Explanation Given data: The length and width of the board are 2 m and 1 m respectively, the mass of the board is 14 kg and the mass of the pot is 5.0 kg . Formula Used: Write the expression for the center of mass of a 2D system with distributed mass. x c m = m 1 x 1 + m 2 x 2 m 1 + m 2 And y c m = m 1 y 1 + m 2 y 2 m 1 + m 2 Here, x c m and y c m are the x and y coordinates of the center of mass of the board-pot system respectively, m 1 is the mass of the board, m 2 is the mass of the pot, x 1 is the horizontal distance of the center of mass of the board along the x-axis from the origin, x 2 is the horizontal distance of the center of mass of the pot along the x-axis from the origin, y 1 is the vertical distance of the center of mass of the board along the y-axis from the origin, and y 2 is the vertical distance of the center of mass of the pot along the y-axis from the origin. Explanation: The figure depicting the situation is give below. Consider the pot to be a point object present at the top right corner of the board. The point O represents the center of mass of the board and the center of mass of the pot is at the top right corner. Assume the point O to be the origin or reference in order to calculate the center of mass of the board-pot system. Referring the figure, the horizontal distance of the center of mass of the board along the x-axis from the origin O is 0 m , the horizontal distance of the center of mass of the pot along the x-axis from the origin O is 1 m , the vertical distance of the center of mass of the board along the y-axis from the origin O is 0 m , and the vertical distance of the center of mass of the pot along the y-axis from the origin is 0.5 m . Recall the expression for the center of mass of a 2D system with distributed mass. x c m = m 1 x 1 + m 2 x 2 m 1 + m 2 Substitute, 14 kg for m 1 , 5 kg for m 2 , 0 m for x 1 , and 1 m for x 2 in the above expression

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ISBN: 9780134601823

Author: ETKINA, Eugenia, Planinšič, G. (gorazd), Van Heuvelen, Alan

Publisher: Pearson,

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Chapter 8, Problem 30P

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The center of mass of a board-pot system, when a pot of mass 5 kg is placed on a 1 m wide and 2 m long board of 14 kg.

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Chapter 8, Problem 29P

Chapter 8, Problem 31P

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The center of mass of a board-pot system, when a pot of mass 5 kg is placed on a 1 m wide and 2 m long board of 14 kg .

Solution Summary: The author explains the formula used to determine the center of mass of a 2D system with distributed mass.

The center of mass of a board-pot system, when a pot of mass 5 kg is placed on a 1 m wide and 2 m long board of 14 kg.

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