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Chapter 9: Problem 6

A homogeneous rope with mass \(m\) hangs between two equally high poles. The angle between the rope and the horizontal at each of the attachment points is \(\alpha\). (a) Find the tension at each end of the rope. (b) Find the tension at the lowest point of the rope. (c) Is it possible to tighten the rope so much that \(\alpha=0\) ?

Short Answer

Expert verified
(a) \( \frac{mg}{2 \sin \alpha} \), (b) \( \frac{mg}{2 \tan \alpha} \), (c) Not possible

Step by step solution

01

Understand the Problem

A uniform rope is hanging between two poles with an angle \( \alpha \) at the attachment points. Need to find the tensions at the ends and the lowest point, and determine if the rope can be tightened to make \( \alpha = 0 \).
02

Draw a Free Body Diagram

Draw the rope with forces acting on it. Tensions at each end act upward/diagonally and gravity acts downward on the rope's center of mass.
03

Resolve Forces Horizontally (Part a)

In equilibrium, the horizontal components of the tensions must balance. Let T be the tension at the ends. Then: \[ T \cos \alpha = T \cos \alpha \]
04

Resolve Forces Vertically (Part a)

The sum of vertical components should equal the weight of the rope \(mg\). This gives: \[ 2T \sin \alpha = mg \]
05

Solve for Tension at Ends (Part a)

Solve for T using the previous equation: \[ T = \frac{mg}{2 \sin \alpha} \]
06

Determine Tension at Lowest Point (Part b)

At the lowest point, the tension is solely horizontal because the vertical forces cancel out. Using equilibrium, the horizontal component T\(cos \alpha\): \[ T_{lowest} = T \cos \alpha = \frac{mg}{2 \sin \alpha} \cos \alpha = \frac{mg}{2 \tan \alpha} \]
07

Consider \( \alpha = 0 \) Case (Part c)

As \( \alpha \) approaches 0, \( \tan \alpha \) approaches 0, which would imply infinite tension. Thus, \( \alpha = 0 \) is not possible.

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Key Concepts

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

mechanics
Mechanics is a branch of physics that deals with the motion and forces acting on objects. In this problem, we're dealing specifically with the mechanics of a hanging rope. A rope's behavior is influenced by its weight, tension, and the forces acting on it. When we talk about a ‘homogeneous’ rope, we mean that its mass is evenly distributed throughout its length. Gravity pulls the rope downward, creating a force that must be balanced by the tension at the points where the rope is attached. Understanding these forces and their interactions is crucial for solving problems in mechanics.
equilibrium of forces
An object is in equilibrium when all the forces acting upon it are balanced. In the case of the hanging rope, equilibrium means the sum of all vertical forces must be zero, and the sum of all horizontal forces must also be zero. To achieve this:
  • Vertically, the upward tension forces at the ends of the rope must balance the downward gravitational force (weight) of the rope.
  • Horizontally, the horizontal components of the tension at each end must cancel each other out.
This balance is what keeps the rope stationary and hanging in a stable position. The key idea here is that in equilibrium, there's no net force causing the rope to move.
free body diagram
A Free Body Diagram (FBD) is a simple sketch that shows an object and all the forces acting on it. Drawing an FBD helps us visualize and analyze the problem more clearly. For the hanging rope:
  • Draw the rope suspended between two poles.
  • Identify the points where the rope is attached and draw arrows representing tension forces acting upwards and diagonally.
  • Show the weight of the rope as a force acting downwards, concentrated at the rope's center of mass.
The FBD helps to set up the equations for equilibrium and resolve the forces into their horizontal and vertical components. This visualization tool is vital for solving mechanics problems accurately.
trigonometry in physics
Trigonometry is the study of the relationships between the angles and sides of triangles. In physics, it's often used to break down forces into components. For the hanging rope problem:
  • We use the angle \(\alpha\) to find the horizontal and vertical components of the tension force.
  • The horizontal component is \( \cos \alpha\), which relates to how much of the tension force acts horizontally.
  • The vertical component is \(\sin \alpha\), which relates to how much of the tension force acts vertically.
By applying trigonometric identities and equations, we can solve for the unknown forces. This approach is essential for problems where forces are not aligned with the standard axes and need to be resolved into components to analyze the equilibrium.

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Most popular questions from this chapter

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