Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
bartleby

Videos

Textbook Question
Book Icon
Chapter 3, Problem 57SP

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 N, (b) 150 N, (c) zero, (d) 196 N.

(a)

Expert Solution
Check Mark
To determine

The magnitude of the acceleration on thecrate when 20 kg mass of the crate hangs at the end of a long rope and tension in the rope is 250 N.

Answer to Problem 57SP

Solution:

2.7 m/s2 upward

Explanation of Solution

Given data:

The mass of the crate is 20 kg.

The tension force in the rope is 250 N.

Formula used:

FromNewton’s second law of motion, the expression of the sum of the force acting upon the crateis as follows:

Fy=may

Here, Fy is the sum of the net force in y-direction, m is the mass of the crate, and ay is the acceleration of the crate.

Explanation:

Draw the free body diagram of the crate:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  1

Here, Ft is a tension force, m is the mass of the crate, a is the acceleration of the crate, and g is acceleration due to gravity.

Recall the expression for Newton’s second law of motion when the motion of the crate is along the vertical direction:

Fy=may

Considering the direction of the forces, upward is taken aspositive and downward direction of the forces is taken as negative. Hence,

Ftmg=ma

Rearrange the value for acceleration, a, of the crate:

a=Ftmgm

Substitute 250 N for Ft, 20 kg for m, and 9.80 m/s2 for g

a=(250 N)(20 kg)(9.80 m/s2)(20 kg)=53.8 N20 kg=2.7 m/s2

Conclusion:

Therefore, the acceleration on the crate is 2.7 m/s2 along the upward direction.

(b)

Expert Solution
Check Mark
To determine

The magnitude of the acceleration on the crate when 20 kg mass of the crate hangs at the end of a long rope and tension in the rope is 150 N.

Answer to Problem 57SP

Solution:

2.3 m/s2 downward

Explanation of Solution

Given data:

Mass of the crate that is hanging is 20 kg.

Tension force is 150 N.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the crateis as follows:

Fy=may

Here, Fy is the sum of the net force in y-direction, m is the mass of the crate, and ay is the acceleration of the crate.

Explanation:

Draw the free body diagram of the crate:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  2

Here, Ft is a tension force, m is the mass of the crate, a is the acceleration of the crate, and g is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

Fy=may

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

Ftmg=ma

Rearrange the value for acceleration a:

a=Ftmgm

Substitute 150 N for Ft, 20 kg for m, and 9.80 m/s2 for g

a=(150 N )(20 kg)(9.80 m/s2)(20 kg)=(46 N)(20 kg)=2.3 m/s2

Here, the value of a is negative. Therefore, the free body diagram of the rope is as follows:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  3

Conclusion:

The acceleration on the crate is 2.3 m/s2 where the negative sign denotes that the direction of the acceleration is downward.

(c)

Expert Solution
Check Mark
To determine

The magnitude of the acceleration on the crate when 20 kg mass of the crate hangs at the end of a long rope and tension in the rope is 0 N .

Answer to Problem 57SP

Solution:

9.80 m/s2 downward

Explanation of Solution

Given data:

Mass of the crateis 20 kg.

Tension force in the rope is 0 N.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the rope is as follows:

Fy=may

Here, Fy is the sum of the net force in y-direction, ay is the acceleration of the crate, and m is the mass of the crate.

Explanation:

Draw the free body diagram of the crate:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  4

Here, Ft is a tension force, m is the mass of the crate, a is the acceleration of the crate, and g is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

Fy=may

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

Ftmg=ma

Rearrange the value for acceleration a:

a=Ftmgm

Substitute 0 N for Ft, 20 kg for m, and 9.80 m/s2 for g

a=(0)(20 kg)(9.80 m/s2)(20 kg)=9.80 m/s2

The negative value of a shows that its direction mentioned in the diagram should be along the downward direction.

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  5

Conclusion:

The acceleration on the crate is 9.80 m/s2 and the direction of the acceleration is downward.

(d)

Expert Solution
Check Mark
To determine

The magnitude of the acceleration on the crate when 20 kg mass of the crate hangs at the end of a long rope and tension in the rope is 196 N.

Answer to Problem 57SP

Solution:

0.0 m/s2

Explanation of Solution

Given data:

The mass of the crateis 20 kg.

The tension force in the rope is 196 N.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the rope is as follows:

Fy=may

Here, FY is the sum of the net force in y-direction, ay is the acceleration of the crate, and m is the mass of the crate.

