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Question Video: Finding the Magnitude of the Moment Formed by a System of Forces Mathematics

𝐴𝐡𝐢𝐷 is a rectangle, in which 𝐴𝐡 = 45 cm, 𝐡𝐢 = 55 cm, and 𝐷𝐸 = 28 cm. Forces of magnitudes 225, 275, 265, and 135 newtons act along 𝐴𝐡, 𝐡𝐢, 𝐢𝐸, and 𝐸𝐴 respectively. If the system of forces is equivalent to a couple, determine the magnitude of the moment of the forces.

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Video Transcript

𝐴𝐡𝐢𝐷 is a rectangle, in which 𝐴𝐡 equals 45 centimeters, 𝐡𝐢 equals 55 centimeters, and 𝐷𝐸 equals 28 centimeters. Forces of magnitudes 225, 275, 265, and 135 newtons act along 𝐴𝐡, 𝐡𝐢, 𝐢𝐸, and 𝐸𝐴 respectively. If the system of forces is equivalent to a couple, determine the magnitude of the moment of the forces.

Okay, so here we have this rectangle 𝐴𝐡𝐢𝐷. And we’re told that side length 𝐴𝐡 equals 45 centimeters, 𝐡𝐢 is 55 centimeters, and that side length 𝐷𝐸 is 28 centimeters. We’re Also told the magnitude of four forces acting on this rectangle and that this system of forces is equivalent to a couple. This means we can effectively replace the four forces with two equal and opposite forces that don’t lie along the same line of action. Knowing all this, we want to determine the magnitude of the moment of these four forces.

To start doing that, let’s clear a bit of space on screen. And let’s consider what it means that these four forces are equivalent to a couple. It means if we were to draw the lines of action of these four forces, then if we were to pick two intersection points of these lines so that these points included all four lines of action, then we could effectively model all four forces as though they originated from those two points. So for example, if we pick the points 𝐢 and 𝐴 in our rectangle, then from point 𝐴 we can consider our 225-newton force to be acting down and our 135-newton force to be acting to the right, likewise with point 𝐢 where we say our 275-newton force acts to the left and our 265-newton force acts up and to the right. Because our system of forces is equivalent to a couple, we can say that the total force acting at point 𝐴 is equal and opposite that total force acting at point 𝐢.

And actually we can go further than this. Since forces can be separated into independent vertical and horizontal components, we can say that the net horizontal force at point 𝐴 is equal and opposite that at point 𝐢, and similarly for the net vertical force at these two points. As a side note, we didn’t need to use the points 𝐴 and 𝐢 as a model for the origins of our four forces. Taken together, these two points meet our condition of overlapping all four lines of action, but, looking back to our rectangle, so did the points, say, 𝐸 and 𝐡. Either of these pairs of points would work for analyzing this scenario.

But anyway, knowing that the total forces at 𝐢 and 𝐴 form a couple, our goal is to solve for the moment created by that couple. We can recall that the moment created by a couple of forces is equal to two times the force component perpendicular to the distance between where the force is applied and where the axis of rotation is located. The idea here is that each of the forces in the couple contributes equally to the moment overall. That’s why if we solve for this perpendicular component of one of those two forces and then multiply it by the distance 𝑑, we need only multiply that result by two to solve for the overall moment.

Using 𝐴 and 𝐢 as the points of origin of the two forces in our couple, we can think of those forces as acting on a line like this that joins at the two points. So the moment we’re solving for acts about this midpoint of that line. In our rectangle, that line and midpoint would look like this. So here’s the idea. If we can find the net force that’s perpendicular to this line acting either at point 𝐴 or point 𝐢, then we will have solved for 𝐹 perpendicular in this equation. We could choose either point 𝐢 or point 𝐴 at which to solve for this force.

And since the forces at point 𝐴 act in what we could call the purely vertical and purely horizontal directions, let’s choose that point. Our goal then, is to solve for the components of the 225- and 135-newton forces that are perpendicular to this dashed orange line. Adding those together will give us 𝐹 perpendicular here. Looking at the right triangle created by our 225-newton force, we can call this interior angle of that triangle πœƒ sub one. We’re interested in this angle because the sin of πœƒ sub one times 225 equals the perpendicular component of this force.

Now, if we go back to our original diagram, this angle in that diagram is also equal to πœƒ sub one. And we see that indeed this is an interior angle in this right triangle drawn in orange. The shorter two sides of the triangle have lengths of 45 and 55 centimeters, respectively. And by the Pythagorean theorem, we can say then that the hypotenuse has a length of the square root of 45 squared plus 55 squared. That’s equal to the square root of 5050.

At this point, we can recall that given a right triangle where πœƒ is another interior angle of that triangle, then the sin of πœƒ equals the ratio of the opposite side length to the hypotenuse. And that means when it comes to our angle of interest, the sin of πœƒ sub one, this is equal to 55 divided by the square root of 5050. For πœƒ sub one, that’s the opposite side length to the hypotenuse length ratio. So then we now have an expression for the perpendicular component of this 225-newton force. And we can start to keep a tally of this using a variable we’ll call 𝐹 perpendicular. This is exactly that variable we see in our equation for 𝑀 sub c. And we now know that it’s equal to this term plus a second term we’ll figure out soon.

That second term is equal to this component of our 135-newton force. To solve for it, we can use a similar approach. Let’s say that this angle interior in this right triangle is πœƒ sub two. On our rectangle, the angle πœƒ sub two would look like this. And notice that this is the same as this angle here in our triangle. Once more, to calculate the component of this force perpendicular to the dashed line, we’ll want to use the sin of this angle. We want to calculate 135 times the sin of πœƒ sub two. The sin of this angle equals the opposite side length, 45 centimeters, divided by the hypotenuse.

We now know the component of the 135-newton force that’s perpendicular to our dashed orange line. This then is the second and final term in our equation for 𝐹 perpendicular. All right, so to find 𝑀 sub c, the only thing that remains for us now is to solve for this distance 𝑑 and then multiply it by 𝐹 perpendicular and two. Looking at our original diagram, that distance is half the distance from point 𝐢 to point 𝐴. In other words, it’s this distance here. We see right away that that equals one-half, our hypotenuse length. In other words, 𝑑 is equal to the square root of 5050 over two.

Knowing all this, we’re now ready to calculate 𝑀 sub c. Plugging in the values we calculated for 𝐹 perpendicular and 𝑑, notice that this leading factor of two will cancel out with this denominator of two and that also this denominator of the square root of 5050 that appears in both terms in 𝐹 perpendicular will be multiplied by that same value and therefore become equal to one. We find then that 𝑀 sub c equals 225 times 55 plus 135 times 45. And this equals 18450. We recall that the units of our forces are newtons and the units of our distances in this example are centimeters. The moment then created by this system of forces, which is equivalent to a couple, equals 18450 newton centimeters.

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