COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 4, Problem 51QAP
To determine

The magnitude of forces F2 and F3

Expert Solution & Answer
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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

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Chapter 4 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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