Video Transcript
A force 𝐅 equals four 𝐢 plus 12𝐣 newtons acts at the point 𝐴 negative four, negative one meters. Calculate the moment 𝐌 zero of this force about the origin and the length of the perpendicular 𝐿 from its line of action to the origin.
As we get started, let’s set up a coordinate frame. Here we have our positive 𝑥-axis pointing off to the right and our positive 𝑦 pointing vertically upward. And on this grid, let’s plot the point 𝐴. We see it has an 𝑥-coordinate of negative four and a 𝑦-coordinate of negative one. The point 𝐴 then is right here. We’re told that from this point a force 𝐅 with given components acts. Since the 𝑥-component of this force is positive four and the 𝑦-component is positive 12, we can say that if we were to sketch in the line of action of this force on our graph, then starting from point 𝐴 for every one unit we move in the positive 𝑥-direction, we’ll move one, two, three units in the positive 𝑦. The line of action of our force then will look like this.
Our question asked us to do two things. We want to calculate the moment 𝐌 sub zero of this force about the origin of our coordinate frame. And we also want to solve for the length of the perpendicular — it’s called 𝐿 — from this line of action we’ve drawn to the origin. Now it turns out that we can actually solve for 𝐿 on the way to solving for the overall moment created by this force. So let’s do that first. Let’s consider what the length is of the perpendicular 𝐿 that goes from our line of action to the origin.
We can start to do that by solving for the equations of our pink dashed line as well as our solid orange line. One general way to write the equation of a line is 𝑦 is equal to 𝑚, the slope of the line, times 𝑥 plus 𝑏, the 𝑦-intercept of the line. Now, if we call the 𝑦-coordinate of our vector’s line of action 𝑦 sub 𝑣, then we know that that coordinate equals some slope, we’ll call it 𝑚 sub 𝑣, times 𝑥 plus some 𝑦-intercept. We’ll call it 𝑏 sub 𝑣. The slope of a line 𝑚 in general is equal to the change in the vertical coordinate of that line over a change in its horizontal coordinate. Because the slope of 𝑚 sub 𝑣 follows the line of action of our force 𝐅, we can say that 𝑚 sub 𝑣 equals the ratio of the 𝑦-value of this force to the 𝑥-value. That is, 𝑚 sub 𝑣 equals three.
Knowing that, we next look to calculate 𝑏 sub 𝑣. This is the 𝑦-intercept of this line. To figure this out, we’re helped by the fact that we know that our line of action passes through point 𝐴. This means we can plug in the 𝑥-coordinate of that point here and the 𝑦-coordinate here. And because this point is on the line, this whole equation must hold true. This implies then that 𝑏 sub 𝑣 equals negative one plus 12 or positive 11. Knowing this, we can now write a completed equation for 𝑦 sub 𝑣. The equation of the line of action of our force is 𝑦 sub 𝑣 equals three 𝑥 plus 11.
Our next goal is to come up with a similar relationship for the solid orange line in our sketch, the one of length capital 𝐿. Because this line is perpendicular to 𝑦 sub 𝑣, its slope will be the negative inverse of the slope of 𝑦 sub 𝑣. Then when we consider what the 𝑦-intercept of this line would be, we can call it 𝑏 sub 𝐿, we recognize that because this line passes through the origin where 𝑦 is zero, its 𝑦-intercept and therefore 𝑏 sub 𝐿 equals zero.
We have completed equations for each of our lines now. What we want to do next is figure out the point in 𝑥𝑦-space where they intersect. At this intersection point, we could say that 𝑦 sub 𝑣 equals 𝑦 sub 𝐿 or in other words three 𝑥 plus 11 equals negative one-third 𝑥. If we work towards solving this equation for 𝑥, we find that ten-thirds 𝑥 equals negative 11. So if we multiply both sides by three-tenths, we find that 𝑥 equals negative 33 over 10. This, then, is the 𝑥-coordinate of the intersection point of our two lines.
