Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 14

A canal lock is \(6 \mathrm{~m}\) wide and has two vertical gates which make an angle of \(120^{\circ}\) with each other. The depths of water on the two sides of the gates are \(9 \mathrm{~m}\) and \(2.7 \mathrm{~m}\) respectively. Each gate is supported on two hinges, the lower one being \(600 \mathrm{~mm}\) above the bottom of the lock. Neglecting the weight of the gates themselves, calculate the thrust between the gates and the height of the upper hinges if the forces on them are to be half those on the lower hinges.

Short Answer

Expert verified
The resultant thrust between the gates is approximately \(472928 \, \text{N}\) and the height of the upper hinge should be \(0.1 \, \text{m}\).

Step by step solution

01

Determine the Water Pressure

The pressure exerted by water at a depth is given by the formula: \( P = \rho g h \) Where: \( \rho \) is the density of water (assumed to be \(1000 \; \text{kg/m}^3 \)), \( g \) is the acceleration due to gravity (\(9.81 \; \text{m/s}^2 \)), and \( h \) is the depth of water.
02

Calculate Water Pressure on Each Gate

For the first side with a depth of \(9 \, \text{m} \): \( P_1 = \rho g h_1 = 1000 \times 9.81 \times 9 = 88290 \, \text{Pa (Pascal)} \). For the second side with a depth of \(2.7 \, \text{m} \): \( P_2 = \rho g h_2 = 1000 \times 9.81 \times 2.7 = 26487 \, \text{Pa (Pascal)} \).
03

Calculate the Horizontal Thrust on Each Gate

The total force acting on a gate due to water pressure is given by: \( F = P \times A \) where \( A \) is the area of the gate. To find the thrust per unit width, we consider the area as height times width (with 1 meter width). For the first gate with height \(9 \, \text{m} \): \( F_1 = \frac{1}{2} \times \rho \times g \times (h_1)^2 \times 1 = \frac{1}{2} \times 1000 \times 9.81 \times 81 \times 1 = 397845 \, \text{N} \). For the second gate with height \(2.7 \, \text{m} \): \( F_2 = \frac{1}{2} \times \rho \times g \times (h_2)^2 \times 1 = \frac{1}{2} \times 1000 \times 9.81 \times 7.29 \times 1 = 35734 \, \text{N} \).
04

Calculate the Resultant Thrust Between Gates

The thrust between the gates primarily comes from the different water levels and their effects on the gates. Since the gates make an angle of \(120^{\circ}\) with each other (which translates to \(60^{\circ}\) per side):The resultant thrust can be given by: \( T = \sqrt{F_1^2 + F_2^2 - 2 \times F_1 \times F_2 \times \cos(60^{\circ})} \). Plugging in the values: \( T = \sqrt{(397845)^2 +(35734)^2 - 2 \times(397845) \times(35734)\times (0.5)} \) \( T = \sqrt{234896600025 + 1276513156 - 14221332310} \) \( T = \sqrt{223565795871} \approx 472928 \, \text{N} \)
05

