, \text{m}^3
\]
2. **Calculate the weight (W)**:
The weight of the water can be calculated using the formula:
\[
W = V \times \rho \times g
\]
where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
Now, substituting the values:
\[
W = 0.00014137167 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2
\]
\[
W \approx 0.00014137167 \times 9810 \approx 1.38643927 \, \text{N}
\]
The maximum tension \( T \) in the string, which is equal to the weight of the water, is approximately:
\[
T \approx 1.39 \, \text{N}
\]
Thus, the formula to calculate the maximum tension a string can support in this case is:
\[
T = \pi r^2 h \times \rho \times g
\]
where \( r \) is the radius of the cylinder, \( h \) is the height of the water, \( \rho \) is the density of water, and \( g \) is the acceleration due to gravity.
### Final Result
The maximum tension the string can support when tied to the cylindrical object filled with water to a height of 5 cm is approximately **1.39 N**.