COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
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, 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
Truck suspensions often have "helper springs" that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as in the figure below. The helper spring engages when the main leaf spring is compressed by distance yo, and then helps to support any additional load. Consider a leaf spring constant of 5.45 × 105 N/m, helper spring constant of 3.60 × 105 N/m, and y = 0.500 m. Truck body Dyo Axle (a) What is the compression of the leaf spring for a load of 4.90 × 105 N? m (b) How much work is done compressing the springs? ]
A skier of mass 75 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 50 m up a 30° slope (assumed frictionless) at a constant speed of 2.8 m/s? KJ (b) What power (expressed in hp) must a motor have to perform this task? hp
A block of mass 1.4 kg is attached to a horizontal spring that has a force constant 900 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest. a x = 0 x b (a) A constant friction force of 4.4 N retards the block's motion from the moment it is released. Using an energy approach, find the position x of the block at which its speed is a maximum. cm (b) Explore the effect of an increased friction force of 13.0 N. At what position of the block does its maximum speed occur in this situation? cm

Chapter 4 Solutions

COLLEGE PHYSICS

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