The portal has been deactivated. Please contact your portal admin.

Question Video: Analysis of Two Coplanar Forces Acting Together to Produce a Resultant Mathematics

Two forces of magnitudes 15 kg-wt and 7 kg-wt act as shown in the diagram. Determine 𝑅, the magnitude of the resultant, and find πœƒ, the measure of the angle between the resultant and due east.

03:37

Video Transcript

Two forces of magnitudes 15 kilogram-weight and seven kilogram-weight act as shown in the diagram. Determine 𝑅, the magnitude of the resultant, and find πœƒ, the measure of the angle between the resultant and due east.

And then we have a diagram that shows the two forces and the angle between them. So let’s begin by reminding ourselves what we mean when we talk about the resultant of a pair of forces. Suppose we have two forces defined by the vectors 𝐅 sub one and 𝐅 sub two. The resultant of these two forces is simply their sum. And thinking about these in terms of vectors can really help us plan what to do next.

Let’s define the force with magnitude 15 kilogram-weight to be the vector 𝐅 sub one and the other force to be the vector 𝐅 sub two. The sum of these two vectors can be represented with the vector of magnitude 𝑅 as shown. This is a triangle of forces. And we can use right triangle and non-right triangle trigonometry to help us find any missing dimensions.

Let’s begin by finding the measure of the angle that we can label 𝛼 between the seven-newton and 15-newton force. We know that the 𝛼 and 30-degree angle sum to 90 degrees. So if we subtract 30 from both sides, we know that 𝛼 must be equal to 60. Then, we now have a non-right triangle for which we know two of its lengths and the angle between those. So we can use the law of cosines to find the measure of the third side. That is, π‘Ž squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. We then label the side that we’re trying to find lowercase π‘Ž and the angle opposite it, the 60-degree angle, uppercase 𝐴. So 𝑅 squared is seven squared plus 15 squared minus two times seven times 15 times cos 60. Seven squared plus 15 squared is 274, and cos of 60 is one-half. So the second part of this is simply seven times 15, which is 105. 274 minus 105 is 169. So our equation is 𝑅 squared equals 169.

Since 𝑅 is a magnitude, it has to be positive. So we can solve this equation by simply taking the positive root of 169, which is 13. So the magnitude of the resultant in this case is 13 kilogram-weight.

So we’re now ready to calculate the angle πœƒ. On our diagram, it’s the measure of the angle between the resultant and due east. Well, in our diagram, the 15-newton force is acting due east. So the angle πœƒ is the angle that the side that we calculated to be 13 makes with the side of 15. We can use the law of sines to calculate this angle. In this case, that’s sin 𝐴 over π‘Ž equals sin 𝐡 over 𝑏. We’ve just calculated 𝑅 to be 13. So that’s sin 60 divided by 13 equals sin πœƒ over seven. sin 60 is root three over two. So when we multiply through by seven, we find that sin πœƒ is seven root three over 26.

To solve for πœƒ, we take the inverse or arcsine of both sides of the equation. So πœƒ is inverse sin of seven root three over 26, which is 27.795 and so on. Next, we’ll multiply the decimal part by 60 so we can get this to the nearest minute. That gives us a minute part of 47.7, which is 48 correct to the nearest minute. And so we found 𝑅 and πœƒ. 𝑅 is 13 kilogram-weight, and πœƒ is 27 degrees and 48 minutes.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.