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Question Video: Analysis of Three Coplanar Forces Acting on a Particle to Produce a Resultant Mathematics • Second Year of Secondary School

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Coplanar forces of magnitudes 𝐹 N, 8√3 N, √3 N, and 9√3 N act on a particle as shown in the diagram. Given that the magnitude of their resultant is 9√3 N, determine the value of 𝐹.

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Video Transcript

Coplanar forces of magnitudes 𝐹 newtons, eight root three newtons, root three newtons, and nine root three newtons act on a particle, as shown in the diagram. Given that the magnitude of their resultant is nine root three newtons, determine the value of 𝐹.

We’re given information about the magnitude of the resultant of four coplanar forces, where of course the resultant is the vector sum of those forces. The problem is, one of those forces is 𝐹, and that’s what we’re trying to find. So what we’re going to do is resolve each of our forces into perpendicular components. And then we’re going to find the vector sum of these and then find the magnitude and equate that to nine root three. So let’s define the direction in which the 𝐹 force is acting to be the positive horizontal direction. And then perpendicular to this and upwards is the positive vertical direction.

Let’s begin by resolving our forces then in a horizontal direction. We know 𝐹 is acting in this direction. But actually, we need to find the horizontal component of the eight root three, nine root three, and root three forces. We add a right-angled triangle to our eight root three force. We know that the horizontal component will be acting to the right and the vertical component this will be acting upwards. And then we can use right angle trigonometry to see that the adjacent side, and that’s the horizontal component, is eight root three cos 60 and the opposite side, the vertical component, is eight root three sin 60.

Let’s repeat this for the root three force. Since angles on a straight line add up to 180 degrees, the included angle this time is 60. And then we see that the side opposite to this, that’s the vertical component acting upwards, is root three sin 60. And the adjacent, which is the horizontal component acting to the left, is root three cos 60. There’s one more triangle that we’re interested in. Finally, we have the nine root three force. We could add a triangle to the left of this, but there’s an awful lot going on there already. So we add a triangle to the right as shown.

We know that this angle here is 60 degrees. So the included angle must be 30. And once again, we resolve it into its horizontal and vertical components. We get nine root three cos of 30 for the vertical component and nine root three sin of 30 for the horizontal component.

Now that we’ve resolved all of these forces, let’s find the horizontal sum. We have 𝐹 acting in a positive direction plus eight root three cos of 60. That’s the component of the eight root three force that acts in the horizontal direction. And then we subtract root three cos of 60 since that’s acting to the left, and nine root three sin of 30. By using the fact that both sin of 30 and cos of 60 are one-half, we simplify this to 𝐹 plus eight root three times one-half minus root three times one-half minus nine root three times one-half. And all that rather complicated working simplifies to just 𝐹 minus the square root of three.

Let’s do the same in the vertical direction. We know that we have eight root three sin 60 acting upwards, so in the positive direction, and root three sin 60. But in the opposite direction, we have nine root three cos of 30. So we’re going to subtract this from our sum. This time, we use the fact that sin of 60 and cos of 30 are root three over two. And so we get eight root three times root three over two plus root three times root three over two minus nine root three times root three over two, which is in fact equal to zero. And that’s really useful because we can actually work out the magnitude of the resultant really easily.

Since our force is only acting in one direction, the magnitude of our resultant must be 𝐹 minus root three. But we were told that the magnitude of the resultant is nine root three. So we can set up a simple equation for 𝐹. 𝐹 minus the square root of three is equal to nine root three. And we’ll solve for 𝐹 by adding root three to both sides. And so we see that 𝐹 is 10 root three or 10 root three newtons.

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