/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 122 A box of weight w is accelerated... [FREE SOLUTION] | 91Ó°ÊÓ

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A box of weight w is accelerated up a ramp by a rope that exerts a tension \(T .\) The ramp makes an angle \(\alpha\) with the horizontal, and the rope makes an angle \(\theta\) above the ramp. The coefficient of kinetic friction between the box and the ramp is \(\mu_{k}\) . Show that no matter what the value of \(\alpha\) , the acceleration is maximum if \(\theta=\arctan \mu_{k}\) (as long as the box remains in contact with the ramp).

Short Answer

Expert verified
The acceleration is maximized when \( \theta = \arctan \mu_{k} \).

Step by step solution

01

Analyzing the forces acting on the box

Start by identifying the forces acting on the box. The forces include the weight of the box, the normal force, the tension in the rope, and the frictional force. The gravitational force acting on the box is given by \( w = mg \), where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity. The tension \( T \) acts at an angle \( \theta \) above the ramp, and its components along the ramp and perpendicular to the ramp can be calculated using trigonometric functions. The normal force is perpendicular to the ramp's surface. The kinetic frictional force is given by \( F_k = \mu_k N \), where \( N \) is the normal force.
02

Setting up the equations of motion

Break the forces into components parallel and perpendicular to the ramp. The force components parallel to the ramp are: \( T \cos \theta - w \sin \alpha - \mu_k N \) and the perpendicular components are: \( N - w \cos \alpha - T \sin \theta = 0 \). Using Newton's second law, the equation for motion along the ramp is \( ma = T \cos \theta - w \sin \alpha - \mu_k N \). For the perpendicular direction, solve for the normal force: \( N = w \cos \alpha + T \sin \theta \).
03

Substituting normal force and solving for acceleration

Substitute \( N = w \cos \alpha + T \sin \theta \) into the equation for parallel components. The new expression is \( ma = T \cos \theta - w \sin \alpha - \mu_k (w \cos \alpha + T \sin \theta) \). Simplify to find the acceleration \( a \): \( a = \frac{T \cos \theta - \mu_k T \sin \theta - w(\sin \alpha + \mu_k \cos \alpha)}{m} \).
04

Optimizing the angle for maximum acceleration

To find the angle \( \theta \) that maximizes acceleration \( a \), take the derivative of \( a \) with respect to \( \theta \) and set it to zero: \( \frac{da}{d\theta} = -T(\sin \theta + \mu_k \cos \theta) = 0 \). Solve for \( \theta \): \( \tan \theta = \mu_k \). Therefore, \( \theta = \arctan \mu_k \). This shows that the acceleration is maximized when the angle \( \theta \) is equal to the arctangent of the coefficient of kinetic friction.

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Key Concepts

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

Forces on Inclined Planes
Imagine you have a box and you place it on a ramp. Forces naturally come into play. Understanding these forces is vital in comprehending how objects move on inclined planes.

There are several forces at work here:
  • The **weight** of the box acts downward due to gravity. If the weight of the box is given by \( w = mg \), where \( m \) is the mass and \( g \) is the gravitational acceleration (approximately \(9.8 \text{ m/s}^2\) on Earth).
  • The **normal force** pushes perpendicular to the ramp’s surface, balancing the component of gravitational force acting perpendicular to the incline.
  • **Tension** in the rope is applied at an angle \( \theta \) above the ramp, helping to pull the box upwards.
  • Finally, there is the force of **kinetic friction** which opposes the motion, acting down the ramp.
The interaction between these forces determines how the box accelerates or moves along the plane.
Kinetic Friction
Kinetic friction comes into play whenever there is movement between surfaces. It resists the sliding motion of two surfaces in contact.

For the box on the ramp, kinetic friction is a crucial player because it opposes the upward pull of tension. It largely depends on:
  • The **normal force**: This is the net force acting perpendicular to the surfaces in contact.
  • The **coefficient of kinetic friction** \( \mu_{k} \): A measure of how much frictional resistance exists between the two surfaces.
Kinetic frictional force can be described by the equation \( F_k = \mu_k N \). This means that if the normal force or the roughness (indicated by \( \mu_{k} \)) of the surfaces changes, so does the kinetic friction.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics that ties together the net force acting on an object, its mass, and its acceleration. According to this law, the acceleration \( a \) of an object is directly proportional to the net external force acting upon it and inversely proportional to its mass. The law can be written as:

\[ F_{ ext{net}} = ma \]
In the context of the box on the incline, we use this law to analyze the box’s motion. We set up equations to represent the forces in both the parallel and perpendicular directions of the incline. In doing so, we can solve for the box's acceleration using:
  • **Components of tension** along the incline.
  • **Weight component** acting downwards the ramp.
  • **Frictional force** opposing the tension force.
Using all these forces, we apply Newton's Second Law to find the net force along the direction of motion and thus determine the box's acceleration.
Trigonometric Components in Physics
When dealing with forces on an inclined plane, trigonometry plays a vital role in breaking down forces into components. Imagine the tension in the rope on the ramp separating into different directions, while mathematics helps us manage these directions smoothly.

For each force like tension or weight, there are often components:
  • **Parallel to the incline**: Helps or opposes motion along the ramp.
  • **Perpendicular to the incline**: Interacts with the normal force.
A common practice is using sine (\( \sin \)) and cosine (\( \cos \)) functions:
  • For example, if tension \( T \) makes an angle \( \theta \) with the incline, the parallel component is \( T\cos(\theta) \) while the perpendicular component is \( T\sin(\theta) \).
  • Similarly, the gravitational force \( mg \) is split using the angle of the incline \( \alpha \). The parallel component is given by \( mg\sin(\alpha) \), opposing the box’s upward motion, and the perpendicular component is \( mg\cos(\alpha) \), contributing to the normal force.
Through these trigonometric relationships, we can understand and solve complex problems involving forces in multiple directions.

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Most popular questions from this chapter

You are working for a shipping company. Your job is to stand at the bottom of a 8.0 -m-long ramp that is inclined at \(37^{\circ}\) above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is \(\mu_{\mathbf{X}}=0.30\) . (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you?

Losing Cargo. A \(12.0-\mathrm{kg}\) box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{s}=0.19\) and \(\mu_{k}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

A 750.0-kg boulder is raised from a quarry 125 \(\mathrm{m}\) deep by a long uniform chain having a mass of 575 \(\mathrm{kg}\) . This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take to be lifted out at maximum acceleration if it started from rest?

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end (Fig. 5.57\()\) . Each arm supports a seat suspended from a cable 5.00 \(\mathrm{m}\) long, the upper end of the cable being fastened to the arm at a point 3.00 \(\mathrm{m}\) from the central shaft. (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of \(30.0^{\circ}\) with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution? figure can't copy

A rock with mass \(m=3.00 \mathrm{kg}\) falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v,\) where \(v\) is the speed in \(\mathrm{m} \mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\) (see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0} \cdot\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{t}\) (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed 0.9\(v_{\mathrm{t}}\)

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