/*! 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 103 To hoist himself into a tree, a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 103

To hoist himself into a tree, a 72.0 -kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of \(358 \mathrm{~N}\). Neglect any friction between the rope and the branch, and determine the man's upward acceleration.

Short Answer

Expert verified
The man's upward acceleration is approximately 0.144 m/s虏.

Step by step solution

01

Understand the Force Components

First, identify the forces involved. The man is pulling down with a force of 358 N, and his weight is acting downward, which is calculated as the gravitational force: \( F_{gravity} = m \times g = 72.0\, \text{kg} \times 9.8\, \text{m/s}^2 \).
02

Calculate the Gravitational Force

Calculate the gravitational force acting on the man:\[F_{gravity} = 72.0\, \text{kg} \times 9.8\, \text{m/s}^2 = 705.6\, \text{N}\]
03

Apply Newton's Second Law of Motion

According to Newton's Second Law, the net force acting on the man is equal to the mass times his acceleration:\[ F_{net} = m \cdot a \]The net force has two components: the force the man exerts and the gravitational force, both acting on him. Since he is pulling the rope and lifting himself, it also adds to the upward direction. Therefore, \[ 2 imes 358\, \text{N} - 705.6\, \text{N} = 72.0\, \text{kg} \cdot a \].
04

Solve for the Acceleration

Rearrange the equation to solve for the acceleration:\[a = \frac{2 \times 358\, \text{N} - 705.6\, \text{N}}{72.0\, \text{kg}}\]Calculate the value of \( a \):\[a = \frac{716\, \text{N} - 705.6\, \text{N}}{72.0\, \text{kg}} = \frac{10.4\, \text{N}}{72.0\, \text{kg}} \approx 0.144\, \text{m/s}^2\]
05

State the Final Result

The upward acceleration of the man is approximately \(0.144\, \text{m/s}^2\).

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.

Gravitational Force
Let's start by understanding the concept of gravitational force. Gravitational force is the pull that the Earth exerts on objects, bringing them toward its center. We calculate this force by using the formula:
  1. The mass of the object, denoted as \( m \).
  2. The acceleration due to gravity, denoted as \( g \), which is approximately \( 9.8 \text{ m/s}^2 \) on Earth.
The formula is simple: \[ F_{gravity} = m \times g \]For our exercise, the man has a mass of \( 72.0 \text{ kg} \), so the gravitational force is:\[ F_{gravity} = 72.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 705.6 \text{ N} \]This is the force pulling the man downwards due to gravity.
Net Force
Next, we'll look at net force, which is the combined effect of all forces acting on an object. According to Newton's Second Law of Motion, the net force determines the object's acceleration.
In our situation, several forces act on the man:
  • The gravitational force pulling him downward (\( 705.6 \text{ N} \)).
  • The force exerted by the man pulling the rope (\( 358 \text{ N} \)), which we count twice as it affects both ends of the rope.
To find the net force:
1. Calculate the total upward force: Since the rope goes over a branch, the man effectively applies his pulling force twice, making the upward force \( 2 \times 358 \text{ N} = 716 \text{ N} \).
2. Apply Newton's Second Law: Subtract the gravitational force from this to get the net force:\[ F_{net} = 716 \text{ N} - 705.6 \text{ N} = 10.4 \text{ N} \]This net force results in the man's upward motion.
Upward Acceleration
Finally, let's talk about upward acceleration, which is what we are ultimately calculating in this problem. Acceleration is how fast the velocity of an object changes with time, and it's calculated using the equation from Newton's Second Law:\[ F_{net} = m \cdot a \]We rearrange this equation to solve for acceleration \( a \), by dividing the net force by the mass:\[ a = \frac{F_{net}}{m} \]Inserting the numbers from our example:\[ a = \frac{10.4 \text{ N}}{72.0 \text{ kg}} \approx 0.144 \text{ m/s}^2 \]This result tells us how quickly the man's speed increases as he hoists himself upward. The small magnitude of this number reflects the balance between the substantial gravitational force and the force he applies to lift himself.

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

The space probe Deep Space 1 was launched on October 24,1998 . Its mass was \(474 \mathrm{~kg}\). The goal of the mission was to test a new kind of engine called an ion propulsion drive. This engine generated only a weak thrust, but it could do so over long periods of time with the consumption of only small amounts of fuel. The mission was spectacularly successful. At a thrust of \(56 \mathrm{mN}\) how many days were required for the probe to attain a velocity of \(805 \mathrm{~m} / \mathrm{s}(1800 \mathrm{mi} / \mathrm{h})\), assuming that the probe started from rest and that the mass remained nearly constant?

A student is skateboarding down a ramp that is \(6.0 \mathrm{~m}\) long and inclined at 18 with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{~m} / \mathrm{s}\). Neglect friction and find the speed at the bottom of the ramp.

The drawing shows box 1 resting on a table, with box 2 resting on top of box \(1 .\) A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2 and the other end is connected to box \(3 .\) The weights of the three boxes are \(W_{1}=55 \mathrm{~N}, W_{2}=35 \mathrm{~N},\) and \(W_{3}=28 \mathrm{~N}\). Determine the magnitude of the normal force that the table exerts on box \(1 .\)

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 \(\mathrm{kg},\) and the coefficient of kinetic friction between the road and him is \(0.870 .\) Find the tension in the cable.

A small sphere is hung by a string from the ceiling of a van. When the van is stationary, the sphere hangs vertically. However, when the van accelerates, the sphere swings backward so that the string makes an angle of \(\theta\) with respect to the vertical. (a) Derive an expression for the magnitude \(a\) of the acceleration of the van in terms of the angle \(\theta\) and the magnitude \(g\) of the acceleration due to gravity. (b) Find the acceleration of the van when \(\theta=10.0^{\circ} .\) (c) What is the angle \(\theta\) when the van moves with a constant velocity?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Kinematics Physics

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

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.