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Chapter 7: Problem 40

A wire is suspended vertically from a rigid support. When loaded with a steel weight in air, the wire extends by \(16 \mathrm{~cm}\). When the weight is completely immersed in water, the extension is reduced to \(14 \mathrm{~cm}\). The relative density of the material of the weight is (1) \(2 \mathrm{~g} / \mathrm{cm}^{3}\) (2) \(6 \mathrm{~g} / \mathrm{cm}^{3}\) (3) \(8 \mathrm{~g} / \mathrm{cm}^{3}\) (4) \(16 \mathrm{~g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
The relative density of the material of the weight is 8 g/cm鲁.

Step by step solution

01

Identify the extensions and forces involved

When the steel weight is in air, the wire extends due to the force of the weight, which we can denote as gravity force \( F_g \). When the weight is immersed in water, the buoyant force \( F_b \) acts upwards and reduces the effective weight, thereby decreasing the extension from \( 16 \text{ cm} \) to \( 14 \text{ cm} \).
02

Determine the change in buoyant force

The difference in wire extension, \(2 \text{ cm} \), is due to the buoyant force exerted by the water. We know the buoyant force can be calculated as \( F_b = V \cdot \rho_{water} \cdot g \), where \( V \) is the volume of the immersed body and \( \rho_{water} \) is the density of water \( (1 \text{ g/cm}^3) \).
03

Write the balance of forces equations

When in air, the force due to gravity is equal to \( k \cdot 16 \text{ cm} \), where \( k \) is the force constant of the wire. When in water, the effective force \( F' = F_g - F_b \) is equal to \( k \cdot 14 \text{ cm} \).
04

Relate the forces to create an equation

We have \( k \cdot 16 = mg \) for when the object is in air, and \( k \cdot 14 = mg - V \cdot \rho_{water} \cdot g \) for when the object is in water. From this, it's known \( mg - F_b = k \cdot 14 \). Simplify to find the relation: \( F_b = k \cdot 2 \).
05

Determine the relative density of the object

Since the volume \( V \) times relative density \( \rho_{r} \) is equal to \( V \cdot \rho_{water} \), the formula can be restructured to solve for \( \rho_{r} \): \[ \rho_r = \frac{F_g}{F_g - F_b} = \frac{16}{2} = 8 \] g/cm鲁. Hence, the relative density is 8 g/cm鲁.

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Key Concepts

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

Buoyant Force
When an object is submerged in a fluid like water, it experiences an upward force known as buoyant force. This magical force is what makes you feel lighter when swimming! Archimedes discovered that this force equals the weight of the fluid displaced by the object. That's why ships stay afloat and why sometimes heavy objects appear lighter when underwater. In the given exercise, when a steel weight is immersed in water, it experiences a buoyant force that lightens the load on the wire, causing the wire to stretch less compared to when it is in air. You can think of it as nature's way of helping hold up the weight! This reduction in the wire's extension, from 16 cm to 14 cm, represents the effect of the buoyant force reducing the weight of the object.
Gravity Force
Gravity force, also known as gravitational force, is the force with which the Earth pulls you and everything else towards its center. It is the reason why we stay grounded and why dropped objects fall. In the context of the exercise, before the weight is immersed, its gravitational force is wholly responsible for the extension of the wire by 16 cm. This is the force we calculate with the formula \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. It's like the invisible hand that pulls things down and, in this exercise, is responsible for the wire extending as much as it does when the weight hangs in the air.
Force Constant
The force constant, often denoted as \( k \), is a measure of a material's stiffness. It's part of Hooke's Law, which states that the extension of a spring or wire is directly proportional to the force applied to it, as long as the deformation is within the elastic limit. In this exercise, the force constant of the wire tells us how much force is needed for each unit of extension. The equation \( F = k \, x \) helps us understand the relationship between force and extension, where \( F \) is the force applied, \( k \) is the force constant, and \( x \) is the extension. Thus, by knowing the force constant, we can see how the same force results in different extensions, such as the 16 cm in air and 14 cm in water, indicating how the buoyant force affects the measures due to the relative elasticity or stiffness of the wire.

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Most popular questions from this chapter

A steel wire of length \(4.7 \mathrm{~m}\) and cross-sectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) stretches by the same amount as a copper wire of length \(3.5 \mathrm{~m}\) and cross-sectional area of \(4 \times 10^{-6} \mathrm{~m}^{2}\) under a given load. The ratio of Young's modulus of steel to that of copper is (1) \(1.8\) (2) \(3.6\) (3) \(0.6\) (4) \(8.7\)

A solid sphere of radius \(R\), made up of a material of bulk modulus \(K\) is surrounded by a liquid in a cylindrical container. A massless piston of area \(A\) floats on the surface of the liquid. When a mass \(M\) is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is (1) \(\frac{M g}{A K}\) (2) \(\frac{M g}{3 A K}\) (3) \(\frac{3 M g}{A K}\) (4) \(\frac{M g}{2 A K}\)

A rod is made of uniform material and has non-uniform cross section. It is fixed at both the ends as shown and heated at mid-section. Which of the following statements are not correct? (1) Force of compression in the rod will be maximum at mid-section. (2) Compressive stress in the rod will be maximum at left end. (3) Since rod in fixed at both the ends, its length will remain unchanged. Hence, no strain will be induced in it. (4) None of the above.

Two wires of the same length and same material but radil in the ratio of \(1: 2\) are stretched by unequal forces to produce equal elongation. The ratio of the two forces is (1) \(1: 1\) (2) \(1: 2\) (3) \(1: 3\) (4) \(1: 4\)

The ratio of diameters of two wires of same material is \(n: 1\). The length of each wire is \(4 \mathrm{~m}\). On applying the same load, the increase in the length of the thin wire will be \((n>1)\) (1) \(n^{2}\) times (2) \(n\) times (3) \(2 n\) times (4) \((2 n+1)\) times
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