Explanation Given information: The masses m i are located at points P i as: m 1 = 4 , m 2 = 3 , m 3 = 6 , m 4 = 3 , and P 1 6 , 1 , P 2 3 , − 1 , P 3 − 2 , 2 , and P 3 − 2 , − 5 respectively. Method/Formula used: If masses m 1 , m 2 , m 3 , … , m n are located in the X-Y coordinate plane at points P 1 x 1 , y 1 , P x 2 , y 2 , P 3 x 3 , y 3 , … , P n x 1 , y 1 respectively, then the moment of mass M x and M y are given by M x = m 1 x 1 + m 2 x 2 + m 3 x 3 + ⋯ + m n x n M y = m 1 y 1 + m 2 y 2 + m 3 y 3 + ⋯ + m n y n ... (1) The coordinates of the center of mass x c m , y c m of this mass distribution system is given by x c m = m 1 x 1 + m 2 x 2 + m 3 x 3 + ⋯ + m n x n m 1 + m 2 + m 3 + ⋯ + m n y c m = m 1 y 1 + m 2 y 2 + m 3 y 3 + ⋯ + m n y n m 1 + m 2 + m 3 + ⋯ + m n … (2) Calculations: The masses m 1 = 4 , m 2 = 3 , m 3 = 6 , m 4 = 3 are located at points P 1 6 , 1 , P 2 3 , − 1 , P 3 − 2 , 2 , and P 3 − 2 , − 5 respectively, i.e., x 1 = 6 , x 2 = 3 , x 3 = − 2 , and x 4 = − 2 . Moment of mass: The moment of mass of the system of particles M x (about y -axis) is, M x = m 1 y 1 + m 2 y 2 + m 3 y 3 … (3) Substitute, m 1 = 4 , m 2 = 3 , m 3 = 6 , m 4 = 3 and y 1 = 1 , y 2 = − 1 , y 3 = 2 , and y 4 = − 5 in result (3), M x = m 1 y 1 + m 2 y 2 + m 3 y 3 = 4 ⋅ 1 + 3 x

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Author: Stewart

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Chapter 8.3, Problem 24E

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To find: The center of mass of the mass distribution (system).

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Chapter 8.3, Problem 23E

Chapter 8.3, Problem 25E

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The center of mass of the mass distribution (system).

Solution Summary: The author explains that the moment of the mass system M_x is -2 and 15 respectively.

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To find: The center of mass of the mass distribution (system).

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Chapter 8 Solutions