/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A vessel of water of total mass ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
The spring balance reads 22.932 N and the parcel balance reads 52.528 N.

Step by step solution

01

Calculate the Volume of the Iron Block

The relative density of iron is given as 7.5. This means the density of iron is 7.5 times the density of water. The density of water is \rho_w = 1000 \frac{kg}{m^3}. Therefore, the density of iron \rho_{iron} is 7.5 \times 1000 = 7500 \frac{kg}{m^3}. Using the mass of the iron block, we can calculate its volume. The formula to calculate volume is: \[ V = \frac{mass}{density} \]Plugging in the given values:\[ V_{iron} = \frac{2.7 \text{ kg}}{7500\frac{kg}{m^3}} = 0.00036 \text{ m}^3 \]
02

Calculate the Buoyant Force Acting on the Iron Block

A block immersed in water experiences an upward buoyant force equal to the weight of the water displaced. The volume of the displaced water is the same as the volume of the iron block. The buoyant force is given by: \[ F_b = \text{volume} \times \text{density of water} \times g \]Here, g is the acceleration due to gravity, which is approximately 9.8 \frac m{s^2}. The density of water is \( \rho_w = 1000 \frac{kg}{m^3} \). Using the volume calculated in Step 1, we get: \[ F_b = 0.00036 \text{ m}^3 \times 1000 \frac{kg}{m^3} \times 9.8 \frac m{s^2} = 3.528 N \]
03

Determine the Reading on the Spring Balance

The spring balance measures the tension in the wire, which is the actual weight of the iron block minus the buoyant force acting on it. The weight of the iron block is given by: \[ W_{iron} = mass \times g = 2.7 \text{ kg} \times 9.8 \frac m{s^2} = 26.46 N \]Therefore, the reading on the spring balance is: \[ T = W_{iron} - F_b = 26.46 N - 3.528 N = 22.932 N \]
04

Determine the Reading on the Parcel Balance

The parcel balance measures the total weight of the vessel and its contents. Initially, the vessel of water has a mass of 5 kg. When the iron block is fully immersed, it displaces its volume of water which doesn't change the weight of the water, but the buoyant force acts on the vessel as an added weight. Hence, the added weight is equal to the buoyant force: \[ 3.528 N \]Converting this force back into mass: \[ \text{additional mass} = \frac{3.528 N}{9.8 \frac{m}{s^2}} = 0.36 kg \]Thus, the total mass on the parcel balance becomes the initial mass of the vessel plus the mass equivalent to the buoyant force: \[ 5 kg + 0.36 kg = 5.36 kg \]Therefore, the reading on the parcel balance in weight is: \[ 5.36 kg \times 9.8 \frac{m}{s^2} = 52.528 N \]

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Key Concepts

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

Buoyant Force
The buoyant force is the upward force that fluids like water exert on objects that are completely or partially immersed in them.
This force is governed by Archimedes' Principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object.
In this exercise, when the iron block is lowered into the water, it displaces a volume of water equal to its own volume.
The weight of this displaced water gives us the buoyant force.
For instance, if the volume of the iron block is 0.00036 m³, and the density of water is 1000 kg/m³, then using the formula for buoyant force \( F_b = \text{volume} \times \text{density of water} \times g \), the buoyant force is calculated as \( 0.00036 \text{ m}^3 \times 1000 \frac{kg}{m^3} \times 9.8 \frac{m}{s^2} = 3.528 N \).
This understanding is crucial in determining how objects behave when submerged in fluids and affects readings on balances.
Relative Density
Relative density, also known as specific gravity, is a measure of the density of a material compared to the density of water at a specified temperature, typically 4°C.
The relative density of the iron block in this exercise is given as 7.5.
This means the iron is 7.5 times denser than water.
To find the density of iron, we simply multiply the density of water by the relative density: \( \rho_{iron} = 7.5 \times 1000 \frac{kg}{m^3} = 7500 \frac{kg}{m^3} \).
Understanding the relative density helps us find the volume of the iron block by using its mass and density.
For example, the volume of a 2.7 kg iron block is calculated using \( V = \frac{mass}{density} = \frac{2.7 \text{ kg}}{7500 \frac{kg}{m^3}} = 0.00036 \text{ m}^3 \).
This is essential for calculating the buoyant force and understanding buoyancy in different fluids.
Spring Balance Reading
A spring balance measures the tension in the wire or string from which an object is suspended.
When the iron block is fully immersed in water, the spring balance no longer reads just the object's weight.
Instead, it reads the weight minus the buoyant force.
To get this reading, we first calculate the actual weight of the iron block: \( W_{iron} = 2.7 \text{ kg} \times 9.8 \frac {m}{s^2} = 26.46 N \).
Then, we subtract the buoyant force from this weight: \( T = 26.46 N - 3.528 N = 22.932 N \).
Hence, the spring balance shows 22.932 N, reflecting the reduced weight due to the upward buoyant force supporting part of the iron block’s weight.
This principle is fundamental in various applications, including underwater weighing and determining the density of objects.
Weight Measurement
Weight measurement involves determining the gravitational force acting on an object’s mass.
Different types of balances measure this force, including parcel balances and spring balances.
In the example, the parcel balance initially measures the vessel and water's combined weight, which totals 5 kg.
When the iron block is immersed, part of its weight is transferred to the water as the buoyant force but does not change the vessel's actual weight.
The added weight due to the buoyant force manifests as an increase in the parcel balance reading.
To reflect this, we add the mass equivalent to the buoyant force to the initial mass: \( 5 \text{ kg} + 0.36 \text{ kg} = 5.36 \text{ kg} \). This converts to a new weight of \( 5.36 \times 9.8 \frac{m}{s^2} = 52.528 N \).
Such accurate weight measurements are crucial for scientific experiments, shipping, and various industrial applications.

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Most popular questions from this chapter

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.

An open channel has a cross-section in the form of an equilateral triangle with \(2.5 \mathrm{~m}\) sides and a vertical axis of symmetry. Its end is closed by a triangular vertical gate, also with \(2.5 \mathrm{~m}\) sides, supported at each corner. Calculate the horizontal thrust on each support when the channel is brim-full of water.

A manometer consists of two tubes \(A\) and \(B\), with vertical axes and uniform cross-sectional areas \(500 \mathrm{~mm}^{2}\) and \(800 \mathrm{~mm}^{2}\) respectively, connected by a U-tube \(C\) of cross-sectional area \(70 \mathrm{~mm}^{2}\) throughout. Tube \(A\) contains a liquid of relative density \(0.8\); tube \(B\) contains one of relative density \(0.9 .\) The surface of separation between the two liquids is in the vertical side of \(\mathrm{C}\) connected to tube \(A\). What additional pressure, applied to the tube \(B\), would cause the surface of separation to rise \(60 \mathrm{~mm}\) in the tube \(C\) ?

An open-topped tank, in the form of a cube of \(900 \mathrm{~mm}\) side, has a mass of \(340 \mathrm{~kg}\). It contains \(0.405 \mathrm{~m}^{3}\) of oil of relative density \(0.85\) and is accelerated uniformly up a long slope at arctan (1/3) to the horizontal. The base of the tank remains parallel to the slope, and the side faces are parallel to the direction of motion. Neglecting the thickness of the walls of the tank, estimate the net force (parallel of the slope) accelerating the tank if the oil is just on the point of spilling.

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.)

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