Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 6, Problem 11P
To determine

To Find: The resultant gravitational force on each particle.

Expert Solution & Answer
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Explanation of Solution

Given:

Mass of each particle, m1=m2=m3=m=2.00kg . d=1.00m

Distance between particles, d=1.0m

Formula used:

The gravitational force between two bodies having mass m and M , d distance away is:

  F=GmMd2

G is the gravitational constant.

Calculation:

Free body diagram:

  Physics Fundamentals, Chapter 6, Problem 11P

Gravitational force on m1 due to m2 and m3 :

  F13=Gm1m3d2F13=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F12=Gm1m2d2F12=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F12=F13

  F13+F12=(F 12sin30°F 13sin30°)i+(F 13+F 12)cos30°jF13+F12=0+2×26.68×1011×32jF13+F12=46.2×1011Nj

The gravitational force on

  m2 due to m1 and m3 :

  F21=Gm1m2d2F21=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F23=Gm2m3d2F23=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F21=F23

  F21+F23=(F 23+F 21sin30°)(i)+F21cos30°(j)F21+F23=(26.68× 10 11+26.68× 10 11×12)i26.68×1011×32jF13+F12=40.02×1011Ni23.11×1011j

Gravitational force on m3 due to m1 and m2 :

  F31=Gm1m3d2F31=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F32=Gm2m3d2F32=(6.67× 10 11) 2.02 1.02=26.68×1011N

  F31+F32=(F 32+F 31sin30°)i+F31cos30°(j)F31+F32=(26.68× 10 11+26.68× 10 11×12)i26.68×1011×32jF31+F32=40.02×1011Ni23.11×1011j

Conclusion:

Thus, resultant gravitational force on m1 is 46.2×1011Nj , on m2 is 40.02×1011Ni23.11×1011j and on m3 is 40.02×1011Ni23.11×1011j .

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