The diagram shows a system of three forces measured in newtons. Given that 𝐴𝐵 is equal to 24 centimeters and 𝐴𝐷 equals 18 centimeters, determine 𝑅, the magnitude of their resultant, and find 𝜃, the angle between their resultant and the positive 𝑥-axis. Give the answer rounded to the nearest minute.
In this question, we’re dealing with three forces which we will call vector 𝐅 sub one, 𝐅 sub two, and 𝐅 sub three, as shown. These forces are acting in two dimensions, and we will define the horizontal unit vector 𝐢 and the vertical unit vector 𝐣. This means that we call vector 𝐅 sub one negative 40𝐢, as it acts in the negative 𝑥-direction. In the same way, as 𝐅 sub two is acting vertically downwards, it is equal to negative 30𝐣.
But what about 𝐅 sub three? We know it has a magnitude of 24 newtons. And we will need to split this into both its horizontal and vertical components. To do this, we will use the fact that 𝐴𝐵 is equal to 24 centimeters and 𝐴𝐷, and therefore 𝐵𝐶, is equal to 18 centimeters. We can draw a right triangle as shown and use this to find the ratio between the hypotenuse and the other two sides. We will let the angle at 𝐴 be 𝛼, and we have the opposite side equal to 18 centimeters and the adjacent equal to 24 centimeters.
Using the tangent ratio, we have tan 𝛼 is equal to 18 over 24. And this simplifies to three-quarters. We know that the triangle stays in proportion with the force 𝐅 sub three. So, we relabel the triangle with a hypotenuse as 24 newtons. This time, let’s label the adjacent side 𝑥 and the opposite side 𝑦. Using the cosine ratio, we have cos 𝛼 is equal to 𝑥 over 24. It’s the adjacent over the hypotenuse.
In the same way, the sine ratio tells us that sin 𝛼 is equal to 𝑦 over 24. We can multiply through by 24 in both equations such that 𝑥 is equal to 24 cos 𝛼 and 𝑦 is equal to 24 sin 𝛼. Recalling that tan 𝛼 is equal to three-quarters, we know that we have a Pythagorean triple. This means that cos 𝛼 is equal to four-fifths and sin 𝛼 is equal to three-fifths.
The horizontal and vertical components of our force are therefore equal to ninety-six fifths and seventy-two fifths, respectively. And since both of these are acting in the positive direction, 𝐅 sub three is equal to 96 over five 𝐢 plus 72 over five 𝐣. We now have expressions for the three forces in terms of the unit vectors 𝐢 and 𝐣.
After clearing some space, we know that the first thing we are trying to calculate is 𝑅, the magnitude of their resultant. The resultant is equal to the sum of all the forces. So, 𝐑 is equal to 𝐅 sub one plus 𝐅 sub two plus 𝐅 sub three. To add the three vectors, we add their horizontal and vertical components. The horizontal components are negative 40 and 96 over five, whereas the vertical components are negative 30 and seventy-two over five. The resultant force is therefore equal to negative 104 over five 𝐢 minus 78 over five 𝐣.
The resultant force therefore acts in the direction shown. In the horizontal direction, it’s 104 over five newtons, and in the vertical direction, 78 over five newtons. The magnitude of this resultant is the length of the vector. And using the Pythagorean theorem, this is equal to the square root of 104 over five squared plus 78 over five squared. Typing this into our calculator gives us an answer of 26. We can therefore conclude that 𝑅 is equal to 26 newtons.
We are also asked to find 𝜃, the angle between this resultant and the positive 𝑥-axis. We will begin by finding the angle 𝛽, which is the angle that the resultant makes with the negative 𝑥-axis. Using the tangent ratio, we have tan 𝛽 is equal to 78 over five divided by 104 over five. That simplifies to 78 over 104. And 𝛽 is therefore equal to the inverse tangent of 78 over 104. Ensuring that our calculator is in degree mode, we can type this in, giving us 𝛽 is equal to 36.869 and so on degrees.
Now we’re measuring 𝜃 from the positive 𝑥-axis, and as usual we move in a counterclockwise direction. This means that we add 180 degrees to our value of 𝛽. 𝜃 is therefore equal to 216.869 and so on degrees. Since we want our answer to the nearest minute, we multiply the decimal part of our answer by 60, giving us 𝜃 is equal to 216 degree 52 minutes to the nearest minute.
We now have answers to both parts of the question. The magnitude of the resultant 𝑅 is equal to 26 newtons, and the angle 𝜃 between the resultant and the positive 𝑥-axis is 216 degrees and 52 minutes.