To solve for the corresponding 𝑦-value, we can take this 𝑥-value and substitute it back into either of our two lines’ equations. Let’s say we make that substitution here. What we find is that the 𝑦-intercept coordinate, we’ll call it 𝑦 sub 𝑖, equals negative one-third times negative thirty-three tenths. That equals positive eleven-tenths. Now that we know both the 𝑥- and 𝑦-coordinates of our point of intersection, we’re getting very close to being able to solve for this length 𝐿.
𝐿, we see, is the straight line distance from this point of intersection to the origin. In other words, we want to calculate a distance 𝑑. In general, the distance between two points on the 𝑥𝑦-plane with coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two is given by this expression here. In our example, those two points are the origin, with coordinates zero, zero, and our point of intersection. Because the distance we want to solve for is called 𝐿, we can say that 𝐿 equals the square root of the quantity negative thirty-three tenths minus zero squared plus the quantity eleven-tenths minus zero squared. That simplifies to this expression.
And squaring negative thirty-three tenths gives us 1089 over 100, while squaring 11 over 10 gives us 121 over 100. Because one one hundredth is common to both of these terms and 100 equals 10 squared, we can factor out 1 over 10. And if we add 1089 and 121, we get 1210. Then we can notice that 1210 is mathematically equal to 121 times 10. Since 121 equals 11 squared, we can further simplify this result to be 11 over 10 times the square root of 10. This is equal to the length of 𝐿. And recall that this length is in units of meters.
We can say then that the perpendicular distance between our force’s line of action and the origin is eleven-tenths times the square root of 10 meters. Knowing 𝐿, we can come back around, so to speak, and solve for the moment created by this force about the origin. Now the magnitude of that moment will equal the magnitude of our force 𝐅 times 𝐿. The magnitude of the force 𝐅 equals the square root of its 𝑥-component squared plus its 𝑦-component squared. Four squared is 16, and 12 squared is 144. And if we add them together we get 160
160, though, can be written as 16 times 10. And 16 we know is equal to four times four. We can write our moment magnitude, then, as four times the square root of 10 times 11 over 10 times the square root of 10. But then the square root of 10 times the square root of 10 equals 10. And we divide that by 10 giving one. So this whole expression equals four times 11 or 44. In terms of units, we’ve seen that the units of our force are newtons and the units of our distance are meters. So the magnitude of 𝐌 zero equals 44 newton meters.
But then notice that 𝐌 zero in our problem statement, what we really want to solve for, is a vector. We’ve solved for its magnitude, but we’ll need to know its direction as well. To figure this out, let’s return to our sketch. And we know that our force 𝐅 is acting along this line in this direction. We can see then that this force would tend to create a clockwise rotation about the origin. To figure out the direction of this moment, we can imagine a right-handed screw being located at the origin and oriented so that the head of the screw is facing up out of the screen at us. The direction of our moment 𝐌 will be the same direction that our screw travels under the influence of this clockwise rotation.
If we turn a right-handed screw clockwise, then we know it will tend to go from this perspective into the screen. So the question is, what direction is that? Let’s recall that our positive 𝑥-axis points to the right and our positive 𝑦-axis vertically up. In a right-handed coordinate system then, if we were to add in a 𝑧-axis, then the positive 𝑧-direction would be out of the screen at us, which tells us that into the screen the direction that our right-handed screw would travel under the influence of this rotation is in the negative 𝑧-direction. And that, as we said, is the true direction of our moment 𝐌 zero.
So when we write our moment 𝐌 zero as a vector with both magnitude and direction, we would say it has a magnitude of 44 newton meters in the negative 𝑧 — that is, 𝐤 hat — direction. And so we’ve now answered both parts of our question. The moment 𝐌 zero as a vector is negative 44𝐤 hat newton meters, and the perpendicular distance between the line of action of our force and the origin 𝐿 is eleven-tenths times the square root of 10 meters.
