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

Physics, Books a la Carte Edition (5th Edition)

5th Edition

ISBN: 9780134020853

Author: James S. Walker

Publisher: PEARSON

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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.

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Chapter 12, Problem 70GP

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

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Chapter 12, Problem 69GP

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

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