To solve this problem, we need to analyze the effects of both horizontal and vertical accelerations on the water inside the box. The box has dimensions of height 2m, width 6m, and length 6m, and it is half-filled with water. Given that the box is subjected to horizontal and vertical accelerations of g/2, we can determine the angle of the free surface of the water and the pressure at the center of the box.
Understanding the Forces at Play
When the box is accelerated, the water inside experiences pseudo-forces due to the acceleration. The effective gravitational force acting on the water can be calculated by combining the effects of the vertical and horizontal accelerations. The vertical acceleration is g/2 (downward), and the horizontal acceleration is g/2 (to the side).
Calculating the Effective Gravitational Force
The effective gravitational force can be represented as a vector sum of the vertical and horizontal components. We can use the Pythagorean theorem to find the resultant acceleration:
- Vertical component: g/2
- Horizontal component: g/2
The resultant acceleration (a) is given by:
a = √((g/2)² + (g/2)²) = √(g²/4 + g²/4) = √(g²/2) = g/√2
Finding the Angle of the Free Surface
The angle (θ) that the free surface makes with the horizontal can be determined using the tangent function:
tan(θ) = (horizontal acceleration) / (vertical acceleration) = (g/2) / (g/2) = 1
Thus, θ = arctan(1) = 45 degrees.
Determining the Length of the Exposed Portion
Since the box is half-filled with water, the initial height of the water is 1m (half of 2m). When the box is accelerated, the water surface tilts, and we need to find the new height of the water at the back of the box.
Using the geometry of the situation, we can find the height (h) at the back of the box:
h = initial height + (length of the box) * tan(θ)
Here, the length of the box is 6m, so:
h = 1m + 6m * tan(45°) = 1m + 6m = 7m
However, since the box is only 2m high, the water cannot rise above this height. Therefore, the exposed portion of the top of the box is:
Exposed height = total height - new height = 2m - 1m = 1m
Pressure at the Center of the Box
To find the pressure at the center of the box, we need to consider the depth of the water at that point. The center of the box is located at a height of 1m from the bottom. The pressure (P) at a certain depth in a fluid is given by the formula:
P = ρgh
Where:
- ρ = density of water (approximately 1000 kg/m³)
- g = acceleration due to gravity (approximately 9.81 m/s²)
- h = depth of the water (1m in this case)
Substituting the values:
P = 1000 kg/m³ * 9.81 m/s² * 1m = 9810 Pa
Summary of Results
In conclusion, the angle made by the free surface with the horizontal is 45 degrees. The length of the exposed portion of the top of the box is 1m, and the pressure at the center of the box is 9810 Pa. This analysis illustrates how acceleration affects fluid behavior in a confined space.
