Question Video: Moment of a Force about a Point in 2D Mathematics

Video Transcript

In the given figure, find the magnitude of the moment of the force 115 newtons about the origin point 𝑂 rounded to two decimal places.

Remember, the moment of the force is a measure of its tendency to cause the body to rotate about a specific point. We calculate the moment of a force by multiplying the magnitude of that force by the perpendicular distance from the line of action of the force to the point about which it’s trying to rotate. So drawing a line between 𝑂 and the line of action of the force, let’s begin by calculating this distance. We’ll define it at the moment to be equal to 𝑥 meters. And then we notice that this forms a right-angle triangle.

We can therefore use the Pythagorean theorem to calculate the measurement 𝑥, since 𝑥 forms the hypotenuse of the right triangle 𝑥 squared equals three squared plus 4.5 squared. This right-hand side simplifies to 117 over four. Then we take the positive square root. Now we’re only interested in the positive square root because this is a dimension. The square root of 117 over four is three root 13 over two. So the distance from 𝑂 to the point at which the force acts is three root 13 over two meters.

Now the force and this distance must be perpendicular to one another. And at the moment, we don’t necessarily know that they are. In fact, it’s unlikely that our 115-newton force is acting perpendicular to this distance. So we need to calculate the component of this force that does act perpendicular to this line. And so we need to begin by working out the measurement of 𝜃. This is the angle between the 115-newton force and the component of this force that’s perpendicular to the line. We’ll begin by working out one of the included angles in our triangle.

Let’s label the angle at the bottom of our diagram 𝛼. Then, since we know all three dimensions, but we were given the opposite and adjacent, we can use the tangent ratio to find the value of 𝛼. tan of 𝛼 is opposite over adjacent. So here tan of 𝛼 is three over 4.5. To solve 𝛼, we take the inverse or our tan of both sides of this equation. So 𝛼 is the inverse tan of three over 4.5, which is 33.69 and so on degrees. This means that the angle that the perpendicular component of our force makes with the horizontal is 33.69 degrees. So 𝜃 must be 33.69 minus 30 which is 3.69 degrees.

And this is useful because we can now form a right triangle using our 115-newton force. We want to work out the component of this force that acts perpendicular to the distance between 𝑂 and the point of which our force is acting. So looking at our right triangle, we have an included angle of 3.69 degrees. And we know the hypotenuse and the adjacent. So let’s clear some space and calculate the value of 𝑦, the adjacent side in our triangle.

This time, we use the cosine ratio. cos of 3.69 is equal to 𝑦 over 115. So 𝑦 is 115 times cos of 3.69. We said remember that this was the component of our force that’s perpendicular to the distance. So we’re now ready to calculate the magnitude of the moment about 𝑂.

Now, when calculating moments, we generally work in a counterclockwise direction as being positive. But we’re looking to find the magnitude of the moment here. And this force is trying to rotate the object in a clockwise direction. So let’s define clockwise to be positive. So the moment is the product of the force and distance. It’s three root 13 over two times 115 cos of 3.69. And that’s 620.668 and so on. Correct two decimal places, that’s 620.67. So the magnitude of the moment is 620.67 newton-meters.