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Chapter 4: Problem 77

A bird is sitting on stretched telephone wires. If its weight is \(W\) then the additional tension produced by it in the wires will be (a) \(T=W\) (b) \(T>W\) (c) \(T

Short Answer

Expert verified
The answer is (a) \(T = W\). The additional tension produced by the bird in the wires is equal to its weight, because the tension needs to counteract the bird's weight to maintain equilibrium.

Step by step solution

01

Understand the forces at play

A bird sitting on stretched telephone wires applies a downward force to the wire. This force is the bird's weight (W), which is the force of gravity acting on it. The wire resists this weight by creating an upward force, that is tension (T), to balance out the downward force applied by the weight of the bird.
02

Analyze the tension in the wire

The wire must produce an amount of tension that is equal to the bird's weight to counteract the weight of the bird so that there is no net vertical force on the system i.e., the system is in equilibrium.
03

Answer the exercise

You can now say that the additional tension (T) produced in the wires by the weight of the bird (W) is equal to the weight of the bird. Therefore, (T = W).

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Key Concepts

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

Equilibrium in Mechanics
In mechanics, equilibrium occurs when all forces acting on a system are balanced. This means there is no net force, so the system remains steady without accelerating in any direction.
When you consider a bird sitting on a wire, the wire system must be balanced. For equilibrium, the upward force (tension) provided by the wire needs to match the downward force (weight) of the bird.
If these forces are balanced, the wire does not move, sag excessively, or change its tension unnecessarily. Therefore, achieving equilibrium in such scenarios ensures stability.
  • Equilibrium is crucial in everyday situations, like buildings remaining stable, vehicles maintaining balance, or bridges holding their structural integrity.
  • It is important for safety and functionality in engineering and physics.
    • Force of Gravity
    Gravity is the force that attracts two bodies towards each other. On Earth, this force is generally what keeps us grounded.
    In physics, when dealing with objects on Earth, gravity usually pulls them towards the center of the Earth. This force is proportional to the mass of the object.
    For the bird on the wire, gravity pulls it downward with a force equal to its weight, which can be expressed mathematically as:
    \( W = mg \)
    where \(m\) is the mass of the bird and \(g\) is the acceleration due to gravity (approximately \(9.8\, ext{m/s}^2\)).
    Gravity is a constant force we often consider in problems related to weight, tension, and other force-related concepts.
    Tension in Wires
    Tension is a pulling force transmitted through a string, cable, or wire when it is pulled tightly by forces from opposite ends.
    In the case of a bird on a telephone wire, the tension in the wire is what balances the bird's weight.
    When the bird sits on the wire, it creates a downward force due to its weight. The wire responds by exerting an upward force, known as tension, to support the bird and maintain balance.
  • The tension in the wire must counteract the force of gravity to keep the wire in equilibrium.
  • Tension increases with the weight of the object 鈥� in this case, the bird 鈥� and the angle of the wire relative to the horizontal.
  • Correct tension ensures that wires, ropes, or cables do not sag or snap under load.
    • Tension is crucial in applications ranging from construction to physics experiments, providing stability and safety.

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    Most popular questions from this chapter

    A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be \(W_{1}\). If the bird flies about inside the cage, the reading of the spring balance is \(W_{2}\). Which of the following is true (a) \(W_{1}=W_{2}\) (b) \(W_{1}>W_{2}\) (c) \(W_{1}

    A force of 50 dynes is acted on a body of mass \(5 g\) which is at rest for an interval of 3 seconds, then impulse is (a) \(0.15 \times 10^{-3} N-s\) (b) \(0.98 \times 10^{-3} N-s\) (c) \(1.5 \times 10^{-3} \mathrm{~N}-\mathrm{s}\) (d) \(2.5 \times 10^{-3} \mathrm{~N}-\mathrm{s}\)

    A monkey of mass \(20 \mathrm{~kg}\) is holding a vertical rope. The rope will not break when a mass of \(25 \mathrm{~kg}\) is suspended from it but will break if the mass exceeds \(25 \mathrm{~kg}\). What is the maximum acceleration with which the monkey can climb up along the rope \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(25 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(5 \mathrm{~m} / \mathrm{s}^{2}\)

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