Question Video: Finding the Magnitudes of Two Unknown Forces out of Group of Forces given the Free Body Diagram | Nagwa Question Video: Finding the Magnitudes of Two Unknown Forces out of Group of Forces given the Free Body Diagram | Nagwa

Question Video: Finding the Magnitudes of Two Unknown Forces out of Group of Forces given the Free Body Diagram Mathematics • Second Year of Secondary School

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Forces of magnitude 𝐅, 16, 𝐊, 18, 9√3 newtons act at a point in the directions shown on the diagram. Their resultant, 𝑅, has a magnitude of 20 N. Find the values of 𝐅 and 𝐊.

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Video Transcript

Forces of magnitude 𝐅, 16, 𝐊, 18, nine root three newtons act at a point in the directions shown on the diagram. Their resultant, 𝑅, has a magnitude of 20 newtons. Find the values of 𝐅 and 𝐊.

We will begin by finding the sum of the forces in the π‘₯- and 𝑦-directions, denoted 𝑅 sub π‘₯ and 𝑅 sub 𝑦, where the positive π‘₯-direction is to the right and the positive 𝑦-direction is up. Four of the forces act in either the π‘₯- or 𝑦-direction. The force 𝐅 acts in the positive π‘₯-direction. The 18-newton force acts in the negative π‘₯-direction. The force 𝐊 acts in the positive 𝑦-direction. And finally, the nine root three newton force acts in the negative 𝑦-direction. The 16-newton force acts in neither the horizontal nor vertical direction. By creating a right triangle as shown, we will be able to find its horizontal and vertical components. Both of these act in the positive direction, and we will call them π‘₯ and 𝑦.

By considering the 30-degree angle given in the diagram, we can use our knowledge of right angle trigonometry. We know that the sin of any angle πœƒ is equal to the opposite over the hypotenuse. This means that from our diagram, the sin of 30 degrees is equal to π‘₯ over 16. We know that the sin of 30 degrees is one-half. And multiplying through by 16, we have π‘₯ is equal to eight. The horizontal component of the 16-newton force is eight newtons.

We also know that the cos of any angle πœƒ is equal to the adjacent over the hypotenuse. This means that the cos of 30 degrees is equal to 𝑦 over 16. And as the cos of 30 degrees is root three over two, 𝑦 is equal to eight root three. The vertical component of the 16-newton force is positive eight root three.

We now have expressions for 𝑅 sub π‘₯ and 𝑅 sub 𝑦. These can be simplified so that 𝑅 sub π‘₯ is equal to 𝐅 minus 10 and 𝑅 sub 𝑦 is equal to 𝐊 minus root three. At this stage, we don’t have enough information to calculate the values of 𝐅 and 𝐊. However, we are told that the magnitude of the resultant is 20 newtons, and this acts at an angle at 30 degrees to the positive π‘₯-axis. We can therefore calculate the values of 𝑅 sub π‘₯ and 𝑅 sub 𝑦 directly from the diagram.

Once again, we have a right triangle. 𝑅 sub 𝑦 is equal to 20 multiplied by the sin of 30 degrees. This means that 𝑅 sub 𝑦 is equal to 20 multiplied by one-half. This is equal to 10 newtons. Likewise, 𝑅 sub π‘₯ is equal to 20 multiplied by the cos of 30 degrees. This is equal to 20 multiplied by root three over two, which is 10 root three.

Substituting these values into our equations, we have 10 root three is equal to 𝐅 minus 10 and 10 is equal to 𝐊 minus root three. We can then solve these equations to find the values of 𝐅 and 𝐊. 𝐅 is equal to 10 root three plus 10, and 𝐊 is equal to 10 plus root three. In order for the five forces to have a resultant of magnitude 20 newtons as shown, 𝐅 is equal to 10 root three plus 10 newtons and 𝐊 is equal to 10 plus root three newtons.

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