### Video Transcript

π΄π΅πΆπ· is a rectangle, where π΄π΅ equals six centimeters and π΅πΆ equals eight centimeters, and forces of magnitudes 24, 30, eight, and 30 newtons are acting along π΅π΄, π΅πΆ, πΆπ·, and πΆπ΄, respectively. If the point πΈ lies along π΅πΆ, where the sum of the moments of the forces about πΈ is 53 newton centimeters in the direction of π΄π΅πΆπ·, determine the length of π΅πΈ.

Okay, so in this situation, we have this rectangle π΄π΅πΆπ·. And weβre told that π΄π΅ equals six centimeters and π΅πΆ is eight. Along with this, there are forces that act along the four side lengths of this shape. First, along the side length π΄π΅ but pointed from π΅ to π΄, weβre told thereβs a force of 24 newtons. Next, along side length π΅πΆ pointed from π΅ to πΆ, thereβs a 30-newton force. And then along the line from πΆ to π·, thereβs an eight-newton force acting. Weβre then told further that, on a line from corner πΆ to corner π΄, we can sketch that in like this, thereβs another 30-newton force acting.

Weβre then told about this point, point πΈ, which lies along the line segment π΅πΆ. We donβt yet know where point πΈ is exactly. But just to represent it, letβs say that itβs at this pink dot. Our problem statement tells us that the sum of the moments due to these four forces about point πΈ is 53 newton centimeters in the direction of π΄π΅πΆπ·. So hereβs the idea. Any force whose line of action does not pass through a given axis of rotation will create a moment capital π about that axis. The magnitude of π equals the component of the force πΉ that is perpendicular to a distance π between where the force is applied and the axis of rotation.

All this to say, when we consider the four forces acting on a rectangle, in sum, they do create a moment about point πΈ. Weβre told the magnitude of that moment as well as its direction. The direction of π΄π΅πΆπ· as weβve drawn it is counterclockwise. And since the given moment in that direction is positive, weβll say that counterclockwise moments are positive, so clockwise are negative. With all this as introduction, what weβre really after in this question is this line segment here, the distance between π΅ and point πΈ.

To start solving for it, letβs record our total moment about point πΈ, weβll call it π sub π‘, and then clear some space on screen to start working. Knowing that there are four forces acting on a rectangle, we can say that this total moment is the sum of the moments from each one. Writing this as an equation, we can say that π sub π‘ equals the sum of these four moments where each of these is created by a force acting along this given line segment. We already know π sub π‘. And one by one, we can solve for these other four moments. In the process, that will let us solve for the length of line segment π΅πΈ.

So beginning with this moment, π sub π΅π΄ is caused by this 24-newton force acting from corner π΅ to corner π΄. We can write an equation for the magnitude of this moment using this general expression here. And note that, in this instance, our force of 24 newtons is already acting perpendicularly to the distance between the line of action of that force and our axis of rotation, point πΈ. Leaving out units then, we can write π sub π΅π΄ as 24 multiplied by the line segment π΅πΈ.

And then letβs consider the sign of this moment. We see that about point πΈ, this force would tend to create a clockwise rotation. By our sign convention, thatβs a negative moment. This then is our complete expression for the moment created by our 24-newton force about point πΈ. Considering the next moment, π π΅πΆ, this is the moment created by the 30-newton force from corner π΅ to corner πΆ. We notice though that the line of action of this force passes through point πΈ. This tells us that our distance π in this equation for the moment created by a force is zero. That is, thereβs no perpendicular distance between the line of action of this force and the axis of rotation. That tells us that π sub π΅πΆ is zero.

Moving on, letβs look at π sub πΆπ·. This is the moment about πΈ created by this eight-newton force. And here, we once again have a situation where our force is acting perpendicularly to the line segment π΅πΆ. Now, the distance by which we want to multiply this force magnitude is given by the length of this orange line here. We could call this line segment πΈπΆ. But thereβs another way we can express this distance. We can call it π΅πΆ, the total distance from corner π΅ to corner πΆ, minus π΅πΈ, the length we want to solve for. And then, as we think about the sign of this moment, notice that this eight-newton force tends to create a counterclockwise rotation about πΈ. Therefore, this moment is positive.

This brings us at last to our final moment. This is the one about point πΈ due to this 30-newton force acting on a line from corner πΆ to corner π΄. The distance from the line of action of this force to our axis of rotation is once again the length of this orange line π΅πΆ minus π΅πΈ. But notice that this force is no longer acting perpendicularly to that line. What weβd like to do then is break this 30-newton force up into its component parts and then use this vertical component of that force in our calculation. The question then is, what is that vertical component?

Well, we know that this triangle here is a right triangle. And we also know that itβs similar to this much larger right triangle that forms the top-right half of our rectangle. For that larger triangle, we know the ratio of the horizontal and vertical side lengths. That same ratio holds true for our similar orange triangle. In terms of relative side lengths then, we can say that this side of our orange triangle is four units long compared to three units long for this side.

We say this because if we can find the relative side length of the hypotenuse of this triangle, which we know corresponds to a force of 30 newtons, then we can use ratios to solve for the vertical component of that force. Since this is a right triangle and the two shorter sides have lengths three and four, we can identify this as a three-four-five triangle. Hereβs what that means. If we call the vertical component of our 30-newton force πΉ sub π£, then πΉ sub π£ over three is equal to three newtons over five. Or leaving out units, πΉ sub π£ equals three-fifths times 30. Thatβs 18 in units of newtons.

Knowing this, we can start to make some progress in solving for π sub πΆπ΄. Itβs equal to our vertical force component, 18 newtons, multiplied by this distance π΅πΆ minus π΅πΈ. Because this force also tends to create a counterclockwise rotation about point πΈ, this moment is positive. We now have a complete expression for all the moments of our forces. We recall that π π‘ is a known value, while π΅πΈ, this line segment length, is what we want to solve for. What we can do then is start to collect terms that include this value π΅πΈ.

Taking this step, our expression becomes π΅πΈ times the quantity negative 24 minus 18 minus eight plus π΅πΆ, unknown side length, times the quantity eight plus 18. This gives us negative 50 π΅πΈ plus 26 π΅πΆ. At this point, letβs recall that this expression equals π sub π‘, which, leaving out the units, is 53. Taking this expression, if we add 50 π΅πΈ to both sides and subtract 53 from both, we find that 26 π΅πΆ minus 53 equals 50 π΅πΈ. And then, lastly, dividing both sides by 50, canceling that factor on the right, we have an expression for the length we want to find, π΅πΈ.

Knowing that line segment π΅πΆ is eight centimeters, we have that π΅πΈ equals 26 times eight minus 53 all divided by 50. Entering this expression on our calculator, it comes out to 3.1. This result is a distance, and it has units of centimeters. This then is our final answer that line segment π΅πΈ is 3.1 centimeters long.