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Chapter 2: Problem 27

A buoy, floating in sea-water of density \(1025 \mathrm{~kg} \cdot \mathrm{m}^{-3}\), is conical in shape with a diameter across the top of \(1.2 \mathrm{~m}\) and a vertex angle of \(60^{\circ}\). Its mass is \(300 \mathrm{~kg}\) and its centre of gravity is \(750 \mathrm{~mm}\) from the vertex. A flashing beacon is to be fitted to the top of the buoy. If this unit is of mass \(55 \mathrm{~kg}\) what is the maximum height of its centre of gravity above the top of the buoy if the whole assembly is not be unstable? (The centroid of a cone of height \(b\) is at \(3 h / 4\) from the vertex.)

Short Answer

Expert verified
The beacon's maximum center of gravity height above the buoy is around 0.75 meters.

Step by step solution

01

Understand the problem

Given a conical buoy with specific dimensions and weights, determine the maximum height at which a beacon can be installed without making the buoy unstable. Key data includes the density of sea-water, buoy dimensions, the mass of the buoy and beacon, and center of gravity locations.
02

Calculate the waterline height for stability

Calculate the height at which the buoy will float stable by equating the weight of the buoy and beacon to the buoyancy force.
03

Determine the volume and buoyancy force

Calculate the total volume of the conical buoy using its height and diameter. Use this to find the buoyancy force by multiplying the volume by the density of sea-water and gravitational acceleration.
04

Set up the instability condition

For the buoy to remain stable, the combined center of mass (buoy + beacon) should not extend beyond the buoy's geometric center. Calculate the combined center of mass and use the given distances to place the buoy and beacon's mass centers correctly.
05

Use moment calculations for stability

Using the formula for moments, equate the moments about the buoy's base to find the allowable maximum height of the beacon above the buoy’s top.
06

Solve for the maximum height

Solve the final expression derived in Step 5 to get the maximum height of the beacon's center of gravity above the buoy's top.

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Key Concepts

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

fluid mechanics
Fluid Mechanics is the branch of physics concerned with the behavior and properties of liquids and gases. It studies how fluids move and the forces on them. In our exercise, fluid mechanics helps us understand how the sea-water supports the buoy. The buoyancy force is derived from fluid mechanics principles. This force needs to counteract the weight of the buoy and the beacon to keep it afloat.

Key concepts in fluid mechanics include:
  • Density: The mass per unit volume of a fluid, here sea-water with a density of 1025 kg/m³.
  • Buoyancy: The force exerted by a fluid on a submerged object, causing it to float or rise.
  • Hydrostatic pressure: The pressure exerted by a fluid at rest due to the force of gravity.
In this problem, the buoyancy force relies on the volume of the buoy submerged in water and the fluid density.
center of gravity
The center of gravity of an object is the point at which its weight is evenly distributed in all directions. For the buoy, the center of gravity influences its stability in water.

Important points to understand:
  • The buoy's center of gravity is 750 mm from its vertex.
  • The center of gravity impacts how the moment forces act on the buoy when additional weight, like the beacon, is added.
The position of the center of gravity must be such that the buoy and beacon combined remain stable. If the center of gravity moves too high, the buoy can tip over.
buoyancy force
The buoyancy force is the upward force exerted by a fluid, counteracting the weight of an object submerged in it.
To calculate the buoyancy force:
  • Determine the volume of the submerged part of the buoy using its geometry.
  • Apply the formula: \[ F_b = \rho \times V \times g \] where \[\rho \] is the fluid density, \[ V \] is the submerged volume, and \[ g \] is the gravitational acceleration (9.81 m/s²).
In this exercise, we calculate the total volume of the conical buoy using the diameter and height values. Multiplying this volume by the density of sea-water and the gravitational acceleration gives us the buoyancy force. This force must equal the weight of the buoy and the beacon for the buoy to float stably.
moment calculation
Moment calculation is essential to determine the balance conditions of the buoy. Moments are the product of the force and the distance from a pivot point. In our problem, we use moments to derive the maximum height of the beacon's center of gravity.

Important steps in moment calculation include:
  • Identify the pivot point (typically, the base of the buoy).
  • Calculate the moments using the masses involved and their distances from the pivot point.
  • Set the sum of clockwise moments equal to the sum of counterclockwise moments for equilibrium.
Using moment calculations, we ensure that placing the beacon does not push the center of gravity too high, making the buoy unstable.
In this case, we calculate the moments created by the buoy's own weight and the beacon. By balancing these moments, we derive the maximum permissible height for the beacon's center of gravity to maintain stability.

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Most popular questions from this chapter

A hollow cylinder with closed ends is \(300 \mathrm{~mm}\) diameter and \(450 \mathrm{~mm}\) high, has a mass of \(27 \mathrm{~kg}\) and has a small hole in the base. It is lowered into water so that its axis remains vertical. Calculate the depth to which it will sink, the height to which the water will rise inside it and the air pressure inside it. Disregard the effect of the thickness of the walls but assume that it is uniform and that the compression of the air is isothermal. (Atmospheric pressure \(=101.3 \mathrm{kPa}\) )

A tank with vertical sides contains water to a depth of \(1.2 \mathrm{~m}\) and a layer of oil \(800 \mathrm{~mm}\) deep which rests on top of the water. The relative density of the oil is \(0.85\) and above the oil is air at atmospheric pressure. In one side of the tank, extending through its full height, is a protrusion in the form of a segment of a vertical circular cylinder. This is of radius \(700 \mathrm{~mm}\) and is riveted across an opening \(500 \mathrm{~mm}\) wide in the plane wall of the tank. Calculate the total horizontal thrust tending to force the protrusion away from the rest of the tank and the height of the line of action of this thrust above the base of the tank.

A rectangular pontoon \(6 \mathrm{~m}\) by \(3 \mathrm{~m}\) in plan, floating in water, has a uniform depth of immersion of \(900 \mathrm{~mm}\) and is subjected to a torque of \(7600 \mathrm{~N} \cdot \mathrm{m}\) about the longitudinal axis. If the centre of gravity is \(700 \mathrm{~mm}\) up from the bottom, estimate the angle of heel.

A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are \(2 \mathrm{~m}\) long and the tank contains a liquid of relative density 1.15. The centre of the aperture is \(1.5 \mathrm{~m}\) below the free surface. Calculate the net hydrostatic thrust on the plate, the moment of this thrust about the hinge and the position of the centre of pressure.

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?
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