Question Video: Finding the Sum of Moments of Five Forces Acting on a Square Mathematics

Given that 𝐴𝐵𝐶𝐷 is a square with side length seven centimeters and forces acting on it as shown in the figure, calculate the algebraic sum of the moments about vertex 𝐵.

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Video Transcript

Given that 𝐴𝐵𝐶𝐷 is a square with side length seven centimeters and forces acting on it as shown in the figure, calculate the algebraic sum of the moments about vertex 𝐵.

Remember, the turning effect of a force is called its moment. To calculate the moment of a force about a point 𝑜, we multiply that force 𝐹 by the perpendicular distance from the point 𝑜 to the line of action of the force. The moment is equal to 𝐹 times 𝑑. We also recall that moments act about a point in either a clockwise or anticlockwise direction.

In this question, we’ve been given a little diagram to show us which direction were seen to be positive. We’re going to assume that an anticlockwise direction is positive here. And so, to calculate the sum of the moments about vertex 𝐵, we’re going to need to work out which forces we’re actually interested in. Now, we’re going to begin by disregarding this force that acts from 𝐴 to 𝐵. Similarly, we’re going to disregard this force that acts from 𝐵 to 𝐶. This is because the line of action of the force cannot pass through the point about which we’re taking moments. If it does, as in the case of the forces from 𝐴 to 𝐵 and 𝐵 to 𝐶, we say the moment is zero.

And so, we’ll begin by looking at this force here, the force that acts from 𝐴 to 𝐷. This force is perpendicular to the line 𝐴𝐵. And so, we multiply this force by the perpendicular distance. Well, we were told that the square has side length seven centimeters. So, the perpendicular distance from 𝐵 to the point at which the force acts, 𝐴, is seven. And the moment is therefore two multiplied by seven.

We’ll now consider the other forces in our diagram. We’ve got this force here. It acts along the diagonal of the square. It’s still, however, acting at the point 𝐴, so it’s still seven centimeters away from 𝐵. But remember, we’re interested in the component of the force that’s perpendicular to the line 𝐴𝐵. Luckily, since we’re working with the diagonal of a square, we can sketch a right-angle triangle to work out the value of 𝑥. That’s the component of the force four root two newtons that acts perpendicular to the line 𝐴𝐵.

We’ll use the standard conventions for labeling right-angled triangles, and we see we’re looking to calculate the value of the adjacent side, and we know its hypotenuse. The included angle is 45 degrees. And so, we’re going to use the cosine ratio. cos of 𝜃 is adjacent over hypotenuse. So, in the case of this right-angled triangle, cos of 45 is equal to 𝑥 over four root two. We solve for 𝑥 by multiplying through by four root two, and we find 𝑥 is equal to four root two times cos of 45. And since cos of 45 degrees is root two over two, we can say 𝑥 is equal to four root two times root two over two.

Then, we’ll simplify a little by dividing through by two. And since root two times root two is two, 𝑥 is two times two, which is four newtons. And so, the moment of this force is four — remember, that’s the component which acts perpendicular to the line 𝐴𝐵 — times seven, and seven since it’s acting at 𝐴 which is seven centimeters away from 𝐵.

We’re not quite finished. There’s one more force that we’re interested in, and it’s this one here; the force acts at 𝐷. So to find the distance from 𝐵 to 𝐷, we need to calculate the length of the diagonal of our square. Once again, we draw a right-angled triangle, but this time we label it with the side length. We know two sides of our triangle are seven centimeters in length, and we’re looking to find the length of its hypotenuse. And so, we’re going to use the Pythagorean theory.

This says the sum of the squares of the shorter two sides in our right-angled triangle is equal to the square of the longest side. So, seven squared plus seven squared equals 𝑥 squared, or 98 equals 𝑥 squared. We’re then going to find the square root of both sides. And we find 𝑥 is equal to root 98 or seven root two. And so, now we know the distance of 𝐷 from 𝐵. It’s seven root two centimeters.

We need to work out the component of the force from 𝐷 to 𝐶 that acts perpendicular to this diagonal. That’s this force here. By drawing another right-angled triangle, once again, the included angle is 45 degrees. But this time labeling the forces, we see we know the length of the hypotenuse to be four, and we’re looking to find the adjacent. So once again, we use the cosine ratio. This time we get cos of 45 degrees equals 𝑥 over four, so 𝑥 is equal to four cos of 45 degrees or four times root two over two, which is two root two or two root two newtons.

This is still acting in a counterclockwise direction, and so we add on the force multiplied by the distance from 𝐵. That’s two root two times seven root two. Simplifying, we get 14 plus 28 plus 28, which is equal to 70. But what are the units here? Well, our force is measured in newtons, and our distance is measured in centimeters. And so, the algebraic sum of the moments about vertex 𝐵, taking the counterclockwise direction to be positive, is 70 newton centimeters.

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