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Water flows into the tank at the rate of \(0.05 \mathrm{~m}^{3} / \mathrm{s}\) from the 100 -mm-diameter pipe. If the tank is \(500 \mathrm{~mm}\) on each side, determine the compression in each of the four springs that support its corners when the water reaches a depth of \(h=1 \mathrm{~m}\). Each spring has a stiffness of \(k=8 \mathrm{kN} / \mathrm{m}\). When empty, the tank compresses each spring \(30 \mathrm{~mm}\).

Short Answer

Expert verified
The compression in each of the four springs when the water reaches a depth of \(h=1 m\) is \(106.6 mm\).

Step by step solution

01

Calculate the weight of water

First, calculate the volume of the water in the tank using the depth given, \(h=1 m\), and the cross-sectional area of the tank, \(A\), which is given as \(500 mm \times 500 mm = 0.25 m^2\). Therefore, the volume of water, \(V=Ah = 0.25 m^2 \times 1 m = 0.25 m^3 \). The density of water, \(\rho\), is typically \(1000 kg/m^3\). The weight of the water, \(W\), is calculated by multiplying the volume with the density of water and acceleration due to gravity, \(g=9.81 m/s^2\). Thus, \(W = \rho V g = 1000 kg/m³ \times 0.25 m³ \times 9.81 m/s² = 2452.5 N\).
02

Compute the compression of springs

Now, knowing the weight of the water, which is also the force applied onto the springs, and the spring constant, \(k=8 kN/m = 8000 N/m\), we can calculate the increase in each springs' compression due the weight of the water. As the force is evenly distributed among the four springs, each spring will experience a force, \(F\), of \(W/4 = 2452.5 N/4 = 613.125 N\). From Hooke's law, \(F=k \Delta x_{1}\), the additional compression, \(\Delta x_{1}\), equals to \(F/k = 613.125 N / 8000 N/m = 0.0766 m = 76.6 mm\). Lastly, the total compression in each spring, \(X\), when the water reaches a depth of \(h=1 m\), is therefore the sum of initial compression (\(30 mm\)) and the additional compression due to weight of the water, thus we get \(X = 30 mm + 76.6 mm = 106.6 mm\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hooke's Law
Understanding Hooke's Law is essential when studying how materials deform under force. It's a principle that describes the relationship between the force applied to a spring and the resulting extension or compression. Hooke's Law can be expressed with the simple equation:
\[ F = k \Delta x \]
where F is the force applied on the spring, k is the spring constant, which represents the stiffness of the spring, and \Delta x\ is the change in length from the spring's original length.

In the context of the exercise, we were given a spring constant, k, and needed to find the spring’s change in length, or compression, \Delta x\. After calculating the weight of the water in the tank acting as the force, we applied Hooke's Law to determine each spring's compression.
It is vital to consider that Hooke's Law only remains valid as long as the material does not surpass what is known as the elastic limit; beyond this point, the material may deform permanently.
Hydrostatic Pressure
The concept of hydrostatic pressure is key to fluid mechanics education, particularly when dealing with fluids at rest. It is defined as the pressure exerted by a fluid due to the force of gravity. The hydrostatic pressure, P, at a specific depth below the surface of the fluid is given by:
\[ P = \rho gh \]
where \rho is the fluid's density, g is the acceleration due to gravity, and h is the depth of the fluid.

Although hydrostatic pressure wasn't directly used in the exercise's calculations, understanding this concept is fundamental in fluid mechanics. In scenarios where force distribution due to fluid pressures on surfaces is analyzed, hydrostatic pressure calculations become essential. In the given problem, the hydrostatic pressure at the bottom of the tank would influence the force applied to the springs supporting the tank.
Force Distribution
Force Distribution involves analyzing how a force is spread over an area or among different supporting structures. When a system is in equilibrium, the total forces and moments must be balanced. In the context of the given exercise, the weight of the water within the tank is evenly distributed across the four supporting springs.

Each spring supports a quarter of the total weight, which is consistent with concepts of force distribution in static equilibrium. This idea is crucial in many engineering applications, where the stability and integrity of structures depend on the correct distribution of loads.
  • Misjudging force distribution can lead to uneven stress, excessive deformation, or even structural failure.
  • The correct calculation ensures that each component of a structure is utilized within its capacity.

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