Question Video: Finding the Position and the Center of Mass of a Uniform Triangular Lamina | Nagwa Question Video: Finding the Position and the Center of Mass of a Uniform Triangular Lamina | Nagwa

Question Video: Finding the Position and the Center of Mass of a Uniform Triangular Lamina Mathematics • Third Year of Secondary School

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In the figure, find the position of the center of mass of the uniform triangular lamina π΄π΅πΆ, considering π΄ to be the origin point.

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Video Transcript

In the figure shown, find the position of the center of mass of the uniform triangular lamina π΄π΅πΆ, considering π΄ to be the origin point.

In this figure, we see the triangle with vertices π΄, π΅, and πΆ positioned on this π₯π¦-coordinate plane. We want to find the center of mass of this triangular lamina, and we knew that that corresponds with the centroid or geometric center of the shape. Just estimating by eye, we might put the geometric center of this triangle here. But to know the π₯- and π¦-coordinates of this point accurately, weβll call them πΆππ π₯ and πΆππ π¦, weβll need to recall a more precise approach. Whenever weβre working with triangles seeking to find their center of mass, the key information for doing this is the coordinates of the triangleβs three vertices.

If we know these, then regardless of the shape of the triangle, we can calculate the π₯- and π¦-coordinates of its center of mass using these relationships. Basically, they involve solving for the average π₯-coordinate and the average π¦-coordinate among the vertices. If we apply these relationships to our scenario with triangle π΄π΅πΆ, then we can note that the coordinates at the vertex π΅ are zero, five π; those at vertex πΆ are four π, zero. And because vertex π΄ is positioned at the origin, those coordinates are zero, zero.

To solve then first for the center of mass π₯-coordinate, weβll add together zero, zero, and four π and divide all that by three, which gives us four π over three. And then to solve for the center of mass π¦-coordinate, weβll add together five π, zero, and zero, the π¦-coordinates of the three vertices of our triangle, and divide all that by three, giving us five-thirds π. These, then, are the coordinates of the center of mass of this uniform triangular lamina.

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