Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 11P
To determine

To Find: The resultant gravitational force on each particle.

Expert Solution & Answer
Check Mark

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 .

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
a) Consider the following function, where A is a constant. y(x,t) = A(x — vt). Can this represent a wave that travels along? Explain. b) Which of the following are possible traveling waves, provide your reasoning and give the velocity of the wave if it can be a traveling wave. e-(a²x²+b²²-2abtx b.1) y(x,t) b.2) y(x,t) = = A sin(ax² - bt²). 2 b.3) y(x,t) = A sin 2π (+) b.4) y(x,t) = A cos² 2π(t-x). b.5) y(x,t) = A cos wt sin(kx - wt)
The capacitor in (Figure 1) is initially uncharged. The switch is closed at t=0. Immediately after the switch is closed, what is the current through the resistor R1, R2, and R3? What is the final charge on the capacitor? Please explain all steps.
Suppose you have a lens system that is to be used primarily for 620-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength? × nm 434
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
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
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:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Gravitational Force (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=pxp1Z91S5uQ;License: Standard YouTube License, CC-BY