Explanation Given info: The length of the side of equilateral triangle is 10.0 m , mass at corner C of triangle is 1.00 kg , mass at the corner A of triangle is 2.00 kg and the mass at corner B of the triangle is 3.00 kg . The diagram of the equilateral triangle with masses is shown below. Figure (1) Write the expression to calculate the gravitational force on mass at corner B due to mass at corner A . F AB = G m A m B l 2 ( − i ^ ) Here, F AB is the gravitational force on mass at corner B due to mass at corner A . G is the Gravitational Constant. m A is the mass at corner A . m B is the mass at corner B . l is the length of side of equilateral triangle. Substitute 6.67 × 10 − 11 Nm 2 / kg 2 for G , 2.00 kg for m A , 3.00 kg for m B and 10 m for l in above expression. F AB = 6.67 × 10 − 11 Nm 2 / kg 2 × 2.00 kg × 3.00 kg ( 1 0 m ) 2 ( − i ^ ) = − 4 × 10 − 12 N ( i ^ ) Thus, the gravitational force on mass at corner B due to mass at corner A is − 4 × 10 − 12 N ( i ^ ) . Write the expression to calculate the gravitational force on corner A due to corner C . F CA = G m A m C l 2 Here, F CA is the gravitational force on corner A due to corner C . G is the Gravitational Constant. m A is the mass at corner A . m C is the mass at corner C . l is the length of side of equilateral triangle. Substitute 6.67 × 10 − 11 Nm 2 / kg 2 for G , 2.00 kg for m A , 1.00 kg for m C and 10 m for l in above expression. F CA = 6.67 × 10 − 11 Nm 2 / kg 2 × 2.00 kg × 1.00 kg ( 10 m ) 2 = 1.334 × 10 − 12 N Thus, the gravitational force on corner A due to corner C is 1.334 × 10 − 12 N . Write the expression to calculate the gravitational force on corner A due to corner C in horizontal direction. F CAx = F CA cos θ ( i ^ ) Here, F CA is the gravitational force on corner A due to corner C . F CAx is the gravitational force on corner A due to corner C in horizontal direction. θ is the angle of the equilateral triangle. Substitute 1.334 × 10 − 12 N for F CA and 60 ° for θ in above expression. F CAx = 1.334 × 10 − 12 N × cos 60 ° ( i ^ ) = 6.67 × 10 − 13 N ( i ^ ) Thus, the gravitational force on corner A due to corner C in horizontal direction is 6.67 × 10 − 13 N ( i ^ ) . Write the expression to calculate the gravitational force on corner A due to corner C in vertical direction. F CAy = F CA sin θ ( − j ^ ) Here, F CA is the gravitational force on corner A due to corner C . F CAy is the gravitational force on corner A due to corner C in vertical direction

5th Edition

ISBN: 9780321976444

Author: James S. Walker

Publisher: PEARSON

See similar textbooks

Concept explainers

Center of Gravity of a system

The centre of gravity is a property that is based on the geometry of a rigid body or any particle. It is described as an equilibrium point or a hypothetical location where the total weight of the body is concentrated.

Question

Chapter 12, Problem 70GP

To determine

The magnitude and direction of net force acting on the 2.00 kg mass.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12, Problem 69GP

Chapter 12, Problem 71GP

Students have asked these similar questions

Forces P, Q and S of magnitudes 6.00 N, 10.00 N and 8.00 N respectively, acts on point O such that the angle made by force P is 50o counter clockwise from the positive x axis, force Q acts at an angle of 38 o counter clockwise from the negative y axis and force S acts at an angle of 32 o counter clockwise from the positive y axis. The x component of force S is

PROBLEM 02 Find the angle between two vector forces whose magnitudes are F and 3F and the magnitude of whose resultant 19 3F. PROBLEM 03 A force T passing through points A and B as shown, has a counterclockwise moment of 400 KN-m about point O. If the coordinates of points A and B are (0,6) and (8,0) in meters, compute the value of force T. PROBLEM 04 The resultant of the concurrent forces shown 15 1OOKN with O, of 60° directed DR. Determine the unknown force. Figure, PROBLEM 05. Refer to problem SP203. Q = 10 KN %3D Change P=25 KN up to the left at 30 degrees w/ x-axis, Q=30 KN, W=35 KN, T=18 KN directed to the right. F = ? 3| 4 P = 30 KN T = 20 KN

1. Determine the resultant vector of the three couples +20 N.m, -30 N.m, and 100 N.m acting respectively in the xy, yz, and xz planes.

Chapter 12 Solutions

Physics (5th Edition)

Ch. 12.1 - Rank the four systems shown in Figure 12-6 in...Ch. 12.2 - Suppose the Sun suddenly collapsed to half its...Ch. 12.3 - Satellite A orbits the Earth at the radius r,...Ch. 12.4 - Prob. 4EYU Ch. 12.5 - Prob. 5EYU Ch. 12.6 - If the radius of the Moons orbit around the Earth...Ch. 12 - It is often said that astronauts in orbit...Ch. 12 - When a person passes you on the street, you do not...Ch. 12 - Imagine bringing the tips of your index fingers...Ch. 12 - Does the radius vector of Mars sweep out the same...

Knowledge Booster

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.

The magnitude and direction of net force acting on the 2.00 kg mass.

Solution Summary: The author explains the magnitude and direction of net force acting on the 2.00kg mass.

Concept explainers

The magnitude and direction of net force acting on the 2.00 kg mass.

Want to see the full answer?

Chapter 12 Solutions