Video: Finding the Perpendicular Distance between Two force Vectors Forming a Couple

Given that two forces 𝐹₁ = βˆ’π‘– + 2𝑗 and 𝐹₂ are acting at the points 𝐴 (2, 2) and 𝐡 (βˆ’2, βˆ’2) respectively to form a couple, find the perpendicular distance between the two forces.

06:23

Video Transcript

Given that two forces 𝐹 sub one equals negative 𝑖 plus two 𝑗 and 𝐹 sub two are acting at the points 𝐴: two, two and 𝐡: negative two, negative two, respectively, to form a couple, find the perpendicular distance between the two forces.

We’re told the first force, 𝐹 sub one and its components. And that there’s a second force involved, 𝐹 sub two, that forms a couple with 𝐹 sub one. We’re also told the locations on an π‘₯𝑦-coordinate plane, where the two forces act. 𝐹 sub one acts at point 𝐴. And 𝐹 sub two acts at point 𝐡. We want to solve for the perpendicular distance between these two forces, 𝐹 sub one and 𝐹 sub two. We’ll call that distance 𝑑. To start on our solution, let’s recall the definition of a force couple.

A couple, we recall, is a pair of parallel forces with equal magnitude and opposite direction which do not lie in the same line of action. We’re told that 𝐹 sub one and 𝐹 sub two meet this definition of a force couple. If we go to graph these two forces on an π‘₯𝑦-coordinate plane, we know that 𝐹 sub one acts at point 𝐴 and 𝐹 sub two acts at point 𝐡. Based on the components of 𝐹 one, we can draw it in on our graph. Even though we weren’t given the components of 𝐹 sub two, because it’s a couple with 𝐹 sub one, we can also infer its coordinates and draw it in on our graph.

In this exercise, we want to solve for the perpendicular distance 𝑑 that separates these two lines of action. If we extend the lines along which 𝐹 sub one and 𝐹 sub two act, then, beginning at the origin, we can draw in our line 𝑑 which runs perpendicularly to the two lines of action of the forces.

At this point, we like to be able to write the equation of the line along which our force 𝐹 sub one acts. If we call the equation of that line, 𝑦 sub 𝐹 one, we know its general form will be π‘š times π‘₯ plus 𝑏, where π‘š is the slope of the line and 𝑏 is its 𝑦-intercept. Looking again at our diagram, we can see what the slope of that line is. The change in 𝑦 divided by the change in π‘₯, for the line along which 𝐹 sub one lies, equals two divided by negative one, or negative two. So the slope of our line, 𝑦 sub 𝐹 one, is equal to negative two.

Again, looking at our diagram, we can extrapolate the line along which 𝐹 one lies out to see where it would intercept the 𝑦-axis. Since this line’s π‘₯-intercept is at π‘₯ equals three, that means, in moving the three units to the left it would take to reach π‘₯ equals zero, its vertical change over that distance would be six units. In other words, its 𝑦-intercept is plus six.

Now we want to do the same thing, but for line 𝑑. That is, we want to solve for the equation of that line in the form π‘šπ‘₯ plus 𝑏. As we look at the slope of 𝑦 sub 𝑑, we can recall that because this line is perpendicular to the line along which 𝐹 sub one lies, its slope will be the negative inverse of the slope of the line 𝑦 sub 𝐹 one. That is, the negative inverse of negative two. This is equal to positive one-half.

Looking back at our diagram, we see the way we’ve drawn in line 𝑑 has it passing through the origin. In other words, its 𝑦-intercept is zero. We now have equations, both for the line along which 𝐹 sub one lies and the line perpendicular to that, that passes through the origin. The next thing we like to solve for is the point in space where these two lines intersect.

To solve for the coordinates of that point, we’ll set these two line equations equal to one another. At that point, they are. If 𝑦 sub 𝐹 one is equal to 𝑦 sub 𝑑, that means that negative two π‘₯ plus six is equal to one-half π‘₯. Rearranging this equation, we find that π‘₯ is equal to 12 fifths when these two lines intersect. So we can write in the π‘₯-coordinate of this point and we’re about to solve for the 𝑦-coordinate. When π‘₯ equals 12 fifths, our equation for 𝑦 sub 𝑑 tells us that 𝑦 is one-half that, or six divided by five.

Now, we’re really doing well. We have the point at which 𝑦 sub 𝐹 one and 𝑦 sub 𝑑 intersect. And we also have another point on the line 𝑑, which is the origin. We can see if we calculate the distance between those two points, that will be one half the total distance 𝑑. As we set out to calculate 𝑑, let’s recall that the distance between two points in the π‘₯𝑦-plane, if we call that capital 𝐷, is equal to the square root of the change in π‘₯ squared plus the change in 𝑦 squared. So in our case, 𝑑 over two is equal to the square root of 12 over five minus zero, that’s the change in π‘₯ quantity squared, plus six over five minus zero, the change in 𝑦 quantity squared. And removing the zeros, we have this simplified form for the equation.

If we multiply both sides of our equation by two and then factor out one over five squared to outside the square root sign, we see that 𝑑 equals two divided by five times the square root of 12 squared plus six squared. 12 squared plus six squared is 180, which we can also write as 36 times five, or six times six times five. Factoring a six outside of the square root sign, we find that 𝑑 is equal to 12 divided by five times the square root of five units of length. That’s the perpendicular distance between the lines of action of these two forces.

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