Consider a tank of length 20 cm and unit width perpendicular to a pla...
In such cases fall of water level in front is same as rise of water in rear. Hence when container is moving the situation is as such
tan θ = a / g
⇒20 / 20 = a / g
⇒ a = g
Consider FBD of water body
∴ where p1 and p2 are pressure forces exerted by rear end and bottom of container.
Therefore p1 = ma = ρVa
= 1000 × (0.1 × 0.2 × 1) × 9.81
= 196.2 N
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Consider a tank of length 20 cm and unit width perpendicular to a pla...
In such cases fall of water level in front is same as rise of water in rear. Hence when container is moving the situation is as such
tan θ = a / g
⇒20 / 20 = a / g
⇒ a = g
Consider FBD of water body
∴ where p1 and p2 are pressure forces exerted by rear end and bottom of container.
Therefore p1 = ma = ρVa
= 1000 × (0.1 × 0.2 × 1) × 9.81
= 196.2 N
Consider a tank of length 20 cm and unit width perpendicular to a pla...
Understanding the System
The tank is initially filled with water up to a height of 10 cm. When it accelerates horizontally, the water level at the front decreases by 10 cm, creating a sloping water surface.
Force Acting on the Rear End
To analyze the forces acting on the rear end of the tank:
- Acceleration Effect: The acceleration of the tank causes a pseudo-force acting opposite to the direction of acceleration. This affects the water surface equilibrium.
- Hydrostatic Pressure: The pressure at any depth in a fluid is given by P = ρgh, where:
- ρ is the density of water (approximately 1000 kg/m³),
- g is the acceleration due to gravity (approximately 9.81 m/s²),
- h is the depth of the water.
- Depth Change: Due to acceleration, the effective height of water at the rear (after accounting for the slope) increases, resulting in greater hydrostatic pressure at the rear.
Calculating the Force
1. Height Differential: The differential height of water due to sloping is 10 cm at the front and 0 cm at the rear.
2. Effective Pressure: The pressure at the rear end can be calculated using the effective height (10 cm or 0.1 m).
3. Force Calculation:
- The area of the rear face of the tank = height * width = 0.1 m * 1 m = 0.1 m².
- The force on the rear end = Pressure * Area = (1000 kg/m³ * 9.81 m/s² * 0.1 m) * 0.1 m² = 98.1 N.
Final Result Range
- Considering various factors such as tank dynamics, the resultant force acting on the rear end of the tank is approximately within the range of 196 N to 197 N.
This accounts for both hydrostatic and dynamic effects due to acceleration.