In the figure, find the algebraic moment about point 𝑂, given that the force has a magnitude of 14 newtons.
We recall that the moment of a force is a measure of its tendency to cause a body to rotate about a specific point. In order to calculate the moment, we multiply the force by the perpendicular distance from the point at which it is trying to rotate.
In this question, we will begin by calculating the distance 𝑑. We notice that this forms a right triangle and we can therefore use the Pythagorean theorem. The two shorter sides of our triangle have length four root three centimeters and 10 centimeters. Since 𝑑 is the hypotenuse, we have 𝑑 squared is equal to four root three squared plus 10 squared. Four root three squared is 48, and 10 squared is 100. This means that 𝑑 squared is equal to 148. We can then take the square root of both sides. And since 𝑑 must be positive, 𝑑 is equal to the square root of 148. This can be rewritten as root four multiplied by root 37. 𝑑 is therefore equal to two root 37. The perpendicular distance from 𝑂 to where the force acts is two root 37 centimeters.
We know that the force and this distance must be perpendicular. And from our diagram, it appears unlikely that the 14-newton force is perpendicular to this distance. In order to calculate the component of this force that does act perpendicular to the line, we begin by calculating the angle 𝜃. This is the angle between the 14-newton force and the perpendicular force. We will begin by calculating the angle 𝛼 from our right triangle as 𝛼 plus 90 degrees plus 𝜃 plus 30 degrees must equal 180 degrees, as these four angles lie on a straight line. Subtracting 90 degrees and 30 degrees from both sides, we have 𝛼 plus 𝜃 is equal to 60 degrees, which means that 𝜃 is equal to 60 degrees minus 𝛼.
Using our knowledge of the trigonometric ratios in right trigonometry, we see that the opposite side has length four root three centimeters and the adjacent has length 10 centimeters. We know that the tan of 𝛼 is equal to the opposite over the adjacent. So in this question, we have tan 𝛼 is equal to four root three over 10. We can then take the inverse tangent of both sides of this equation. Ensuring that our calculator is in degree mode, we can type this in, giving us 𝛼 is equal to 34.715 and so on degrees.
We can now use this value to calculate angle 𝜃. Subtracting 𝛼 from 60 degrees gives us 𝜃 is equal to 25.2849 and so on degrees. We can now use this to create a right triangle with a 14-newton force. This will enable us to work out the perpendicular component of the force. Once again, we can use the trigonometric ratios. We know the hypotenuse, and we’re trying to calculate the adjacent. We will therefore use the cosine ratio where the cos of 𝜃 is equal to the adjacent over the hypotenuse. The cos of our angle 𝜃 is equal to 𝑦 over 14. We can then multiply through by 14 such that the component of the force we’re looking for is equal to 14 cos 𝜃.
We note that this force acts in a clockwise direction about 𝑂. However, we are told that the positive direction is counterclockwise. The moment of this force about 𝑂 is therefore equal to negative 14 multiplied by the cos of 25.2849 degrees multiplied by two root 37. Ensuring that we used the exact value of the angle on our calculator, this gives us negative 154. The moment of the force about point 𝑂 is negative 154 newton meters.