/*! 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 81 A 205-kg log is pulled up a ramp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 81

A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at \(30.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900 , and the log has an acceleration of magnitude \(0.800 \mathrm{m} / \mathrm{s}^{2}\). Find the tension in the rope.

Short Answer

Expert verified
The tension in the rope is approximately 2734.23 N.

Step by step solution

01

Identify Forces

First, identify the forces acting on the log: gravitational force, normal force, frictional force, and tension in the rope. The gravitational force can be split into two components: parallel and perpendicular to the ramp.
02

Calculate Gravitational Force Components

The gravitational force parallel to the ramp is given by:\[ F_{\text{gravity, parallel}} = mg \sin(\theta) \]\[ = 205 \times 9.8 \times \sin(30^{\circ}) \]\[ = 1004.5 \, \text{N} \]The perpendicular component is:\[ F_{\text{gravity, perpendicular}} = mg \cos(\theta) \]\[ = 205 \times 9.8 \times \cos(30^{\circ}) \]\[ = 1739.7 \, \text{N} \]
03

Determine the Normal Force

The normal force \( F_{\text{normal}} \) is equal to the gravitational force component perpendicular to the ramp:\[ F_{\text{normal}} = 1739.7 \, \text{N} \]
04

Calculate Frictional Force

The frictional force \( F_{\text{friction}} \) can be calculated using the coefficient of kinetic friction:\[ F_{\text{friction}} = \mu_k F_{\text{normal}} \]\[ = 0.9 \times 1739.7 \]\[ = 1565.73 \, \text{N} \]
05

Apply Newton's Second Law

Apply Newton's second law along the ramp to find the tension \( T \):\[ F_{\text{net}} = T - F_{\text{gravity, parallel}} - F_{\text{friction}} \]We know that \( F_{\text{net}} = ma \):\[ ma = T - mg \sin(\theta) - \mu_k mg \cos(\theta) \]\[ 205 \times 0.8 = T - 1004.5 - 1565.73 \]
06

Solve for the Tension

Solve the equation from the previous step for \( T \):\[ 164 = T - 2570.23 \]\[ T = 2570.23 + 164 \]\[ T = 2734.23 \, \text{N} \]
07

Final Answer

The tension in the rope is approximately \( 2734.23 \, \text{N} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the movement of two surfaces sliding past each other. It is crucial in this exercise as the log is being pulled up a ramp. The coefficient of kinetic friction, 渭k, represents how much frictional force acts between the surfaces. Here, it鈥檚 given as 0.9.
To calculate the kinetic frictional force, use the equation:
  • \( F_{\text{friction}} = \mu_k \cdot F_{\text{normal}} \)
This means the frictional force depends not only on the coefficient but also on the normal force, which is the perpendicular force acting on the log.
In this case, the log's frictional force is calculated to be approximately 1565.73 N, impeding its upward motion on the ramp.
Gravitational Force Components
Gravitational force can be split into components when dealing with inclined surfaces like ramps. The full gravitational force, acting straight down, is given by:
  • \( F_{\text{gravity}} = mg \)
However, since the ramp is inclined, we decompose this force into two parts: parallel and perpendicular to the ramp.
  • Parallel component: \( F_{\text{gravity, parallel}} = mg \sin(\theta) \)
  • Perpendicular component: \( F_{\text{gravity, perpendicular}} = mg \cos(\theta) \)
The angle \( \theta \) is vital as it determines the distribution of gravitational force.
In the example, with a log of 205 kg and an angle of 30掳, the parallel component is 1004.5 N, and the perpendicular component is 1739.7 N, affecting how the log will slide down or stay on the ramp.
Newton's Second Law
Newton鈥檚 second law is fundamental in understanding the dynamics of objects. It relates the net force acting on an object to its mass and acceleration:
  • \( F_{\text{net}} = ma \)
This principle helps determine the tension in the rope pulling the log. The net force is calculated based on all existing forces, including gravitational, frictional, and the rope's tension.
In this case, net force along the ramp can be set as:
  • \( T - F_{\text{gravity, parallel}} - F_{\text{friction}} = ma \)
Given the acceleration of the log is 0.8 m/s2, this equation allows us to solve for the tension, ultimately resulting in approximately 2734.23 N.
Normal Force
The normal force is an important concept, especially when dealing with inclines. It acts perpendicular to the contact surface, countering the gravitational pull. In this setup, it aids in calculating the frictional force.
  • \( F_{\text{normal}} = F_{\text{gravity, perpendicular}} \)
For the log on the ramp, this means the perpendicular component of gravitational force 鈥� 1739.7 N 鈥� is also the normal force.
This force is vital as it directly affects the frictional force through the coefficient of kinetic friction, being crucial for determining resistance against the movement up the ramp.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A space traveler weighs 540.0 N on earth. What will the traveler weigh on another planet whose radius is twice that of earth and whose mass is three times that of earth?

Mars has a mass of \(6.46 \times 10^{23} \mathrm{kg}\) and a radius of \(3.39 \times 10^{6} \mathrm{m} .\) What is the acceleration due to gravity on Mars? (b) How much would a \(65-\)kg person weigh on this planet?

Two forces \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) are applied to an object whose mass is \(8.0 \mathrm{kg} .\) The larger force is \(\overrightarrow{\mathbf{F}}_{\mathrm{A}} .\) When both forces point due east, the object's acceleration has a magnitude of \(0.50 \mathrm{m} / \mathrm{s}^{2} .\) However, when \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) points due east and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) points due west, the acceleration is \(0.40 \mathrm{m} / \mathrm{s}^{2},\) due east. Find \((\mathrm{a})\) the magnitude of \(\overrightarrow{\mathbf{F}}_{\mathbf{A}}\) and (b) the magnitude of \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\)

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. Find the tension in the cable.

Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical down-force, which helps the cars hug the track more securely. The coefficient of static friction between the track and the tires of a 690-kg race car is 0.87. What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a 4060-N downforce and an 1190-N horizontal-air- resistance force act on it?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Particle Model of Matter

Read Explanation

Solid State Physics

Read Explanation

Physics of Motion

Read Explanation

Nuclear Physics

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.