/*! 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 55 A 20 -kg crate hangs at the end ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 55

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is \((a) 250 \mathrm{~N},(b) 150 \mathrm{~N},(c)\) zero, \((d) 196 \mathrm{~N}\).

Short Answer

Expert verified
a) 2.7 m/s虏 up, b) 2.3 m/s虏 down, c) 9.8 m/s虏 down, d) 0 m/s虏.

Step by step solution

01

Identify the Forces Acting on the Crate

The crate is subject to two forces: the gravitational force (weight) and the tension in the rope. The gravitational force can be calculated using the equation: \( F_g = m \cdot g \), where \( m \) is the mass of the crate (20 kg) and \( g \) is the acceleration due to gravity (9.8 m/s虏).
02

Calculate the Gravitational Force

Calculate the gravitational force acting on the crate: \( F_g = 20 \times 9.8 = 196 \text{ N} \). This is the force exerted by gravity on the crate.
03

Apply Newton's Second Law

According to Newton's second law, the net force \( F_{net} \) acting on the crate can be expressed as \( F_{net} = m \cdot a \) where \( m = 20 \) kg and \( a \) is the acceleration. The net force is the difference between the tension in the rope (\( T \)) and the gravitational force: \( F_{net} = T - F_g \).
04

Solve for Acceleration for Each Tension Value

**(a) When \( T = 250 \text{ N} \):**\[ 250 - 196 = 20 \times a \]\[ 54 = 20 \times a \]\[ a = \frac{54}{20} = 2.7 \text{ m/s虏} \text{ upwards} \]**(b) When \( T = 150 \text{ N} \):**\[ 150 - 196 = 20 \times a \]\[ -46 = 20 \times a \]\[ a = \frac{-46}{20} = -2.3 \text{ m/s虏} \text{ downwards} \]**(c) When \( T = 0 \text{ N} \):**\[ 0 - 196 = 20 \times a \]\[ -196 = 20 \times a \]\[ a = \frac{-196}{20} = -9.8 \text{ m/s虏} \text{ downwards} \] (free fall)**(d) When \( T = 196 \text{ N} \):**\[ 196 - 196 = 20 \times a \]\[ 0 = 20 \times a \]\[ a = 0 \text{ m/s虏} \text{ (no motion)} \]
05

Summarize the Results

The accelerations for each tension value are: - (a) 2.7 m/s虏 upwards - (b) 2.3 m/s虏 downwards - (c) 9.8 m/s虏 downwards (free fall) - (d) 0 m/s虏 (no motion)

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
Gravitational force is one of the fundamental forces in nature. It is the force of attraction between two masses. In most everyday scenarios, this force is due to the attraction between objects and the Earth.
In our exercise, gravitational force acts on the crate because of Earth鈥檚 gravity pulling it downward. To compute this force, we use the formula: \[ F_g = m \cdot g \] where:
  • \( F_g \) is the gravitational force,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
For the 20 kg crate, the gravitational force is calculated as \( F_g = 20 \times 9.8 = 196 \text{ N} \). This means the Earth pulls the crate towards its center with a force of 196 newtons.
Understanding gravitational force helps us recognize how weight differs from mass; mass remains constant while weight (force experienced due to gravity) can vary depending on the gravitational pull.
Tension in a Rope
Tension in a rope is a type of force that opposes gravitational force in our scenario. This force occurs whenever an object is hanging or being lifted by a rope, cable, or string. It鈥檚 essentially the force transmitted through the rope when it is pulled tight by forces acting from opposite ends.
In our problem, tension opposes the crate's weight due to gravity. Hence, if the tension is greater than the gravitational force, the object accelerates upwards. Conversely, if it's less, the object accelerates downwards.
  • When tension is 250 N, it is greater than gravity鈥檚 pull of 196 N, resulting in an upward acceleration.
  • If tension is 150 N or lower, it means gravity wins, and the crate accelerates downwards.
Understanding tension is vital for analyzing situations where forces resist gravity, such as construction cranes lifting loads or elevators moving between floors.
Net Force
Net force is the total force acting on an object, determining the object's acceleration according to Newton鈥檚 Second Law of Motion. It鈥檚 the combination of all forces, whether they are opposite or aligned.
For the crate in this exercise, net force is the difference between the tension in the rope and the gravitational force:
\[ F_{\text{net}} = T - F_g \]
Newton鈥檚 Second Law states that this net force equals mass times acceleration \( (F_{\text{net}} = m \cdot a) \). Here are the scenarios from our exercise:
  • When tension is 250 N, the net force is 54 N upwards, leading to a positive acceleration of 2.7 m/s虏 upwards.
  • When tension is 150 N, the net force is 46 N downwards, causing a negative acceleration of -2.3 m/s虏.
  • When tension is zero, the only force is gravity, and the crate accelerates downward at 9.8 m/s虏, equivalent to free fall.
  • When tension is equal to the gravitational force, there is no net force, thus no acceleration.
Grasping the concept of net force is key for predicting an object's motion based on the forces applied to it. It connects the physical causes (forces) and effects (the motion of the object).

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 constant force acts on a \(5.0 \mathrm{~kg}\) object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0 \mathrm{~s}\). Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

A 700-N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of ( \(a\) ) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

A \(20.0 \mathrm{~kg}\) object that can move freely is subjected to a resultant force of \(45.0 \mathrm{~N}\) in the \(-x\) -direction. Find the acceleration of the object. We make use of the second law in component form, \(\sum F_{x}=m a_{x}\), with \(\sum F_{x}=-45.0 \mathrm{~N}\) and \(m=20.0 \mathrm{~kg}\). Then $$ a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2} $$ where we have used the fact that \(1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}\). Because the resultant force on the object is in the \(-x\) -direction, its acceleration is also in that direction.

Floating in space far from anything else are two spherical asteroids, one having a mass of \(20 \times 10^{10} \mathrm{~kg}\) and the other a mass of \(40 \times 10^{10} \mathrm{~kg}\). Compute the force of attraction on each one due to gravity when their center-to-center separation is \(10 \times 10^{6} \mathrm{~m}\).

How large a horizontal force in addition to \(F_{T}\) must pull on block- \(A\) in Fig. \(3-21\) to give it an acceleration of \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) toward the left? Assume, as in Problem \(3.31\), that \(\mu_{k}=0.20, m_{A}=25 \mathrm{~kg}\), and \(m_{B}=15 \mathrm{~kg}\). Redraw Fig 3-21 for this case, including a force \(F\) pulling toward the left on \(A\). In addition, the retarding friction force \(F_{\mathrm{f}}\) must be reversed in direction. As in Problem 3.31, \(F_{\mathrm{f}}=49.1 \mathrm{~N}\). Write \(F=m a\) for each block in turn, taking the direction of motion (to the left and up) to be positive. We have $$ \begin{aligned} \pm \sum F_{x A}=F-F_{T}-49.1 \mathrm{~N} &=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \text { and }+\uparrow \sum F_{y B}=F_{T}-(15)(9.81) \mathrm{N} \\ &=(15 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ Solve the last equation for \(F_{T}\) and substitute in the previous equation. Then solve for the single unknown \(F\), and find it to be \(226 \mathrm{~N}\) or \(0.23 \mathrm{kN}\).
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Rotational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Scientific Method Physics

Read Explanation

Geometrical and Physical Optics

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.