Video: Finding the Resultant of Three Forces in Two Dimensions

The diagram shows a system of three forces measured in newtons. Given that 𝐴𝐡 = 24 cm and 𝐴𝐷 = 18 cm, determine R, the magnitude of their resultant, and find πœƒ, the angle between their resultant and the positive π‘₯-axis. Give to the nearest minute.

04:51

Video Transcript

The diagram shows a system of forces measured in newtons. Given that 𝐴𝐡 is equal to 24 centimeters and 𝐴𝐷 is equal to 18 centimeters, determine 𝑅, the magnitude of their resultant, and find πœƒ, the angle between their resultant and the positive π‘₯-axis. Give to the nearest minute.

We’re dealing with three forces here. Let’s call these 𝐅 one, 𝐅 two, and 𝐅 three. We know that we’re working with two directions, so we’re going to label these as vector forces. We’re also going to define the horizontal unit vector 𝐒 and the vertical unit vector 𝐣. This means we can call 𝐅 one negative 40𝐒. Now, we’ve defined it as negative 40 because it’s acting in the negative direction. Similarly, 𝐅 two is acting downwards, so it’s negative 30𝐣. But what do we do about force three? We know its magnitude is 24 newtons. We’re going to need to split it into both its horizontal and vertical components.

Now, to do so, we’ll use the fact that 𝐴𝐡 is equal to 24 centimeters and 𝐴𝐷 is 18 centimeters. Now, of course, if 𝐴𝐷 is 18 centimeters, 𝐡𝐢 must also be 18 centimeters, so we draw a little right-angle triangle as shown. We’re essentially looking to find the ratio between the hypotenuse and the other two sides. And so, let’s call this angle πœƒ. We have the opposite side of this triangle as 18 centimeters and the adjacent as 24. And so, we use the tan ratio. We get tan πœƒ equals opposite over adjacent, 18 over 24, which simplifies to three-quarters.

We know the triangle stays in proportion with force three, so we relabel our triangle with the hypotenuse as being 24 newtons. This time, let’s label the adjacent π‘₯ and the opposite 𝑦. We can use the cos ratio to find expressions linking πœƒ and π‘₯. When we do, we get cos πœƒ is adjacent over hypotenuse, π‘₯ over 24. And so, π‘₯ is 24 cos πœƒ. Similarly, we also find that sin πœƒ is 𝑦 over 24. And we can rearrange and write 𝑦 equals 24 sin πœƒ.

Now, we go back to the fact that tan πœƒ is equal to three-quarters. In this case, we actually have a Pythagorean triple. We know that if the opposite is three centimeters or three units, the adjacent is four units, then the hypotenuse is five units. Then, cos πœƒ is four-fifths; it’s adjacent over hypotenuse. And sin πœƒ is three-fifths; it’s opposite over hypotenuse. So, we find π‘₯ is equal to 24 times four-fifths. And 𝑦 is equal to 24 times three-fifths. This gives us ninety-six fifths and seventy-two fifths as the horizontal and vertical components, respectively. And since both the horizontal and vertical components are acting in the positive direction, force three is 96 over five 𝐒 plus 72 over five 𝐣.

We’re now going to clear some space for the next step. We’re told to find 𝑅, the magnitude of the resultant of these forces. Now, the resultant is the vector sum of all forces. So, we can say that the resultant vector is 𝐅 one plus 𝐅 two plus 𝐅 three. To add the vectors, we add their horizontal and vertical components. Well, the horizontal components are negative 40 and 96 over five. And the vertical components are negative 30 and 72 over five. This simplifies to negative one hundred and four fifths 𝐒 minus seventy-eight fifths 𝐣. And so, the resultant force acts in this direction shown. In the horizontal direction, it’s one hundred and four fifths newtons. And in the vertical direction, it’s seventy-eight fifths newtons.

The magnitude of the resultant is basically the length of the vector, so we use the Pythagorean theorem. It’s the square root of seventy-eight fifths squared plus one hundred and four fifths squared. That’s 26 or 26 newtons. Now, we’re also looking to find πœƒ, the angle between the resultant and the positive π‘₯-axis. So, we’ll begin by finding the angle 𝛼. That’s the angle that 𝑅 makes with the negative π‘₯-axis. And so, we’re going to use the tan ratio. Tan 𝛼 is seventy-eight fifths over one hundred and four fifths. That simplifies to 78 over 104. And therefore, 𝛼 is the inverse tan of 78 over 104, which is 36.86 and so on degrees.

Now, of course, we’re measuring this from the positive π‘₯-axis, and so we as usual move in a counterclockwise direction. We therefore need to add 180 degrees to our value for 𝛼. πœƒ is 180 plus 36.86, which is 216.86 and so on. Correct to the nearest minute, that’s 216 degrees and 52 minutes.

And so, we found the magnitude of the resultant of these forces 𝑅 to be 26 newtons and the angle πœƒ it makes with the positive π‘₯-axis to be 216 degrees and 52 minutes.

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