Explanation:

Draw the free body diagram of the crate:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 3, Problem 57SP , additional homework tip  6

Here, Ft is a tension force, m is the mass of the crate, a is the acceleration of the crate, and g is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

FY=maY

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

Ftmg=ma

Rearrange the value for acceleration of the crate:

a=Ftmgm

Substitute 196 N for Ft, 20 kg for m, and 9.80 m/s2 for g

a=(196)(20 kg)(9.80 m/s2)(20 kg)=0 m/s2

Conclusion:

The acceleration on the crate is 0 m/s2 when the tension force in the rope is 196 N.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
An elevator filled with passengers has a mass of 1.70 × 103 kg . (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?
A block of mass m1 = 3.0 kg on a smooth inclined plane of angle 30◦ is connected, by a cord over a small frictionless pulley to a second block of mass m2 = 2.0 kg hanging vertically as in Figure.(a) What is the acceleration of each body, a1 and a2? (Should have two values for acceleration)(b) What is the tension T in the cord?
An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

Chapter 3 Solutions

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

Ch. 3 - 3.52 [I] A force acts on a 2-kg mass and gives...Ch. 3 - 3.53 [I] An object has a mass of 300 g. (a)...Ch. 3 - 3.54 [I] A horizontal cable pulls a 200-kg cart...Ch. 3 - 3.55 [II] A 900-kg car is going 20 m/s along a...Ch. 3 - 3.56 [II] A 12.0-g bullet is accelerated from rest...Ch. 3 - 3.57 [II] A 20-kg crate hangs at the end of a long...Ch. 3 - 3.58 [II] A 5.0-kg mass hangs at the end of a...Ch. 3 - 3.59 [II] A 700-N man stands on a scale on the...Ch. 3 - 3.60 [II] Using the scale described in Problem...Ch. 3 - 3.61 [II] A cord passing over a frictionless,...Ch. 3 - 3.62 [II] An elevator starts from rest with a...Ch. 3 - 3.63 [II] Just as her parachute opens, a 60-kg...Ch. 3 - 3.64 [II] A 300-g mass hangs at the end of a...Ch. 3 - 3.65 [II] A 20-kg wagon is pulled along the level...Ch. 3 - 3.66 [II] A 12-kg box is released from the top of...Ch. 3 - 3.67 [I] A wooden crate weighing 1000 N is at...Ch. 3 - 3.68 [I] Someone wearing rubber-soled shoes is...Ch. 3 - 3.69 [I] A standing 580-N woman wearing climbing...Ch. 3 - 3.70 [II] For the situation outlined in Problem...Ch. 3 - 3.71 [II] An inclined plane makes an angle of ...Ch. 3 - 3.72 [II] A horizontal force F is exerted on a...Ch. 3 - 3.73 [II] An inclined plane making an angle of ...Ch. 3 - 3.74 [III] Repeat Problem 3.73 if the coefficient...Ch. 3 - 3.75 [III] A horizontal force of 200 N is required...Ch. 3 - 3.76 [II] Find the acceleration of the blocks in...Ch. 3 - 3.77 [III] Repeat Problem 3.76 if the coefficient...Ch. 3 - 3.78 [III] How large a force F is needed in Fig....Ch. 3 - 3.79 [III] In Fig. 3-28, how large a force F is...Ch. 3 - 3.80 [III] (a) What is the smallest force parallel...Ch. 3 - 3.81 [III] A 5.0-kg block rests on a incline. The...Ch. 3 - 3.82 [III] Three blocks with masses 6.0 kg, 9.0...Ch. 3 - 3.83 [I] Floating in space far from anything...Ch. 3 - 3.84 [I] Two cannonballs that each weigh 4.00...Ch. 3 - 3.85 [I] Imagine a planet and its moon...Ch. 3 - 3.86 [I] Two NASA vehicles separated by a...Ch. 3 - 3.87 [I] Suppose you are designing a small,...Ch. 3 - Prob. 88SPCh. 3 - Prob. 89SPCh. 3 - 3.90 [II] A space station that weighs 10.0 MN on...Ch. 3 - 3.91 [II] An object that weighs 2700 N on the...Ch. 3 - 3.92 [II] Imagine a planet having a mass twice...Ch. 3 - 3.93 [II] The Earth’s radius is about 6370 km. An...Ch. 3 - 3.94 [II] A man who weighs 1000 N on Earth stands...Ch. 3 - 3.95 [II] The radius of the Earth is about 6370...Ch. 3 - 3.96 [II] The fabled planet Dune has a diameter...Ch. 3 - 3.97 [III] An astronaut weighs 480 N on Earth. She...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Drawing Free-Body Diagrams With Examples; Author: The Physics Classroom;https://www.youtube.com/watch?v=3rZR7FSSidc;License: Standard Youtube License