Question Video: Finding the Center of Gravity of a System | Nagwa Question Video: Finding the Center of Gravity of a System | Nagwa

Question Video: Finding the Center of Gravity of a System Mathematics • Third Year of Secondary School

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In the given figure, three weights of magnitudes 2 N, 5 N, and 3 N are placed on the vertices of an equilateral triangle of side length 8 cm. Find the center of gravity of the system.

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Video Transcript

In the given figure, three weights of magnitudes two newtons, five newtons, and three newtons are placed on the vertices of an equilateral triangle of side length eight centimeters. Find the center of gravity of the system.

In order to solve this problem, we’re first going to need to find the 𝑥- and the 𝑦-coordinates of each of the weights. We have been given that the weights lie on the vertices of an equilateral triangle. Hence, each of these side lengths will also be eight centimeters. We can already see that the weight of two newtons lies at the origin. So its coordinates will be zero, zero. The weight of five newtons lies on the 𝑦-axis and is eight centimeters from the weight of two newtons. So, its coordinates will be zero, eight.

In order to find the position of the weight of three newtons, we can add in this horizontal line to help us. This line will make a right angle with the vertical axis. Also, since this is an equilateral triangle, this line will bisect the vertical side. So we can see that this length will be four centimeters. So this is the 𝑦-coordinate of the weight of three newtons.

In order to find the 𝑥-coordinate, we’re going to use the Pythagorean theorem. The Pythagorean theorem tells us that this 𝑥-coordinate is equal to the square root of eight squared minus four squared, which is equal to the square root of 64 minus 16, which is also equal to the square root of 48. And this simplifies to give four root three. So the coordinate of our weight of three newtons is four root three four.

We can now write a table which will show us the positions of each of the weights. We’ll also need the formulas for finding the 𝑥- and 𝑦-coordinates of the center of gravity. Let’s start by finding the 𝑥-coordinate. In the numerator, we need to sum the products of the weight and their 𝑥-coordinates. So that’s two multiplied by zero plus five multiplied by zero plus three multiplied by four root three. And then in the denominator, we simply need to sum the weights. This gives us two multiplied by zero plus five multiplied by zero plus three multiplied by four root three over two plus five plus three. The first two terms in the numerator are multiplied by zero. So we can ignore these. Simplifying the rest of the fraction, we find that 𝑥 is equal to 12 root three over 10 centimeters.

Let’s now move on to find the 𝑦-coordinate. We follow a similar process to finding the 𝑥-coordinate, except this time in the numerator, we need to sum the products of the weights with their 𝑦-coordinates. So that′s two multiplied by zero plus five multiplied by eight plus three multiplied by four. Then, we need to divide this by the sum of the weights, giving us that 𝑦 is equal to two multiplied by zero plus five multiplied by eight plus three multiplied by four over two plus five plus three. Since the first term in the numerator is multiplied by zero, we can again ignore this term. Simplifying the rest of the fraction, and we′re left with 𝑦 is equal to 52 over 10 centimeters. Now that we have both the 𝑥- and 𝑦-coordinates, we found the center of gravity of the system. And it’s located at the coordinates 12 root three over 10, 52 over 10.

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