Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 82P

(a)

To determine

The acceleration of each of the blocks.

(a)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

Newton’s third law states that for every action, there is an equal and opposite reaction. It means for every interaction between two objects, there is a pair of forces that act on both the objects.

The free-body diagram for both the block is shown in Figure 1

  Physics for Scientists and Engineers, Chapter 4, Problem 82P

Figure 1

In the above diagram is shown that the force on the block m1 is the gravitational force and the force due to tension on the string. In the same way, the force on the block of mass m2 is the gravitational force and the force due to tension on the string.

Conclusion:

Therefore, the free body diagram for both blocks is shown in Figure 1

(b)

To determine

The expression for the acceleration of the block and the tension in the string.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The expression for the acceleration is (m1m2)gm1+m2 and the expression for the tension in the string is 2m1m2m1+m2g .

Explanation of Solution

Calculation:

The equation of motion for the block m1 is given by,

  m1a=m1gTT=m1gm1a

The equation of motion for the block m2 is given by,

  m2a=Tm2gT=m2a+m2g

The expression for the acceleration of the block is evaluated as,

  m1gm1a=m2am2gm1gm2g=m1a+m2aa=( m 1 m 2 )gm1+m2

The expression for the tension in the string is evaluated a,

  T=m1gm1a=m1gm1( m 1 m 2 )gm1+m2=m1g[ m 1+ m 2 m 1+ m 2 m 1+ m 2]=2m1m2m1+m2g

Conclusion:

Therefore, the expression for the acceleration is (m1m2)gm1+m2 and the expression for the tension in the string is 2m1m2m1+m2g .

(c)

To determine

Whether the expressions for the tension and the acceleration gives the plausible results if m1=m2 in the limit that m1m2 and in the limit that m1m2 .

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The value of tension for the condition m1=m2 is m1g and the value of acceleration is 0 . For the value of m1m2 the plausible value of the tension is 2m2g and the value of acceleration is g . For the value of m1m2 , the plausible value of the tension is 2m1g and the value of acceleration is g .

Explanation of Solution

Calculation:

For the value of m1=m2 the value of m2 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=0

For the value of m1=m2 the value of m2 is negligible and the expression for the tension can be evaluated as,

  T=2m1m2m1+m2g=m1g

For the value of m1m2 the value of m2 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=m1m2g=g

For the value of m1m2 the value of m2 is negligible and the expression for the tension can be evaluated as,

  T=2m1m2m1+m2g=2m2g

For the value of m1m2 the value of m1 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=m2m2g=g

For the value of m1m2 the value of m1 is negligible and the expression for the acceleration can be evaluated as,

  T=2m1m2m1+m2g=2m1g

Conclusion:

Therefore, the value of tension for the condition m1=m2 is m1g and the value of acceleration is 0 . For the value of m1m2 the plausible value of the tension is 2m2g and the value of acceleration is g . For the value of m1m2 , the plausible value of the tension is 2m1g and the value of acceleration is g .

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
A ball of mass m is conncted to a string of length and swings in a vertical circular path. a) Draw the free body diagram of the block at the instant shown in the figure. b) If the block has the speed v at the bottom of the path, find the tension in the string at the top of the path. Present your answer sbmolically.
Please include the full soln. TYSM
Problem Two-block system is given below. The mass of block one is m, = 4 kg and the mass of block two is m = 8 kg. A force is applied on m2 to start the motion of block two knowing that block 1 doesn't move. The coefficient of static friction for all surfaces is qiven as 0.4 (Take g-9.8 m/s). Tension force Fapplied a) Choose the correct answer tor the following question How many torces along the x-axis are acting on block I with mass m? How many forces along the y-axis are acting on block I with mans m? How many forces along the x-axis are acting on block 2 with mass m7 How many torces along the y axis are acting on block 2 with mass m? b) Calculate the magnitade of normal force | FNIlexerted by block 2 on block 1 Caleulate the magnitude of the muxinans atatic tnetien force (Falaated berween banck I anst hiloek 2

Chapter 4 Solutions

Physics for Scientists and Engineers

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
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY