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 51QAP
To determine

The magnitude of forces F2 and F3

Expert Solution & Answer
Check Mark

Answer to Problem 51QAP

The magnitude of forces are the magnitudes of forces are F2=1.388N and F3=1.638N

Explanation of Solution

Given:

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 51QAP

  θ1= 40.0o with respect to +xθ2= 60.0o with respect to +xθ3=20.0owith respect to +xF1=1.00 N

Acceleration =1.50 ms-2

Formula used:

Newton's first law; In equilibrium Fext=0

Newton's second law F=ma

Calculation:

In positive xdirection, from Newton's second law F=ma

  Fx=maxF1x+F2x+F3x=maxF1cos(40o)+F2x+F3x=2 kg×1.50 ms2F2cos(60o)+F3cos(20o)=(30.766)F2×0.5+F3×0.94=2.234(p)

In positive y (perpendicular to x direction) as the acceleration is zero, considering equilibrium in y direction.

Newton's first law; In equilibrium Fext=0

  Fext=0F1y+F2y+F3y=0F1sin(40o)+F3sin(20o)=F2sin(60o)1 N×0.642=F2×0.866F3×0.342(q)

  F2×0.5+F3×0.94=2.234N(p)0.642 N=F2×0.866F3×0.342(q)(p)×0.866(q)×0.5;2.234N×0.8660.642 N×0.5=(F2×0.5+F3×0.94)×0.866(F2×0.866F3×0.342)×0.51.9340.321=F3×0.814+F3×0.176F3=1.638N__(p)×0.342+(q)×0.94;2.234N×0.3420.642 N×0.94=(F2×0.5+F3×0.94)×0.342(F2×0.866F3×0.342)×0.940.764+0.603=F2×0.176+F2×0.814F2=1.388N__

Conclusion:

Thus, the magnitudes of forces are F2=1.388N and F3=1.638N

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
••63 SSM www In the circuit of Fig. 27-65, 8 = 1.2 kV, C = 6.5 µF, R₁ S R₂ R3 800 C H R₁ = R₂ = R3 = 0.73 MQ. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i̟ in resistor 1, (b) current 2 in resistor 2, and (c) current i3 in resistor 3? At t = ∞o (that is, after many time constants), what are (d) i₁, (e) i₂, and (f) iz? What is the potential difference V2 across resistor 2 at (g) t = 0 and (h) t = ∞o? (i) Sketch V2 versus t between these two extreme times. Figure 27-65 Problem 63.
Thor flies by spinning his hammer really fast from a leather strap at the end of the handle, letting go, then grabbing it and having it pull him. If Thor wants to reach escape velocity (velocity needed to leave Earth’s atmosphere), he will need the linear velocity of the center of mass of the hammer to be 11,200 m/s. Thor's escape velocity is 33532.9 rad/s, the angular velocity is 8055.5 rad/s^2. While the hammer is spinning at its maximum speed what impossibly large tension does the leather strap, which the hammer is spinning by, exert when the hammer is at its lowest point? the hammer has a total mass of 20.0kg.
If the room’s radius is 16.2 m, at what minimum linear speed does Quicksilver need to run to stay on the walls without sliding down?  Assume the coefficient of friction between Quicksilver and the wall is 0.236.

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
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
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
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: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
Newton's Third Law of Motion: Action and Reaction; Author: Professor Dave explains;https://www.youtube.com/watch?v=y61_VPKH2B4;License: Standard YouTube License, CC-BY