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Chapter 8: Problem 16

One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof, and a weight \(W_{1}\) is suspended from its lower end. If \(S\) is the area of cross-section of the wire, the stress in the wire at a height ( \(3 L / 4\) ) from its lower end is (A) \(W_{1} / S\) (B) \(\left[W_{1}+(W / 4)\right] / S\) (C) \(\left[W_{1}+(3 W / 4)\right] / S\) (D) \(\left[W_{1}+W\right] / S\)

Short Answer

Expert verified
The stress in the wire at a height (3L/4) from its lower end is (B) \(\left[W_{1}+(W / 4)\right] / S\).

Step by step solution

01

Understand the forces acting on the wire

There are two forces acting on the wire: the gravitational force due to the wire itself (W) and the downward force resulting from the weight W1. Since the wire is uniform, the gravitational force W is evenly distributed along its length.
02

Find the portion of the wire's weight acting above the point (3L/4)

To find the portion of the wire's weight acting above the point (3L/4), we can consider the wire as split into two sections: the portion above the point and the portion below the point. As the wire is uniform, the weight above the point is proportional to the length above the point. The upper section is L - (3L/4) = L/4. Since the total weight of the wire is W, the weight of the upper portion is (W * (L/4)) / L = W/4.
03

Determine the total force at the point (3L/4)

The total force at the point (3L/4) is the sum of the forces acting on the wire above this point. This includes the force due to the weight W1 and the force due to the portion of the wire's weight above this point (W/4). Total force = W1 + (W/4)
04

Divide the total force by the area of cross-section

Finally, to find the stress in the wire at the height (3L/4), we need to divide the total force by the area of cross-section, S. Stress = (W1 + (W/4)) / S So, the correct answer is (B) \(\left[W_{1}+(W / 4)\right] / S\).

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Key Concepts

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

Understanding Uniform Wire
A uniform wire is a type of wire where both its density and cross-sectional area are consistent along its entire length. This means that the wire's weight is distributed evenly throughout its length.

  • Uniform wires have consistent physical and mechanical properties regardless of the section.
  • This consistency allows for predictable behavior when forces, such as tension or gravity, act on the wire.
  • Knowing that the wire is uniform simplifies the calculation of forces acting upon different sections.
For example, if a wire has a total weight of \( W \), and you want to know the weight of a specific portion of the wire, you can calculate this by looking at the proportional length of that portion. In our problem, this understanding of uniformity helps us determine how the wire's weight contributes to the stress calculated at a specific point, namely \( \frac{3L}{4} \) up from its lower end.
Cross-Section Area and Its Relevance
The cross-section area \( S \) of a wire refers to the thickness or the area of the slice perpendicular to its length. This property is crucial in understanding how stress distributes across the material.

  • Stress is defined as the force acting on an object divided by the cross-section area. It is measured in units of pressure (e.g., Pascals).
  • A larger cross-section area means that the same amount of force results in less stress on the material.
  • On the other hand, a smaller cross-section area results in higher stress for the same force.
In the context of the problem, the cross-section area \( S \) ensures that we can understand how the force, consisting of the suspended weight \( W_1 \) and the partial wire weight \( W/4 \), translates into stress at the calculated height within the wire.
Impact of Gravitational Force
Gravitational force is an essential concept in understanding mechanical stress, particularly when dealing with wires and weights. Gravitational force acts to pull objects towards the center of the Earth, creating what we perceive as weight.

  • In the case of the wire, the gravitational force is represented by the wire's weight \( W \).
  • When an additional weight \( W_1 \) is suspended from a wire, this adds additional gravitational force at the lower end.
  • The total gravitational force at any point is the sum of the forces from all sections above that point.
Thus, in the exercise, the gravitational force from the wire itself contributes just a portion of the total force at point \( \frac{3L}{4} \). Together with the suspended weight, it determines the stress experienced by the wire at that height.

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Most popular questions from this chapter

A uniform rod of length \(L\) has a mass per unit length \(\lambda\) and area of cross-section \(A .\) The elongation in the rod is \(l\) due to its own weight if it is suspended from the ceiling of a room. The Young's modulus of the rod is (A) \(\frac{2 \lambda g L^{2}}{A l}\) (B) \(\frac{\lambda g L^{2}}{2 A l}\) (C) \(\frac{2 \lambda g L}{A l}\) (D) \(\frac{\lambda g l^{2}}{A L}\)

The volume of a liquid flowing per sec out of an orifice at the bottom of a tank does not depend upon (A) the height of the liquid above the orifice. (B) the acceleration due to gravity. (C) the density of the liquid. (D) the area of the orifice.

A spherical balls of a radius \(R\) is falling in a viscous fluid of viscosity \(\eta\) with a velocity \(v\). The retarding viscous force acting on the spherical ball is [2004](A) directly proportional to \(R\) but inversely proportional to \(v\). (B) directly proportional to both radius \(R\) and velocity \(v\). (C) inversely proportional to both radius \(R\) and velocity \(v\). (D) inversely proportional to \(\mathrm{R}\) but directly proportional to velocity \(\underline{v}\)

On putting a capillary tube in a pot filled with water, the level of water rises up to a height of \(4 \mathrm{~cm}\) in the tube. If a tube of half the diameter is used, the water will rise to the height of nearly (A) \(2 \mathrm{~cm}\) (B) \(5 \mathrm{~cm}\) (C) \(8 \mathrm{~cm}\) (D) \(11 \mathrm{~cm}\)

A metal wire of length \(L\), area of cross-section \(\mathrm{A}\), and Young's modulus \(Y\) is stretched by a variable force \(F\) such that \(F\) is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is \(\ell\) (A) The work done by \(F\) is \(\frac{Y A \ell^{2}}{2 L}\) (B) The work done by \(F\) is \(\frac{Y A \ell^{2}}{L}\) (C) The elastic potential energy stored in the wire is \(\frac{Y A \ell^{2}}{2 L}\) (D) No heat is produced during the elongation
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