Question Video: Finding the Moment of a Couple Equivalent to Three Couples Acting on a Square | Nagwa Question Video: Finding the Moment of a Couple Equivalent to Three Couples Acting on a Square | Nagwa

# Question Video: Finding the Moment of a Couple Equivalent to Three Couples Acting on a Square Mathematics • Third Year of Secondary School

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𝐴𝐵𝐶𝐷 is a square having a side length of 85 cm. Forces of magnitudes 30, 55, 30, and 55 newtons are acting along the square’s sides, and two equal forces of magnitude 25√2 newtons are acting at 𝐴 and 𝐶 in the directions shown in the figure. Find the couple equivalent to the system.

05:20

### Video Transcript

𝐴𝐵𝐶𝐷 is a square having a side length of 85 centimeters. Forces of magnitudes 30, 55, 30, and 55 newtons are acting along the square’s sides, and two equal forces of magnitude 25 root two newtons are acting at 𝐴 and 𝐶 in the directions shown in the figure. Find the couple equivalent to the system.

Alright, so looking at our square 𝐴𝐵𝐶𝐷, we see a number of forces acting on it. The idea here is that all of these forces, and there are six of them, are equivalent to a couple. That means they have the same effect on the square as two equal and opposite forces that don’t lie along the same line of action. In solving for the couple equivalent to this system of six forces, we’ll be solving for the effect that couple would have on the square. Knowing that the length of each side of the square is 85 centimeters, let’s clear some space on screen and start analyzing the forces acting on our square, knowing that they’re equivalent to a couple.

A moment ago, we mentioned this term lines of action. This is the line along which a given force acts. So, for example, the line of action of this 55-newton force here would look like this. It lies along this side of the square. If we sketch in the lines of action for all of our forces, then we see that this line, this line, and this line all intersect at corner 𝐴 of our square. Likewise, the line of action of this 55-newton force, this 30-newton force, and this 25 root two newton force intersect at point 𝐶. Thinking of these forces altogether being equivalent to a couple, we can say that all the forces whose lines of action intersect at point 𝐴 effectively originate at that point.

This means that if we draw in point 𝐴, we can say that the 30-newton force moves like this from that point, the 55-newton force points like this, and the 25 root two newton force like this. We could say then that these three forces exert their influence on the square from point 𝐴. The same thing is true for the three forces whose lines of action intersect at point 𝐶. In this case, we have our 30-newton force acting down, our 55-newton force acting to the right, and our 25 root two newton force acting up and to the right. Notice that the forces originating at point 𝐶 are equal and opposite to those originating at point 𝐴. This confirms to us that, indeed, all six of these forces considered together are equivalent to a couple.

To calculate the effect of this couple on our square, we locate the midpoint of the square, which would be here relative to 𝐴 and 𝐶. And we’ll calculate the total moment created about this point by our couple of forces. To do that, we’ll consider these three pairs of forces one by one. First, we’ll consider the 30-newton forces that act up and down on opposite sides of the square. If our square looks like this, then the lines of action of the 30-newton forces would look this way, meaning that each one has a perpendicular distance to the center of the square of 85 over two centimeters.

To start solving then for the total moment created by this effective couple of forces, we can write down the contribution of these two 30-newton forces to our overall moment. Notice that this contribution is negative because these two forces would tend to create a clockwise moment about our center. And, by convention, rotation in the counterclockwise direction is considered positive. So our two 30-newton forces contribute a negative overall moment of this magnitude. Next, we’ll look at the contribution to the overall moment of our 55-newton forces. Once again, the perpendicular distance between these two forces’ lines of action and the center of our square is 85 over two centimeters. This time, though, these forces tend to create a moment in the counterclockwise direction, which is therefore positive.

We’ve now accounted for two of our three force pairs. The last to consider is the pair with our 25 root two newton forces. For these forces and their lines of action, the perpendicular distance between these two lines is equal to the diagonal of our square. Because all the interior angles of a square are 90 degrees, we can use the fact that each of our side lengths is 85 centimeters along with the Pythagorean theorem to see that the length of the hypotenuse of this triangle, and therefore the diagonal of our square, equals the square root of 85 squared plus 85 squared. We can write this simply as the square root of two times 85 squared.

Once we know that, we can say that the perpendicular distance between the lines of action of our two 25 root two newton forces and the center of our square is one-half the square root of two times 85 squared. And we see that these forces also tend to create a counterclockwise rotation for our square and therefore a positive moment. What we have then for the overall movement is this expression, which we’ve seen is created by six forces that are equivalent to a couple.

Before we go to evaluate this expression, we can see that some cancellation occurs. For example, the leading factors of two in each term cancel with a factor of one-half. Altogether, we get a net moment of 6375. And recall here that we’re multiplying forces in units of newtons by distances in units of centimeters. This then is the couple equivalent of the forces acting on our square.

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