Determine Height of the Upper Hinges

Given that forces on the upper hinges should be half of those on the lower hinges. Let's assign the height positions: The sum of moments around the bottom hinge should be zero for equilibrium: \( 3 \times h_U = \frac{1}{2} \times h_L \) Given that \(h_L\) = 600 mm above the bottom: \( 3 h_U = \frac{1}{2} \times 0.6} \) Solving for \(h_U\): \( h_U = \frac{0.3}{3} = 0.1 \, \text{m} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Fluid Mechanics
Fluid mechanics is the study of fluids and how they move. Fluids are substances like liquids and gases that can flow. Understanding fluid mechanics helps us determine forces and movements in systems containing fluids, such as canals, pipes, and even weather patterns.
When studying canal locks, fluid mechanics principles are used to calculate the forces on the gates caused by the water. These forces are crucial for engineers to design safe and efficient systems.
In our problem, we use fluid mechanics to determine the water pressure and the resultant thrust on the canal lock gates. This helps ensure that the lock operates correctly without structural damage.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth. The formula to calculate hydrostatic pressure is:
\( P = \rho g h \), where: \( \rho \) is the density of the fluid, \( g \) is the gravitational acceleration, and \( h \) is the depth of the fluid.
In the context of our canal lock problem, we need to find the pressure on the gates at different water depths. This helps us understand how much force the water pushes against the gates. For instance, for a depth of 9 meters, we calculated the pressure to be 88,290 Pascals (Pa). As the pressure pushes against the gates evenly, it causes significant force that needs to be managed.
Key Takeaways:
- Hydrostatic pressure increases linearly with depth.
- Knowing the pressure helps in understanding the force distribution on submerged surfaces.
Thrust Force
Thrust force refers to the force exerted by water on the canal gates due to its pressure. To calculate this force, we use the equation:
\( F = \frac{1}{2} \times \rho \times g \times (h^2) \).
This calculation involves finding the force per unit width on the gates given their height and water pressure.
For a gate height of 9 meters, the thrust force calculated is 397,845 Newtons (N). This force acts horizontally as water pushes against the gate. By determining the thrust force for both sides of the gates, we can evaluate the resulting force that the gates must withstand.
Important Points:
- Higher water levels generate more thrust force.
- Thrust calculation helps in designing the structural strength of the gates to handle the water pressure.
Moment Equilibrium
Moment equilibrium ensures that the forces acting on the canal gates are balanced. For the gates to be stable, the sum of moments (torques) about any hinge should be zero. This means that the clockwise and counterclockwise moments cancel each other out.
Given the problem's condition, we need to ensure that the upper hinge experiences half the force of the lower hinge. Using the calculated thrust, we achieve moment equilibrium by positioning the hinges correctly.
The calculation here involves balancing the moments around the bottom hinge:
\( 3 \times h_U = \frac{1}{2} \times h_L \)
Given the height of the lower hinge, we solve for the upper hinge height. Using the provided dimensions, we determine that the upper hinge should be positioned 0.1 meters above the bottom for equilibrium.
Key Insights:
- Moment equilibrium is essential for maintaining the structural integrity.
- Correct hinge positioning helps distribute the forces effectively, minimizing the risk of gate failure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A test vehicle contains a U-tube manometer for measuring differences of air pressure. The manometer is so mounted that, when the vehicle is on level ground, the plane of the U is vertical and in the fore-and-aft direction. The arms of the \(\mathrm{U}\) are \(60 \mathrm{~mm}\) apart, and contain alcohol of relative density \(0.79 .\) When the vehicle is accelerated forwards down an incline at \(20^{\circ}\) to the horizontal at \(2 \mathrm{~m} \cdot \mathrm{s}^{-2}\) the difference in alcohol levels (measured parallel to the arms of the \(\mathrm{U}\) ) is \(73 \mathrm{~mm}\), that nearer the front of the vehicle being the higher. What is the difference of air pressure to which this reading corresponds?

At the top a mountain the temperature is \(-5^{\circ} \mathrm{C}\) and \(\mathrm{a}\) mercury barometer reads \(566 \mathrm{~mm}\), whereas the reading at the foot of the mountain is \(749 \mathrm{~mm}\). Assuming a temperature lapse rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\) and \(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\), calculate the height of the mountain. (Neglect thermal expansion of mercury.)

A vessel of water of total mass \(5 \mathrm{~kg}\) stands on a parcel balance. An iron block of mass \(2.7 \mathrm{~kg}\) and relative density \(7.5\) is suspended by a fine wire from a spring balance and is lowered into the water until it is completely immersed. What are the readings on the two balances?

A vertical partition in a tank has a square aperture of side \(a\), the upper and lower edges of which are horizontal. The aperture is completely closed by a thin diaphragm. On one side on the diaphragm there is water with a free surface at a distance \(b(>a / 2)\) above the centre-line of the diaphragm. On the other side there is water in contact with the lower half of the diaphragm, and this is surmounted by a layer of oil of thickness \(c\) and relative density \(\sigma\). The free surfaces on each side of the partition are in contact with the atmosphere. If there is no net force on the diaphragm, determine the relation between \(b\) and \(c\), and the position of the axis of the couple on the diaphragm.

A solid cylinder, \(1 \mathrm{~m}\) diameter and \(800 \mathrm{~mm}\) high, is of uniform relative density \(0.85\) and floats with its axis vertical in still water. Calculate the periodic time of small angular oscillations about a horizontal axis.
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Work Energy and Power

Read Explanation

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free