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Chapter 4: Problem 64

A wagon is being pulled at constant speed up a slope \(\theta\) by a rope that makes an angle \(\phi\) with the vertical. (a) Assuming negligible friction, show that the tension in the rope is given by the equation $$ T=\frac{\sin \theta}{\sin (\theta+\phi)} m g $$ (b) Interpret this equation in the special cases of \(\phi=0\) and \(\phi=180^{\circ}-\theta\).(solution in the pdf version of the book)

Short Answer

Expert verified
(a) The tension is \( T = \frac{\sin \theta}{\sin(\theta + \phi)} mg \). (b) For \( \phi = 0 \), \( T = mg \); for \( \phi = 180^\circ - \theta \), tension becomes undefined.

Step by step solution

01

Understanding the Problem

We have a wagon being pulled up a slope at angle \( \theta \) by a rope inclined at angle \( \phi \) to the vertical. We need to find the tension, \( T \), in the rope assuming there is no friction.
02

Identifying Forces

The forces acting on the wagon are its weight, \( mg \), and the tension in the rope, \( T \). The weight of the wagon can be resolved into two components: one parallel to the slope, \( mg \sin \theta \), and one perpendicular to the slope, \( mg \cos \theta \).
03

Analyzing Tension

Since the wagon moves with constant speed, the net force along the slope must be zero. Therefore, the tension in the rope must balance the component of gravitational force along the slope: \( T \cos(\theta + \phi) = mg \sin \theta \).
04

Solving for Tension

Rearrange the equation \( T \cos(\theta + \phi) = mg \sin \theta \) to find \( T \). Dividing both sides by \( \cos(\theta + \phi) \), we have: \[ T = \frac{mg \sin \theta}{\cos(\theta + \phi)}. \] Since \( \cos(\theta + \phi) = \sin(90^\circ - (\theta + \phi)) = \sin(\theta + \phi) \) assuming angles between \( 0\degree \) to \( 90\degree \), we get: \[ T = \frac{\sin \theta}{\sin(\theta + \phi)} mg. \]
05

Special Case \( \phi = 0 \)

If \( \phi = 0 \), the rope is vertical. Substituting this into the tension formula, we get: \( T = \frac{\sin \theta}{\sin \theta} mg = mg \). This means the tension equals the gravitational force.
06

Special Case \( \phi = 180^\circ - \theta \)

If \( \phi = 180^\circ - \theta \), the angles add up to \( 180^\circ \), so \( \sin(\theta + \phi) = 0 \). Since z cannot be divided by zero this case creates an undefined tension, indicating it is not possible to pull with these angles while maintaining equilibrium.

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Key Concepts

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

Understanding Tension
Tension is the pulling force exerted by a string, cable, or, in this case, a rope. When a wagon is pulled, the rope becomes tight, and this tightness represents the tension. Key points to remember include:
  • Tension always pulls along its length and away from the object.
  • It is measured in newtons (N).
  • In problems like the wagon on the slope, tension helps counteract gravity and facilitate movement.
Understanding tension is crucial in mechanics, as it helps balance forces and maintain motion or equilibrium in systems like the one in our problem.
Each situation that involves tension might require different calculations based on angles or incline angles. In this exercise, we work with tension by establishing its relationship with other forces like gravity and friction (when not negligible).
Inclined Plane Dynamics
An inclined plane is a flat surface tilted at an angle, relative to the horizontal. It's a simple machine that makes work easier by allowing objects to be moved up or down with less force over a greater distance. In our context, the incline helps to better illustrate forces at play when the wagon is pulled:
  • The angle of the slope, \( \theta \), influences how much of the gravitational force acts along the incline.
  • As gravity acts on the wagon, its weight can be divided into components: parallel and perpendicular to the inclined plane.
  • The angle \( \phi \), which the rope makes with the vertical, further breaks down the tension force into parts that effectively balance out these components of gravity.
Even without factoring in friction, analyzing an inclined plane is vital because it helps visualize how forces and their components are directed and balanced.
Thus, understanding inclines allows us to simplify complex forces into more manageable calculations.
Forces Analysis in Mechanics
The crux of mechanics lies in breaking down forces, as we do in our wagon and rope problem. Forces analysis involves understanding different forces acting on a body in mechanics and how they interact:
  • Weight (mg), which acts downward due to gravity, can be broken into components. For instance, one parallel (m g \sin(\theta)) and one perpendicular (m g \cos(\theta)) to the incline.
  • Tension in the rope acts in another direction, at angle \( \phi \) to the vertical, providing a component that acts against gravity's pull along the slope.
  • Balanced forces in equilibrium: here, despite forces acting on the wagon, it moves at constant speed, indicating a balance between forward (or upward) tension and downward weight component.
Learning force analysis allows students to dissect and comprehend why objects move or stay still. It informs students how to use equations like those for tension or force components to predict and explain real-world physics scenarios. Analyzing forces aids in grasping the essence of equilibrium and steady movement.
Using Geometry in Physics
In physics, geometry is essential to resolving forces and understanding angles, especially in problems with inclines and ropes. Geometry steps in when:
  • Determining how to split forces into components: trigonometric functions like \(\sin\), \(\cos\), and \(\tan\) are vital here, especially in dealing with inclined planes.
  • Angles like \(\theta\) and \(\phi\) are fundamental in calculating how forces will interact. For example, clarity on whether angles are measured from vertical or horizontal impacts calculations.
    This understanding helps in simplifying the interpretation of tension or weight components, as seen in the formula \(T=\frac{\sin \theta}{\sin (\theta+\phi)} m g\) derived by resolving components coinciding with the geometry of the scenario.
Using geometry, physics students can make sense of abstract forces through visualization and diagrammatic representation, essential for solving multi-stranded physics problems.

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Most popular questions from this chapter

A gun is aimed horizontally to the west. The gun is fired, and the bullet leaves the muzzle at \(t=0\). The bullet's position vector as a function of time is \(\mathbf{r}=b \hat{\mathbf{x}}+c t \hat{\mathbf{y}}+d t^{2} \hat{\mathbf{z}}\), where \(b, c\), and \(d\) are positive constants. (a) What units would \(b, c\), and \(d\) need to have for the equation to make sense? (b) Find the bullet's velocity and acceleration as functions of time. (c) Give physical interpretations of \(b, c, d, \hat{\mathbf{x}}, \hat{\mathbf{y}}\), and \(\hat{\mathbf{z}}\).

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