The specific weight of a fluid is 11 200 N=m3. Calculate the mass contained in 2 m3
(a) Using the standard gravity.
(b) Using the maximum gravity on the earth’s surface.
(c) Using the minimum gravity on the earth’s surface.
The correct answer and explanation is :
To calculate the mass contained in 2 m³ of fluid, we need to use the relationship between weight, mass, and gravitational acceleration. The specific weight ((\gamma)) is defined as the weight per unit volume. Specific weight is related to mass and gravity by the following equation:
[
\gamma = \frac{W}{V}
]
Where:
- (\gamma) is the specific weight (N/m³),
- (W) is the weight (N),
- (V) is the volume (m³).
Weight is related to mass ((m)) and gravitational acceleration ((g)) by:
[
W = m \cdot g
]
So, the specific weight can also be expressed as:
[
\gamma = m \cdot g / V
]
Solving for mass:
[
m = \frac{\gamma \cdot V}{g}
]
Now, we will calculate the mass for each scenario based on the given specific weight of 11,200 N/m³.
(a) Using standard gravity:
The standard gravitational acceleration is (g = 9.81 \, \text{m/s}^2).
Substituting the known values:
[
m = \frac{11,200 \, \text{N/m}^3 \times 2 \, \text{m}^3}{9.81 \, \text{m/s}^2}
]
[
m = \frac{22,400}{9.81}
]
[
m \approx 2,282.97 \, \text{kg}
]
(b) Using the maximum gravity on Earth’s surface:
The maximum gravitational acceleration on Earth’s surface occurs at the poles and is (g_{\text{max}} = 9.83 \, \text{m/s}^2).
Substituting the known values:
[
m = \frac{11,200 \, \text{N/m}^3 \times 2 \, \text{m}^3}{9.83 \, \text{m/s}^2}
]
[
m = \frac{22,400}{9.83}
]
[
m \approx 2,277.28 \, \text{kg}
]
(c) Using the minimum gravity on Earth’s surface:
The minimum gravitational acceleration on Earth’s surface occurs at the equator and is (g_{\text{min}} = 9.78 \, \text{m/s}^2).
Substituting the known values:
[
m = \frac{11,200 \, \text{N/m}^3 \times 2 \, \text{m}^3}{9.78 \, \text{m/s}^2}
]
[
m = \frac{22,400}{9.78}
]
[
m \approx 2,286.88 \, \text{kg}
]
Conclusion:
- (a) Using standard gravity ((g = 9.81 \, \text{m/s}^2)): (m \approx 2,283 \, \text{kg})
- (b) Using maximum gravity ((g = 9.83 \, \text{m/s}^2)): (m \approx 2,277 \, \text{kg})
- (c) Using minimum gravity ((g = 9.78 \, \text{m/s}^2)): (m \approx 2,287 \, \text{kg})
Explanation:
The mass of the fluid depends on the specific weight and the gravitational acceleration in the area. Since specific weight is the weight per unit volume, it is directly affected by gravity. Gravity varies slightly across the Earth, being stronger at the poles and weaker at the equator due to Earth’s shape and rotation. Thus, the mass of the fluid changes depending on the value of gravity used in the calculation. The higher the gravitational acceleration, the lower the resulting mass, and vice versa.