Question Video: Locating a Point on a Rectangle given the Sum of Moments of Forces About It Mathematics

𝐴𝐡𝐢𝐷 is a rectangle, where 𝐴𝐡 = 6 cm and 𝐡𝐢 = 8 cm, and forces of magnitudes 24, 30, 8, and 30 newtons are acting along 𝐡𝐴, 𝐡𝐢, 𝐢𝐷, and 𝐢𝐴, respectively. If the point 𝐸 ∈ 𝐡𝐢, where the sum of the moments of the forces about 𝐸 is 53 N β‹… cm in the direction of 𝐴𝐡𝐢𝐷, determine the length of 𝐡𝐸.

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Video Transcript

𝐴𝐡𝐢𝐷 is a rectangle, where 𝐴𝐡 equals six centimeters and 𝐡𝐢 equals eight centimeters, and forces of magnitudes 24, 30, eight, and 30 newtons are acting along 𝐡𝐴, 𝐡𝐢, 𝐢𝐷, and 𝐢𝐴, respectively. If the point 𝐸 lies along 𝐡𝐢, where the sum of the moments of the forces about 𝐸 is 53 newton centimeters in the direction of 𝐴𝐡𝐢𝐷, determine the length of 𝐡𝐸.

Okay, so in this situation, we have this rectangle 𝐴𝐡𝐢𝐷. And we’re told that 𝐴𝐡 equals six centimeters and 𝐡𝐢 is eight. Along with this, there are forces that act along the four side lengths of this shape. First, along the side length 𝐴𝐡 but pointed from 𝐡 to 𝐴, we’re told there’s a force of 24 newtons. Next, along side length 𝐡𝐢 pointed from 𝐡 to 𝐢, there’s a 30-newton force. And then along the line from 𝐢 to 𝐷, there’s an eight-newton force acting. We’re then told further that, on a line from corner 𝐢 to corner 𝐴, we can sketch that in like this, there’s another 30-newton force acting.

We’re then told about this point, point 𝐸, which lies along the line segment 𝐡𝐢. We don’t yet know where point 𝐸 is exactly. But just to represent it, let’s say that it’s at this pink dot. Our problem statement tells us that the sum of the moments due to these four forces about point 𝐸 is 53 newton centimeters in the direction of 𝐴𝐡𝐢𝐷. So here’s the idea. Any force whose line of action does not pass through a given axis of rotation will create a moment capital 𝑀 about that axis. The magnitude of 𝑀 equals the component of the force 𝐹 that is perpendicular to a distance 𝑑 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 𝐸. We’re told the magnitude of that moment as well as its direction. The direction of 𝐴𝐡𝐢𝐷 as we’ve drawn it is counterclockwise. And since the given moment in that direction is positive, we’ll say that counterclockwise moments are positive, so clockwise are negative. With all this as introduction, what we’re really after in this question is this line segment here, the distance between 𝐡 and point 𝐸.

To start solving for it, let’s record our total moment about point 𝐸, we’ll call it 𝑀 sub 𝑑, 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 𝑀 sub 𝑑 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 𝑀 sub 𝑑. 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 𝐡𝐸.

So beginning with this moment, 𝑀 sub 𝐡𝐴 is caused by this 24-newton force acting from corner 𝐡 to corner 𝐴. 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 𝐸. Leaving out units then, we can write 𝑀 sub 𝐡𝐴 as 24 multiplied by the line segment 𝐡𝐸.

And then let’s consider the sign of this moment. We see that about point 𝐸, this force would tend to create a clockwise rotation. By our sign convention, that’s a negative moment. This then is our complete expression for the moment created by our 24-newton force about point 𝐸. Considering the next moment, 𝑀 𝐡𝐢, this is the moment created by the 30-newton force from corner 𝐡 to corner 𝐢. We notice though that the line of action of this force passes through point 𝐸. This tells us that our distance 𝑑 in this equation for the moment created by a force is zero. That is, there’s no perpendicular distance between the line of action of this force and the axis of rotation. That tells us that 𝑀 sub 𝐡𝐢 is zero.

Moving on, let’s look at 𝑀 sub 𝐢𝐷. This is the moment about 𝐸 created by this eight-newton force. And here, we once again have a situation where our force is acting perpendicularly to the line segment 𝐡𝐢. 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 𝐸𝐢. But there’s another way we can express this distance. We can call it 𝐡𝐢, the total distance from corner 𝐡 to corner 𝐢, minus 𝐡𝐸, 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 𝐸. Therefore, this moment is positive.

This brings us at last to our final moment. This is the one about point 𝐸 due to this 30-newton force acting on a line from corner 𝐢 to corner 𝐴. The distance from the line of action of this force to our axis of rotation is once again the length of this orange line 𝐡𝐢 minus 𝐡𝐸. But notice that this force is no longer acting perpendicularly to that line. What we’d 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 it’s 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. Here’s what that means. If we call the vertical component of our 30-newton force 𝐹 sub 𝑣, then 𝐹 sub 𝑣 over three is equal to three newtons over five. Or leaving out units, 𝐹 sub 𝑣 equals three-fifths times 30. That’s 18 in units of newtons.

Knowing this, we can start to make some progress in solving for 𝑀 sub 𝐢𝐴. It’s equal to our vertical force component, 18 newtons, multiplied by this distance 𝐡𝐢 minus 𝐡𝐸. Because this force also tends to create a counterclockwise rotation about point 𝐸, this moment is positive. We now have a complete expression for all the moments of our forces. We recall that 𝑀 𝑑 is a known value, while 𝐡𝐸, 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 𝐡𝐸.

Taking this step, our expression becomes 𝐡𝐸 times the quantity negative 24 minus 18 minus eight plus 𝐡𝐢, unknown side length, times the quantity eight plus 18. This gives us negative 50 𝐡𝐸 plus 26 𝐡𝐢. At this point, let’s recall that this expression equals 𝑀 sub 𝑑, which, leaving out the units, is 53. Taking this expression, if we add 50 𝐡𝐸 to both sides and subtract 53 from both, we find that 26 𝐡𝐢 minus 53 equals 50 𝐡𝐸. 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, 𝐡𝐸.

Knowing that line segment 𝐡𝐢 is eight centimeters, we have that 𝐡𝐸 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 𝐡𝐸 is 3.1 centimeters long.

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