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 5, Problem 60P

(a)

To determine

The maximum force which is applied so that there is no relative motion between block and bracket.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the block is 10.0kg .

Mass of the bracket is 5.0kg .

Value of the static friction is 0.40 .

Value of the kinetic friction is 0.30 .

Formula used:

Draw free body diagram of block and bracket.

  Physics for Scientists and Engineers, Chapter 5, Problem 60P , additional homework tip  1

Write expression for net force in x -direction on block.

  Fx=0

Substitute fsFm1a1 for Fx in above expression.

  fsFm1a1=0

Here, fs is the force due to static friction, m1 is the mass of block and a1 is the acceleration of the block.

Substitute μsm1g for fs in above expression and solve for a1 .

  μsm1gFm1a1=0

Here, μs is the coefficient static friction and g is the acceleration due to gravity.

Rearrange above expression for F .

  F=μsm1gm1a1.......(1)

Write expression for net force in x -direction on block.

  Fx=0

Substitute 2Fμsm1gm2a2 for Fx in above expression.

  2Fμsm1gm2a2=0

Here, m2 is the mass of bracket and a2 is the acceleration of bracket.

Substitute a1 for a2 and μsm1gm1a1 for F in above expression and solve for a1 .

  a1=μsm1gm2+2m1.......(2)

Calculation:

Substitute 0.40 for μs , 10kg for m1 , 9.81m/s2 for g and 5.0kg for m2 in equation (2).

  a1=( 0.40)( 10kg)( 9.81m/ s 2 )5kg+2( 10kg)=1.56m/s2

Substitute 0.40 for μs , 10kg for m1 , 9.81m/s2 for g and 1.56m/s2 for a1 .

  F=(0.40)(10kg)(9.81m/ s 2)(10kg)(1.56m/ s 2)=23.64N

Conclusion:

Thus, the maximum force is 23.64N .

(a)

To determine

The maximum force which is applied so that there is no relative motion between block and bracket.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of the block is 10.0kg .

Mass of the bracket is 5.0kg .

Value of the static friction is 0.40 .

Value of the kinetic friction is 0.30 .

Formula used:

Draw free body diagram of block and bracket.

  Physics for Scientists and Engineers, Chapter 5, Problem 60P , additional homework tip  2

Write expression for net force in x -direction on block.

  Fx=0

Substitute fsFm1a1 for Fx in above expression.

  fsFm1a1=0

Substitute μsm1g for fs in above expression and solve for a1 .

  μsm1gFm1a1=0

Rearrange above expression for F .

  F=μsm1gm1a1

Write expression for net force in x -direction on block.

  Fx=0

Substitute 2Fμsm1gm2a2 for Fx in above expression.

  2Fμsm1gm2a2=0

Substitute a1 for a2 and μsm1gm1a1 for F in above expression and solve for a1 .

  a1=μsm1gm2+2m1.......(1)

Calculation:

Substitute 0.40 for μs , 10kg for m1 , 9.81m/s2 for g and 5.0kg for m2 in equation (1).

  a1=( 0.40)( 10kg)( 9.81m/ s 2 )5kg+2( 10kg)=1.56m/s2

Conclusion:

The acceleration of the block is equal to the acceleration of the bracket. Thus, the acceleration of the bracket is 1.56m/s2 .

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
Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC
In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?
Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor  µC 6.00 µF capacitor  µC 3.00 µF capacitor  µC capacitor C  µC

Chapter 5 Solutions

Physics for Scientists and Engineers

Ch. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 65PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 101PCh. 5 - Prob. 102PCh. 5 - Prob. 103PCh. 5 - Prob. 104PCh. 5 - Prob. 105PCh. 5 - Prob. 106PCh. 5 - Prob. 107PCh. 5 - Prob. 108PCh. 5 - Prob. 109PCh. 5 - Prob. 110PCh. 5 - Prob. 111PCh. 5 - Prob. 112PCh. 5 - Prob. 113PCh. 5 - Prob. 114PCh. 5 - Prob. 115PCh. 5 - Prob. 116PCh. 5 - Prob. 117PCh. 5 - Prob. 118PCh. 5 - Prob. 119PCh. 5 - Prob. 120PCh. 5 - Prob. 121PCh. 5 - Prob. 122PCh. 5 - Prob. 123PCh. 5 - Prob. 124PCh. 5 - Prob. 125PCh. 5 - Prob. 126PCh. 5 - Prob. 127PCh. 5 - Prob. 128PCh. 5 - Prob. 129PCh. 5 - Prob. 130PCh. 5 - Prob. 131PCh. 5 - Prob. 132PCh. 5 - Prob. 133PCh. 5 - Prob. 134PCh. 5 - Prob. 135PCh. 5 - Prob. 136PCh. 5 - Prob. 137PCh. 5 - Prob. 138PCh. 5 - Prob. 139P
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
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
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
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Gravitational Force (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=pxp1Z91S5uQ;License: Standard YouTube License, CC-BY