COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 62QAP
To determine

(a)

The magnitude of F2

Expert Solution
Check Mark

Answer to Problem 62QAP

The magnitude of F2=|F2|=10.7N

Explanation of Solution

Given info:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 62QAP , additional homework tip  1

Mass, M=3.00kg

Acceleration, a=7.00m/s2 in +x direction

And θ1=30.0

Formula Used:

  Fext,x=F1x+F2x

  Fext,y=F1y+F2y

  F2=F2x2+F2y2

  F=Ma

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope 1 applies a force with magnitude F1 at an angle of 30.0.

We need to determine the magnitude and direction of rope 2 's force, F2.

We are told that the box accelerates only in the x-direction at a rate of 7.00m/s2, which means the acceleration in the y-direction is equal to 0.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force 2.

  Fext,x=F1x+F2x=F1cos(30.0)+F2x=M(7.00m/s2)=>F2x=M(7.00m/s2)F1cos(30.0)

  Fext,y=F1y+F2y=F1sin(30.0)+F2y=0=>F2y=F1sin(30.0)

  F2=F2x2+F2y2=(M(7.00 m/s 2) F 1cos( 30.0 ))2+( F 1sin( 30.0 ))2=M2(49.0m2/s4)(14.0m/s2)MF1cos(30.0)+F12cos2(30.0)+F12sin2(30.0)=M2(49.0m2/s4)(14.0m/s2)MF1cos(30.0)+F12

  F2=(3.00kg)2(49.0m2/s4)(14.0m/s2)(3.00kg)(20.0N)cos(30.0)+(20.0N)2F2=10.7N

Conclusion:

The magnitude of F2 is 10.7N

To determine

(b)

A careful drawing to show direction of F2.

Expert Solution
Check Mark

Answer to Problem 62QAP

The below diagram shows the direction of F2.

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 62QAP , additional homework tip  2

Explanation of Solution

Given info:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 62QAP , additional homework tip  3

Mass, M=3.00kg

Acceleration, a=7.00m/s2 in +x direction

And θ1=30.0

Formula Used:

  Fext,x=F1x+F2x

  Fext,y=F1y+F2y

  F2=F2x2+F2y2

  F=Ma

  θ2=tan1(F2yF2x)

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope 1 applies a force with magnitude F1 at an angle of 30.0.

We need to determine the magnitude and direction of rope 2 's force, F2.

We are told that the box accelerates only in the x-direction at a rate of 7.00m/s2, which means the acceleration in the y-direction is equal to 0.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force 2.

  Fext,x=F1x+F2x=F1cos(30.0)+F2x=M(7.00m/s2)=>F2x=M(7.00m/s2)F1cos(30.0)

  Fext,y=F1y+F2y=F1sin(30.0)+F2y=0=>F2y=F1sin(30.0)

On substituting given values we have: -

  F2x=(M(7.00m/s2)F1cos(30.0))=((3.00kg)(7.00m/s2)(20.0N)cos(30.0))F2x=3.68NF2y=F1sin(30.0)=(20.0N)sin(30.0)F2y=(10.0N)θ2=tan1(10.0N3.68N)=69.8=290.2

290.20

Conclusion:

Thus, we obtained the direction of F2 as below: -

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 62QAP , additional homework tip  4

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
suggest a reason ultrasound cleaning is better than cleaning by hand?
Checkpoint 4 The figure shows four orientations of an electric di- pole in an external electric field. Rank the orienta- tions according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the di- pole, greatest first. (1) (2) E (4)
What is integrated science. What is fractional distillation What is simple distillation

Chapter 4 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

Ch. 4 - Prob. 11QAPCh. 4 - Prob. 12QAPCh. 4 - Prob. 13QAPCh. 4 - Prob. 14QAPCh. 4 - Prob. 15QAPCh. 4 - Prob. 16QAPCh. 4 - Prob. 17QAPCh. 4 - Prob. 18QAPCh. 4 - Prob. 19QAPCh. 4 - Prob. 20QAPCh. 4 - Prob. 21QAPCh. 4 - Prob. 22QAPCh. 4 - Prob. 23QAPCh. 4 - Prob. 24QAPCh. 4 - Prob. 25QAPCh. 4 - Prob. 26QAPCh. 4 - Prob. 27QAPCh. 4 - Prob. 28QAPCh. 4 - Prob. 29QAPCh. 4 - Prob. 30QAPCh. 4 - Prob. 31QAPCh. 4 - Prob. 32QAPCh. 4 - Prob. 33QAPCh. 4 - Prob. 34QAPCh. 4 - Prob. 35QAPCh. 4 - Prob. 36QAPCh. 4 - Prob. 37QAPCh. 4 - Prob. 38QAPCh. 4 - Prob. 39QAPCh. 4 - Prob. 40QAPCh. 4 - Prob. 41QAPCh. 4 - Prob. 42QAPCh. 4 - Prob. 43QAPCh. 4 - Prob. 44QAPCh. 4 - Prob. 45QAPCh. 4 - Prob. 46QAPCh. 4 - Prob. 47QAPCh. 4 - Prob. 48QAPCh. 4 - Prob. 49QAPCh. 4 - Prob. 50QAPCh. 4 - Prob. 51QAPCh. 4 - Prob. 52QAPCh. 4 - Prob. 53QAPCh. 4 - Prob. 54QAPCh. 4 - Prob. 55QAPCh. 4 - Prob. 56QAPCh. 4 - Prob. 57QAPCh. 4 - Prob. 58QAPCh. 4 - Prob. 59QAPCh. 4 - Prob. 60QAPCh. 4 - Prob. 61QAPCh. 4 - Prob. 62QAPCh. 4 - Prob. 63QAPCh. 4 - Prob. 64QAPCh. 4 - Prob. 65QAPCh. 4 - Prob. 66QAPCh. 4 - Prob. 67QAPCh. 4 - Prob. 68QAPCh. 4 - Prob. 69QAPCh. 4 - Prob. 70QAPCh. 4 - Prob. 71QAPCh. 4 - Prob. 72QAPCh. 4 - Prob. 73QAPCh. 4 - Prob. 74QAPCh. 4 - Prob. 75QAPCh. 4 - Prob. 76QAPCh. 4 - Prob. 77QAPCh. 4 - Prob. 78QAPCh. 4 - Prob. 79QAPCh. 4 - Prob. 80QAPCh. 4 - Prob. 81QAPCh. 4 - Prob. 82QAPCh. 4 - Prob. 83QAPCh. 4 - Prob. 84QAPCh. 4 - Prob. 85QAPCh. 4 - Prob. 86QAPCh. 4 - Prob. 87QAPCh. 4 - Prob. 88QAPCh. 4 - Prob. 89QAPCh. 4 - Prob. 90QAPCh. 4 - Prob. 91QAPCh. 4 - Prob. 92QAPCh. 4 - Prob. 93QAPCh. 4 - Prob. 94QAPCh. 4 - Prob. 95QAPCh. 4 - Prob. 96QAPCh. 4 - Prob. 97QAPCh. 4 - Prob. 98QAPCh. 4 - Prob. 99QAPCh. 4 - Prob. 100QAPCh. 4 - Prob. 101QAPCh. 4 - Prob. 102QAPCh. 4 - Prob. 103QAP
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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
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
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:9781305952300
Author:Raymond A. Serway, Chris Vuille
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